\section{Singular Homology}  \label{sectSingHom}

We present the singular approach to homology theory.  We generally
follows the approach of \cite{GH} with a few exceptions.

\subsection{Homotopy Modules}

\begin{defn}
For $k \geq 0$, the
{\bfseries standard $\mathbf{k}$-simplex}\index{Standard $k$-simplex},
denoted $\Delta_k$, is defined by\\
$\displaystyle \Delta_k = \left\{ \sum_{j=0}^k a_j \VEC{e}_j :
a_j \geq 0 \ \text{and} \ \sum_{j=0}^k a_j = 1 \right\}$
where $\displaystyle \VEC{e}_0 = \VEC{0} \in \RR^\infty$ and
$\displaystyle \VEC{e}_j \in \RR^\infty$ for
$1\leq j \leq k$ is the unit vector with $1$ for the
$\displaystyle j^{th}$ component and $0$ everywhere else.
\end{defn}

Using the notation of the section on simplicial homology, we have that
$\Delta_k$ is the closed $k$-simplex
$[\VEC{e}_0,\VEC{e}_1,\VEC{e}_2,\ldots, \VEC{e}_k]$.

\begin{defn}
For $k\geq 0$, a
{\bfseries singular $\mathbf{k}$-simplex}\index{Singular $k$-Simplex}
in a topological space $X$ is a continuous map\\
$\sigma:\Delta_k \to X$.  Given an integral domain $R$, 
the free module over $R$ generated by all the
singular $k$-simplices in $X$ is denoted $S_k(X;R)$.
The elements of $S_k(X;R)$ are called
{\bfseries singular $\mathbf{k}$-chain}\index{Singular $k$-Chain}.
The $R$-module $S_k(X;R)$ is simply denoted $S_k(X)$ when the choice
of the integral domain is clear.  It is also convenient to set
$S_k(X;R) = \{0\}$ for $k<0$.
\end{defn}

The two integral domains that we will use in all our applications
are $R= \RR$ or $R = \ZZ$.

By definition of a module over an integral domain generated by a set
of elements, a singular $k$-chain $c \in S_k(X;R)$ can be written as
$\displaystyle c = \sum_{j \in J} a_j \sigma_j$
where $J \subset \NN$ is a finite set, $a_j \in R$ and
$\sigma_j$ is a singular $k$-simplex for $j \in J$.  The sum of two
singular $k$-chains
$\displaystyle c_1 = \sum_{j \in J} a_j \sigma_j$ and
$\displaystyle c_2 = \sum_{j \in J} b_j \sigma_j$ is therefore the
singular $k$-chain given by
$\displaystyle c_1 + c_2 = \sum_{j \in J} (a_j + b_j) \sigma_j$.
We may always assume that the two singular $k$-chains $c_1$ and $c_2$
are sums over the same set $J$ by adding terms of the form $0
\sigma_j$ to the sum where they are not present.  The product of a
singular $k$-chain $\displaystyle c = \sum_{j \in J} a_j \sigma_j$ by
$r \in R$ is the singular $k$-chain given by 
$\displaystyle r c = \sum_{j \in J} (r a_j) \sigma_j$.

The next known definition is important to review since it is going to
play a very important role in this section.

\begin{defn}
An {\bfseries affine space}\index{Affine Space} is a subset $X$
of $\displaystyle \RR^n$ such that $t \VEC{x}_1 + (1-t)\VEC{x}_2 \in X$
for all $\VEC{x}_i,\VEC{x}_2 \in X$ and $0 \leq t \leq 1$.
A function $\displaystyle f:X\to Y$ between two affine spaces $X$ and $Y$ is
{\bfseries affine}\index{Affine Function} if
$f(t\VEC{x}_1 + (1-t)\VEC{x}_2) = tf(\VEC{x}_1) + (1-t)f(\VEC{x}_2)$ for all
$\displaystyle \VEC{x}_1,\VEC{x}_2 \in X$ and $0 \leq t \leq 1$.
\end{defn}

There is a more general definition of affine space than the one that
we have given.  We will only use affine spaces that are subsets
$\displaystyle \RR^n$ for some $n>0$.  More precisely, we will basically only
consider the affine space $\Delta_k$ for $k\in \NN$. 

Suppose that $X$ and $Y$ are two affine spaces.
It follows from the definition of an affine function that if
$\displaystyle f:X \to Y$ is an affine function, then
$\displaystyle f\Big(\sum_{j\in J} a_j \VEC{x}_j\Big) =
\sum_{j\in J} a_j f(\VEC{x}_j)$ for all $\VEC{x}_j \in X$ 
and $a_j \in \RR$ such that $\displaystyle \sum_{j\in J} a_j = 1$
with $a_j \geq 0$.

\begin{prop}
If $\displaystyle f:\RR^n\to \RR^m$ is affine, then there exist a
linear mapping $\displaystyle L:\RR^n \to \RR^m$ and
$\displaystyle \tilde{\VEC{x}} \in \RR^m$ such that $f(\VEC{x}) = L(\VEC{x})
+ \tilde{\VEC{x}}$ for all $\displaystyle \VEC{x} \in \RR^n$.
\end{prop}

\begin{proof}
Let $\tilde{\VEC{x}} = f(\VEC{0})$ and
$L(\VEC{x}) = f(\VEC{x}) - \tilde{\VEC{x}}$ for
$\displaystyle \VEC{x} \in \RR^n$.  It
suffices to prove that $\displaystyle L:\RR^n \to \RR^m$ is a linear
map.  We have
\begin{align}
L(a\VEC{x}) &= f(a\VEC{x}) - \tilde{\VEC{x}}
= f(a \VEC{x} +(1-a)\,\VEC{0}) - \tilde{\VEC{x}}
= a f(\VEC{x}) +(1-a)f(\VEC{0}) - \tilde{\VEC{x}} \nonumber \\
&= a f(\VEC{x}) +(1-a) \tilde{\VEC{x}} - \tilde{\VEC{x}} 
= a (f(\VEC{x}) - \tilde{\VEC{x}}) = a L(\VEC{x}) \label{singHEq1}
\end{align}
for all $a\in \RR$ and $\displaystyle \VEC{x} \in \RR^n$.  Moreover
\begin{align*}
L(\VEC{x}_1+\VEC{x}_2) &=
L\left(2\left( \frac{1}{2}\VEC{x}_1+ \frac{1}{2}\VEC{x}_2\right)\right)
= 2 L\left(\frac{1}{2}\VEC{x}_1+ \frac{1}{2}\VEC{x}_2\right)
= 2 \left( f\left( \frac{1}{2}\VEC{x}_1+ \frac{1}{2}\VEC{x}_2\right)
- \tilde{\VEC{x}} \right) \\
&= 2 \left( \frac{1}{2} f(\VEC{x}_1)+ \frac{1}{2}f(\VEC{x}_2)
- \tilde{\VEC{x}} \right)
= (f(\VEC{x}_1) - \tilde{\VEC{x}}) + (f(\VEC{x}_2) - \tilde{\VEC{x}})
= L(\VEC{x}_1) + L(\VEC{x}_2)
\end{align*}
for all $\displaystyle \VEC{x}_1,\VEC{x}_2 \in \RR^n$, where we have used
(\ref{singHEq1}) to get the second equality.
\end{proof}

It follows from the previous proposition that an affine function
$\displaystyle f:\RR^n \to \RR^m$ is completely determined by it values at
$\VEC{v}_j$ for $0 \leq j \leq n$ if
$\displaystyle \RR^n = \Span \{ \VEC{v}_j: 0\leq j \leq n\}$.
Moreover, we have that $\displaystyle f\left(\sum_{j=0}^n a_j \VEC{v}_j\right)
= \sum_{j=0}^n a_j f(\VEC{v}_j)$ if $\displaystyle \sum_{j=0}^n a_j = 1$
for $a_j \in \RR$.

For $k>0$ and $0 \leq j \leq k$, let
$\displaystyle F_k^j:\Delta_{k-1} \to \Delta_k$ be the
affine function defined by
\[
F_k^j(\VEC{e}_i) = \begin{cases}
\VEC{e}_i & \quad \text{if} \ i < j \\
\VEC{e}_{i+1} & \quad \text{if} \ i \geq j
\end{cases}
\]
for $0 \leq i \leq k-1$.

\begin{defn}
Let $X$ be a topological space and $R$ be an integral domain.
Given a singular $k$-simplex $\sigma:\Delta_k \to X$ in $X$ with
$k>0$.  The {\bfseries $\mathbf{j}$-face}\index{$j$-face} of
$\sigma$ is the singular $(k-1)$-simplex defined by
$\displaystyle \sigma^{(j)} = \sigma \circ F_k^j:\Delta_{k-1} \to X$.
The {\bfseries boundary}\index{Boundary} of $\sigma$ is defined by
$\displaystyle \partial_k (\sigma) = \sum_{j=0}^k (-1)^j \sigma^{(j)}$.

The {\bfseries boundary operator}\index{Boundary Operator}
$\partial_k : S_k(X;R) \to S_{k-1}(X;R)$ for $k>0$ is
defined by linearity; namely, given a singular $k$-chain
$\displaystyle c = \sum_{j \in J} a_j \sigma_j$, we set
$\displaystyle \partial_k c = \sum_{j \in J} a_j \partial_k(\sigma_j)$.
It is convenient to define $\partial_k = 0$ for $k\leq 0$.
\end{defn}

\begin{prop} \label{propdkmqdk0}
Let $X$ be a topological space.  Then
$\partial_{k-1} \circ \partial_k = 0$ for all $k\in \ZZ$.
\end{prop}

\begin{proof}
The result is obviously true for $k \leq 1$.  Thus, we may assume that
$k>1$.

\stage{i} We first note that
$\displaystyle F_k^i \circ F_{k-1}^j = F_k^j \circ F_{k-1}^{i-1}$
for $0 \leq j < i \leq k$.  To prove this
statement, it suffices to prove it for all $\VEC{e}_q$ for
$1 \leq q \leq k-1$ because we are working with affine functions.  We
have from
\[
F_{k-1}^j(\VEC{e}_q) =
\begin{cases}
\VEC{e}_q & \quad \text{if} \ q < j \\
\VEC{e}_{q+1} & \quad \text{if} \ j \leq q
\end{cases}
\]
that
\[
F_k^i(F_{k-1}^j(\VEC{e}_q)) =
\begin{cases}
\VEC{e}_q & \quad \text{if} \ q < j \\
\VEC{e}_{q+1} & \quad \text{if} \ j \leq q < i -1 \\
\VEC{e}_{q+2} & \quad \text{if} \ i-1 \leq q
\end{cases}
\]
Moreover, we have from
\[
F_{k-1}^{i-1}(\VEC{e}_q) =
\begin{cases}
\VEC{e}_q & \quad \text{if} \ q < i-1 \\
\VEC{e}_{q+1} & \quad \text{if} \ i-1 \leq q
\end{cases}
\]
that
\[
F_k^j(F_{k-1}^{i-1}(\VEC{e}_q)) =
\begin{cases}
\VEC{e}_q & \quad \text{if} \ q < j \\
\VEC{e}_{q+1} & \quad \text{if} \ j \leq q < i -1 \\
\VEC{e}_{q+2} & \quad \text{if} \ i-1 \leq q  
\end{cases}
\]
Thus $\displaystyle F_k^i(F_{k-1}^j(\VEC{e}_q))
= F_k^j(F_{k-1}^{i-1}(\VEC{e}_q))$.

\stage{ii} By linearity, it is enough to proof
$\partial_{k-1} \circ \partial_k = 0$ for singular $k$-simplices.
Given a singular $k$-simplex $\sigma$, we have that
\begin{align*}
\partial_{k-1}(\partial_k(\sigma))
&= \partial_{k-1}\left( \sum_{i=0}^k (-1)^i \sigma \circ F_k^i \right)
= \sum_{j=0}^{k-1} \sum_{i=0}^k (-1)^j (-1)^i
\sigma \circ F_k^i \circ F_{k-1}^j \\
&= \sum_{i=1}^k \sum_{j=0}^{i-1}  (-1)^{j+i}
\sigma \circ F_k^i \circ F_{k-1}^j
+ \sum_{i=0}^{k-1} \sum_{j=i}^{k-1}  (-1)^{j+i}
\sigma \circ F_k^i \circ F_{k-1}^j \\
&= \sum_{i=1}^k \sum_{j=0}^{i-1}  (-1)^{j+i}
\sigma \circ F_k^j \circ F_{k-1}^{i-1}
+ \sum_{j=0}^{k-1} \sum_{i=j+1}^k  (-1)^{j+i-1}
\sigma \circ F_k^j \circ F_{k-1}^{i-1} = 0
\end{align*}
where, for the fourth equality, we have use (i) to get the first sum
and the substitution $(i,j) \mapsto (j, i-1)$ to get the second sum.
Note that each term in the first sum appears with the opposite sign
in the second sum, and we have $0 \leq j < i \leq k$ in both sums.
\end{proof}

\begin{defn}
Let $X$ be a topological space.  Two $k$-cycles $c_1,c_2 \in S_k(X;R)$ are
{\bfseries homologous}\index{Homologous} if there exists
$\tilde{c} \in S_{k+1}(X;R)$ such that $c_1 - c_2 = \partial_{k+1} \tilde{c}$.
We write $c_1 \hsim c_2$.
\end{defn}

The reader should verify that the homologous relation defines an
equivalence relation on $S_k(X;R)$.

\begin{defn}
Let $X$ be a topological space.
A {\bfseries $\mathbf{k}$-cycle}\index{$k$-Cycle}
in $S_k(X;R)$ is a singular $k$-chain $c$ such that $\partial_k c = 0$.
The set of all $k$-cycles in $S_k(X;R)$ is a $R$-module denoted
$Z_k(X;R)$.

A {\bfseries $\mathbf{k}$-boundary}\index{$k$-Boundary}
in $S_k(X;R)$ is a singular $k$-chain $c$ with the property that there
exists a singular $(k+1)$-chain $\tilde{c}$ such that
$\partial_{k+1} \tilde{c} = c$. 
The set of all $k$-boundaries in $S_k(X;R)$ is a $R$-module denoted
$B_k(X;R)$.

The {\bfseries singular $\displaystyle \mathbf{k^{th}}$ homology
module}\index{Homology Module!Singular $k^{th}$ Homology Module} of
$X$ is the $R$-module defined as
$H_k(X;R) = Z_k(X;R) / B_k(X;R)$.
The equivalence class of $H_k(X;R)$ associated to $c \in Z_k(X;R)$ is
denoted $[c]_X$.
\end{defn}

\begin{egg}
The first example is a simple but important example.   \label{eggHk1}
Let $X = \{x\}$, a set with a single element.  A singular $k$-simplex
$\sigma:\Delta_k \to X$ is therefore a constant function defined by
$\sigma(\VEC{x}) = x$ for all $\VEC{x} \in \Delta_k$.  Hence
\[
\partial_k \sigma = \sum_{j=0}^k (-1)^j \sigma^{(i)} =
\begin{cases}
0 & \quad \text{if} \ k \ \text{is odd} \\
\sigma^{(0)} & \quad \text{if} \ k \ \text{is even}
\end{cases}
\]
because $\displaystyle \sigma^{(j)}: \Delta_{k-1} \to X$ is the
constant function defined by
$\displaystyle \sigma^{(j)}(\VEC{x}) = x$ for all $\VEC{x} \in \Delta_{k-1}$
for $0 \leq j \leq k$.

It follows that
\[
Z_k(X;R) = \begin{cases}
S_k(X;R) & \quad \text{if} \ k \ \text{is odd or} \ k = 0 \\
\{0\} & \quad  \text{if} \ k \ \text{is even}
\end{cases}
\]
and
\[
B_k(X;R) = \begin{cases}
S_k(X;R) & \quad \text{if} \ k \ \text{is odd} \\
\{0\} & \quad  \text{if} \ k \ \text{is even or} \ k = 0
\end{cases}
\]
Therefore
\[
H_k(X;R) = \begin{cases}
S_0(X;R) & \quad \text{if} \ k = 0 \\
\{[0]_X\} & \quad \text{if} \ \ k > 0
\end{cases}
\]
From now on, we adopt the traditional convention of writing
$H_k(X;R) = 0$ instead of $H_k(X;R) = \{[0]_X\}$.
\end{egg}

\begin{prop} \label{propHkeHkJ}
If the topological space $X$ is the union of distinct (connected)
components $X_j$ for $j \in J$, then
$\displaystyle H_k(X;R) = \bigoplus_{j\in J}H_k(X_j;R)$.
\end{prop}

\begin{proof}[Proof (Sketch)]
If $\sigma \in S_k(X;R)$, then $\sigma(\Delta_k) \subset X_j$ for some
$j\in J$ because $\sigma(\Delta_k)$ is connected.  Therefore
$\displaystyle S_k(X;R) = \bigoplus_{j\in J} S(X_j;R)$.  Moreover
$\partial_q :S_q(X_j;R) \to S_{q-1}(X_j;R)$ for all $q \in \ZZ$ and all
$j \in J$.  Thus $\displaystyle Z_k(X;R) = \bigoplus_{j\in J} Z(X_j;R)$
and $\displaystyle B_k(X;R) = \bigoplus_{j\in J} B(X_j;R)$.  The
conclusion of the proof follows from a standard result about the
quotient space of direct sums of modules in algebra.
\end{proof}

\begin{prop} \label{propH0XR}
Let $X$ be a path-connected topological space.  Then $H_0(X;R) \cong R$.
\end{prop}

\begin{proof}
Let $e_x:\Delta_0 \to X$ be the singular $0$-simplex defined by
$e_x(\VEC{e}_0) = x$ for $x \in X$.

\stage{i} We define a map $h: S_0(X;R) \to R$ as it follows.
Given $c \in S_0(X;R)$, we have that
$\displaystyle c = \sum_{j \in J} a_j e_{x_j}$ where $J \subset \NN$
is a finite set, $a_j \in R$ and $x_j \in X$ for all $j \in J$.
We set $\displaystyle \epsilon(c) = \sum_{j \in J} a_j$.
We prove that if $c_1$ and $c_2$ are two homologous singular $0$-chains,
then $\epsilon(c_1) = \epsilon(c_2)$.  If $c_1$ and $c_2$ are two homologous
singular $0$-chains, then $c_1 - c_2 = \partial_1 c$ for some
$c \in S_1(X;R)$.  Hence
$\epsilon(c_1) - \epsilon(c_2) = \epsilon(c_1-c_2) = \epsilon(\partial_1 c)$.
To prove that $\epsilon(\partial_1 c) = 0$, it suffices to prove that
$\epsilon(\partial_1 \sigma) = 0$ for any singular $1$-simplex in $X$.
We effectively have that
$\displaystyle \epsilon(\partial_1 \sigma)
= \epsilon(\sigma \circ F^0_1 - \sigma \circ F^1_1)
= \epsilon(e_{\sigma(\VEC{e}_1)} - e_{\sigma(\VEC{e}_0)}) = 1 - 1 = 0$.

We may therefore define a map $\tilde{\epsilon}: H_0(X;R) \to R$ as it follows.
Given a singular $0$-chain $\displaystyle c = \sum_{j \in J} a_j e_{x_j}$
where $J \subset \NN$ is a finite set, $a_j \in R$ and $x_j \in X$ for
all $j \in J$. we set
$\displaystyle \tilde{\epsilon}([c]_X) = \sum_{j \in J} a_j$.

\stage{ii} Let $c$ be a singular $0$-chain.  We have that
$\displaystyle c = \sum_{j \in J} a_j e_{x_j}$ where $J \subset \NN$
is finite, $a_j \in R$ and $x_j \in X$ for all $j \in J$.
Choose $\tilde{x} \in X$.  Given any $x \in X$, there exists a path
$\tau_x:[0,1] \to X$ such that $\tau_x(0) = \tilde{x}$ and $\tau_x(1) = x$
because $X$ is path-connected.
Let $\hat{\tau}_x:\Delta_1 \to X$ be the singular $1$-simplex defined
by $\hat{\tau}_x((1-t)\VEC{e}_0 + t \VEC{e}_1) = \tau_x(t)$ for $t \in [0,1]$.
Since $\partial_1 \hat{\tau}_x = e_x - e_{\tilde{x}}$,
we have that $e_x \hsim e_{\tilde{x}}$.
Therefore $\displaystyle c \hsim \Big( \sum_{j \in J} a_j \Big) e_{\tilde{x}}$.

\stage{iii} The map $\tilde{\epsilon}$ is a module homomorphism.  If
$r \in R$ and $\displaystyle c = \sum_{j \in J} a_j e_{x_j}$
for some $a_j \in R$ and $x_j \in X$, then
$\displaystyle rc = \sum_{j \in J} ra_j e_{x_j}$.
Hence $\displaystyle \tilde{\epsilon}([rc]_X) = \sum_{j \in J} ra_j =
r \sum_{j \in J} a_j = r \tilde{\epsilon}([c]_X)$.
If $\displaystyle c_1 = \sum_{j \in J} a_j e_{x_j}$
and
$\displaystyle c_2 = \sum_{j \in J} b_j e_{x_j}$ for some $a_j, b_j \in R$
and $x_j \in X$, then
$\displaystyle c_1 + c_ 2 = \sum_{j \in J} (a_j+b_j) e_{x_j}$.
Therefore
$\displaystyle \tilde{\epsilon}([c_1+c_2]_X) = \sum_{j \in J} (a_j+b_j)
= \sum_{j \in J} a_j+ \sum_{j \in J}b_j = \tilde{\epsilon}([c_1]_X) +
\tilde{\epsilon}([c_2]_X)$.

The map $\tilde{\epsilon}$ is obviously onto.  It is also injective.
Suppose that
$\displaystyle c_1 = \sum_{j \in J} a_j e_{x_j}$ and
$\displaystyle c_2 = \sum_{j \in J} b_j e_{x_j}$ for some $a_j, b_j \in R$
and $x_j \in X$.  If $\tilde{\epsilon}([c_1]_X) = \tilde{\epsilon}([c_2]_X)$,
then $\displaystyle \sum_{j \in J} a_j = \sum_{j \in J} b_j$.
Hence, we get from (ii) that
$\displaystyle c_1 \hsim \sum_{j \in J} a_j e_{\tilde{x}}
= \sum_{j \in J} b_j e_{\tilde{x}} \hsim c_2$.  Thus $[c_1]_X = [c_2]_X$.
\end{proof}

The proof of the next proposition is left to the reader.

\begin{prop}
Let $X$ and $Y$ be two topological spaces, and $f : X \to Y$ be a
continuous function.  The map
$S_k(f): S_k(X;R) \to S_k(Y;R)$ defined by
$\displaystyle S_k(f)\Big( \sum_{j \in J} a_j \sigma_j \Big) =
\sum_{j \in J} a_j\, f\circ \sigma$ for
is a module homomorphism.
\end{prop}

\begin{prop} \label{propSkPropr}
Let $X$, $Y$ and $Z$ be three topological spaces.  If $f : X \to Y$
and $g:Y \to Z$ are two continuous functions, and $k \in \ZZ$, then
\begin{enumerate}
\item $S_k(g\circ f) = S_k(g) \circ S_k(f)$ and
\item $\partial_k \big( S_k(f) c \big) = S_{k-1}(f) (\partial_k c)$ for all
$c \in S_k(X;R)$.
\end{enumerate}
\end{prop}

\begin{proof}
The proof of (1) is left to the reader.  As usual, we only need to
prove (2) for singular $k$-simplices.  Let $\sigma$ be a singular
$k$-simplex, then
\begin{align*}
S_{k-1}(f) (\partial_k(\sigma))
&= S_{k-1}(f) \left( \sum_{i=0}^k (-1)^i \sigma \circ F_k^i \right)
= \sum_{i=0}^k (-1)^i f \circ \sigma \circ F_k^i \\
&= \partial_k ( f\circ \sigma)
= \partial_k ( S_k(f) (\sigma)) \ .  \qedhere
\end{align*}
\end{proof}

\begin{cor} \label{corDefHkf}
Let $X$ and $Y$ be two topological spaces, and $f : X \to Y$ be a
continuous function.  Then
$S_k(f) (Z_q(X;R)) \subset Z_k(Y;R)$ and
$S_k(f) (B_q(X;R)) \subset B_k(Y;R)$.  Moreover,
$H_k(f) : H_k(X;R) \to H_k(Y;R)$ defined by
$H_k(f)([c]_X) = [S_k(f)(c)]_Y$ for all $c \in H_k(X;R)$ is a module
homomorphism.
\end{cor}

\begin{proof}
Since $\partial_k \big( S_k(f) c \big) = S_{k-1}(f) (\partial_k c)$ for all
$c \in S_k(X;R)$, we have that $S_k(f) (Z_q(X;R)) \subset Z_k(Y;R)$ and
$S_k(f) (B_q(X;R)) \subset B_k(Y;R)$.

To prove that $H_k(f)([c]_X)$ is well defined, we need to show that
$[S_k(f)(c)]_Y$ is independent of the representative $c$ of the
equivalence class $[c]_X$ that is used.
Suppose that $\tilde{c} \hsim c$.  Then $\tilde{c} = c + \partial_{k+1} b$
for some $b \in S_{k+1}(X;R)$.  Hence
$S_k(f)(\tilde{c}) = S_k(f)(c) + S_k(f)(\partial_{k+1} b)
= S_k(f)(c) + \partial_{k+1} (S_k(f)(b))$ with
$S_k(f)(b) \in S_{k+1}(Y;R)$.  Thus
$[S_k(f)(\tilde{c})]_Y = [S_k(f)(c)]_Y$.

It is easy to verify that
$H_k(f)([c_1]_X +[c_2]_X) = H_k(f)([c_1]_X) + H_k(f)([c_2]_X)$ and \\
$H_k(f)(\alpha[c]_X) = \alpha H_k(f)([c]_X)$ for
$[c]_X, [c_1]_X,[c_2]_X \in H_k(X;R)$
and $\alpha \in R$.  This proves that $H_k(f)$ is a module
homomorphism.
\end{proof}

The following results follow from Proposition~\ref{propSkPropr}.

\begin{cor} \label{corCompHgHf}
If $X$, $Y$ and $Z$ are topological spaces and 
$f:X\to Y$ and $g:Y\to Z$ are two continuous functions, then
$H_k(g \circ f) = H_k(g) \circ H_k(f)$.
\end{cor}

\begin{cor}  \label{corIsoHkXY}
Suppose that $X$ and $Y$ are two topological spaces.  If $X$ and $Y$ are
homeomorphic, then $H_k(X;R) \cong H_k(Y;R)$.
\end{cor}

\subsection{Exact Sequences}

\begin{defn}
Let $X_j$ be modules over an integral domain $R$ and
$f_j:X_j\to X_{j+1}$ be continuous functions.  We say that the
sequence
\[
\xymatrix{ \ar[r]^(0.4){f_{j-1}} & X_j \ar[r]^{f_j}
& X_{j+1} \ar[r]^{f_{j+1}} & X_{j+2} \ar[r]^-{f_{j+2}} & }
\]
is a {\bfseries (long) exact sequence}\index{Exact Sequence}\index{Long
Exact Sequence|see{Exact Sequence}}
if the kernel of a map in the sequence equal the image of the
previous map in the sequence.
\end{defn}

We often have to deal with the following type of sequences.

\begin{defn} \label{defnSESequ}
Let $X$, $Y$ and $Z$ be modules over an integral domain $R$.  Moreover
let $f:X\to Y$ and $g:Y\to Z$ be continuous functions.  We say that the
sequence
\[
\xymatrix{ 0 \ar[r] & X \ar[r]^{f} & Y \ar[r]^{g} & Z \ar[r] & 0 }
\]
is a {\bfseries short exact sequence}\index{Short Exact Sequence} if
the kernel of a map in the sequence equal the image of the
previous map (if any) in the sequence.
\end{defn}

By tradition, the expression $0 \rightarrow X$ denotes the map
from the set $\{0\}$ to $X$ that maps $0$ to the null element in $X$.
Similarly, the expression $Z \rightarrow 0$ denotes the map from
$Z$ to $\{0\}$ that map all elements in $Z$ to $0$.

\begin{defn}
Let $X$, $Y$ and $Z$ be modules over an integral domain $R$.
We say that a short exact sequence
\[
\xymatrix{ 0 \ar[r] & X \ar[r]^{f} & Y \ar[r]^{g} & Z \ar[r] & 0 }
\]
is {\bfseries split}\index{Short Exact Sequence!Split} if there
exists a function $h:Y\to X$ such that $h \circ f = \Id_X$.
\end{defn}

The next result is kind of surprising.

\begin{prop} \label{propSplit2Cond}
Let $X$, $Y$ and $Z$ be modules over an integral domain $R$.
A short exact sequence
\[
\xymatrix{ 0 \ar[r] & X \ar[r]^{f} & Y \ar[r]^{g} & Z \ar[r] & 0 }
\]
is split if and only if there exists a function $k:Z \to Y$ such that
$g \circ k = \Id_Z$.
\end{prop}

\begin{proof}
\stage{$\mathbf{\Rightarrow}$} Since the short exact sequence is
split, there exists $h:Y\to X$ such that $h \circ f = \Id_X$.
To define $k:Z \to Y$, consider $z \in Z$.  Since $g$ is onto, there
exists $y \in Y$ such that $g(y) = z$.  Let $k(z) = y - f(h(y))$.

We first show that $k$ is well defined.  Suppose that $\tilde{y} \in Y$
is such that $g(\tilde{y}) = z$.  Then
$g(y - \tilde{y}) = g(y) - g(\tilde{y}) = 0$.  So
$y - \tilde{y} \in \KE(g) = \IMG(f)$.  Thus there exists $x\in X$ such
that $f(x) = y - \tilde{y}$.  We get
\begin{align*}
\big(y - f(h(y))\big) - \big(\tilde{y} - f(h(\tilde{y}))\big)
&= y - \tilde{y} - f(h(y - \tilde{y})) \\
&= f(x) - f(h(f(x))) = f(x) - f(x) = 0
\end{align*}
because $h \circ f = \Id_X$.

We also have that
$g(k(z)) = g(y - f(h(y))) = g(y) - g(f(h(y))) = g(y) = z$
for all $z \in Z$ because $\KE(g) = \IMG(f)$, where $y\in Y$ is the $y$
in the definition of $k$ at $z \in Z$.  Thus $g \circ k = \Id_Z$.

We must not forget to verify that $k$ is a homomorphism of modules.
Given $z, \tilde{z} \in Z$.  Suppose that $y,\tilde{y}$ are such that
$g(y) = z$ and $g(\tilde{y}) = \tilde{z}$.  Then
$g(y + \tilde{y}) = z + \tilde{z}$ and
$k(z + \tilde{z}) = y + \tilde{y} + f(h(y + \tilde{y}))
= \big(y + f(h(y))\big) + \big(\tilde{y} + f(h(\tilde{y}))\big)
= k(z) + k(\tilde{z})$.
Proceeding similarly, we also get that
$k(rz) = rk(z)$ for all $z\in Z$ and $r \in R$.

\stage{$\mathbf{\Leftarrow}$} We assume that there exists $k:Z\to Y$
such that $k \circ g = \Id_Z$.  To define $h:Y\to X$, consider $y\in Y$.
Since $g(y - k(g(y))) = g(y) - g(k(g(y))) = g(y) - g(y) = 0$ because
$k \circ g = \Id_z$, we have that $y-k(g(y)) \in \KE(g)$.  Since
$\IMG(f) = \KE(g)$, there exists $x \in X$ such that $f(x) = y -k(g(y))$.
We set $h(y) = x$.  The function $h$ is well defined because $f$ is
one-to-one.  We also have that $h(f(x)) = x$ for all $x\in X$ because
$f(x) - k(g(f(x))) = f(x)$ since $\KE(g)= \IMG(f)$ and $f$ is one-to-one.
Thus $h \circ f = \Id_X$.  Finally, it is easy to show that $h$ is an
homomorphism of modules because $f$ is one.
\end{proof}

\begin{rmk}
It should be pointed out that $h$ and $k$ in the previous proposition
yields the following short exact sequence
\[
\xymatrix{ 0 & X \ar[l] & Y \ar[l]_{h} & Z \ar[l]_{k} & 0 \ar[l] }
\]
We have that $h \circ f = \Id_X$ ensures that $h$ is onto and
$g \circ k = \Id_Z$ ensures that $k$ is one-to-one.
We also have that $\IMG(k) = \KE(h)$.  To prove that
$\IMG(k) \subset \KE(h)$, consider $\tilde{y} = k(z)$ for $z \in Z$.
Then $\tilde{y} = y - f(h(y))$ for some $y \in Y$.  Hence
$h(\tilde{y}) = h(y - f(h(y))) = h(y) - h(f(h(y))) = h(y) - h(y) = 0$
because $h\circ f = \Id_X$.  Thus $\tilde{y} \in \KE(h)$.
To prove that $\IMG(k) \supset \KE(h)$, suppose that $y \in \KE(h)$.
Then $k(g(y)) = y - f(h(y)) = y$ because $h(y) = 0$.  Thus $y \in \IMG(k)$.
\end{rmk}

\begin{prop} \label{propSplitOplus}
A short exact sequence
\begin{equation} \label{SplitOplusEq1}
\xymatrix{ 0 \ar[r] & X \ar[r]^{f} & Y \ar[r]^{g} & Z \ar[r] & 0 }
\end{equation}
is split if and only if $Y \cong X \oplus Z$.
\end{prop}

\begin{proof}
\stage{i} Suppose that (\ref{SplitOplusEq1}) is split.  We then have
that
\[
\xymatrix{ 0 \ar[r] & X \ar@/^/[r]^{f} & Y \ar@/^/[l]^{h}\ar@/^/[r]^{g}
& Z \ar@/^/[l]^{k}\ar[r] & 0 }
\]
is exact in both direction.

Given $y \in f(X) \cap k(Z)$, then
$y = f(x) = k(z)$ for some $x\in X$ and $z \in Z$.  Thus
$0 = g(f(x)) = g(y) = g(k(z)) = z$ because $\KE(g) = \IMG(f)$.
Hence $y = k(z) = k(0) = 0$.
We get that $f(X) \cap k(Z) = \{0\}$.

Given $y \in Y$, it follows from
$g(y - k(g(y)) = g(y) - g(k(g(y))) = g(y) - g(y) = 0$ that
$y - k(g(y)) = f(x)$ for some $x \in X$ because $\KE(g) = \IMG(f)$.
Thus $y = f(x) + k(g(y)) \in f(X) + k(Z)$.
We get that $Y \subset f(X) + k(Z)$.

Therefore $Y = f(X) \oplus k(Z)$.  Since $X \cong f(X)$ and
$Z \cong k(Z)$ because $f$ and $k$ are isomorphism, we have that
$Y \cong X \oplus Z$.

\stage{ii} Suppose that $Y \cong X \oplus Z$.  So there exists an
isomorphism $\Phi:Y \to X \oplus Z$.  Let\\
$\tilde{h}:X \oplus Z \to X$ be the function defined by
$\tilde{h}(x,z) = x$ for all $(x,z) \in X \oplus Z$,\\
$\tilde{g}:X \oplus Z \to Z$ be the function defined by
$\tilde{g}(x,z) = z$ for all $(x,z) \in X \oplus Z$,\\
$\tilde{f}:X \to X \oplus Z$ be the function defined by
$\tilde{f}(x) = (x,0)$ for all $x \in X$ and\\
$\tilde{k}:Z \to X \oplus Z$ be the function defined by
$\tilde{k}(z) = (0,z)$ for all $z \in Z$.
Then
\[
\xymatrix@C+2ex{ 0 \ar[r] & X \ar@/^/[r]^-{\tilde{f}}
& X \oplus Z \ar@/^/[l]^-{\tilde{h}}\ar@/^/[r]^-{\tilde{g}}
& Z \ar@/^/[l]^-{\tilde{k}}\ar[r] & 0 }
\]
is exact in both direction.  Let
$\displaystyle f = \Phi^{-1}\circ \tilde{f}$, $g = \tilde{g} \circ \Phi$ and
$h = \tilde{h} \circ \Phi$.  We get that
\[
\xymatrix{ 0 \ar[r] & X \ar[r]^{f} & Y \ar[r]^{g} & Z \ar[r] & 0 }
\]
is a short exact sequence and it is split because
$h \circ f = (\tilde{h} \circ \Phi) \circ (\Phi^{-1}\circ \tilde{f})
= \tilde{h} \circ \tilde{f} = \Id_X$.
\end{proof}

\subsection{Chain Complexes}

\begin{defn}
Let $R$ be an integral domain.  A
{\bfseries chain complex}\index{Chain Complex} is a sequence
$\C = \{ (C_k,\partial_k) \}_{k\in \ZZ}$ where the sets $C_k$ are
$R$-modules and the maps $\partial_k:C_k \to C_{k-1}$ are homomorphism
such that $\partial_k \circ \partial_{k+1} = 0$ for $k \in \ZZ$.
\end{defn}

A diagram representing a chain complex is given below.
\[
\xymatrix{
\ar[r]^-{\partial_{k+2}} & C_{k+1} \ar[r]^{\partial_{k+1}}
& C_k \ar[r]^{\partial_k} & C_{k-1} \ar[r]^-{\partial_{k-1}} &
}
\]

\begin{rmk}
Unless otherwise stated, we always assume that $C_k = \{0\}$ for
$k<0$.  This is sufficient for our applications.
\end{rmk}

We have already worked with chain complexes before.  One such example is
given by the sequence of pairs
$(C_k,\partial_k) = (S_k(X;R), \partial_k)$ for $k \in \ZZ$
defined at the beginning of the previous section.

\begin{defn}
A {\bfseries chain map}\index{Chain Map} between two chain complexes
given by $\C = \{(C_k,\partial_k)\}_{k\in \ZZ}$ and
$\tilde{\C} = \{ (\tilde{C}_k,\tilde{\partial}_k)\}_{k\in \ZZ}$
is a sequence $\F = \{f_k\}_{k\in \ZZ}$ where the maps
$f_k:C_k \to \tilde{C}_k$ are homomorphism such that
$\tilde{\partial}_k \circ f_k = f_{k-1} \circ \partial_k$ for $k \in \ZZ$.
\end{defn}

The previous definition yields the following commutative diagram.
\[
\xymatrix{
\ar[r]^-{\partial_{k+2}} & C_{k+1}  \ar[r]^-{\partial_{k+1}} \ar[d]^-{f_{k+1}}
& C_k \ar[r]^-{\partial_k} \ar[d]^-{f_k} & C_{k-1} \ar[r]^-{\partial_{k-1}} 
\ar[d]^-{f_{k-1}} & \\
\ar[r]_-{\tilde{\partial}_{k+2}} & \tilde{C}_{k+1}
\ar[r]_-{\tilde{\partial}_{k+1}} & \tilde{C}_k \ar[r]_-{\tilde{\partial}_k}
& \tilde{C}_{k-1} \ar[r]_-{\tilde{\partial}_{k-1}} &
}
\]

As we did before, we may define $k$-cycles, $k$-boundaries and
$k$-singular homology modules.

\begin{defn}
Consider a chain complex given by $\C = \{(C_k,\partial_k)\}_{k\in \ZZ}$.
A {\bfseries $\mathbf{k}$-cycle}\index{$k$-Cycle}
in $\C$ is an element $c \in C_k$ such that $\partial_k(c) = 0$.
The set of all $k$-cycles in $C_k$ is a $R$-module denoted $Z_k(\C)$.

A {\bfseries $\mathbf{k}$-boundary}\index{$k$-Boundary}
in $\C$ is an element $c \in C_k$ with the property that there exists an
element $\tilde{c} \in C_{k+1}$ such that $\partial_{k+1}(\tilde{c}) = c$.
The set of all $k$-boundaries in $C_k$ is a $R$-module denoted
$B_k(\C)$.

The {\bfseries singular $\displaystyle \mathbf{k^{th}}$ homology
module}\index{Homology Module!Singular $k^{th}$ Homology Module} of
$\C$ is the quotient space $H_k(\C) = Z_k(\C) / B_k(\C)$.  The
equivalence class of $H_k(\C)$ associated to $c \in Z_k(\C)$ is denoted $[c]$.
\end{defn}

It follows from the definition that a chain map
$\F = \{f_k\}_{k\in\ZZ}$ between two chain complexes
$\C = \{(C_k,\partial_k)\}_{k\in \ZZ}$ and
$\tilde{\C} = \{ (\tilde{C}_k,\tilde{\partial}_k)\}_{k\in \ZZ}$
sends $k$-cycles in $\C$ to $k$-cycles in $\tilde{\C}$, and
$k$-boundaries in $\C$ to $k$-boundaries in $\tilde{\C}$.
Hence, as we did in Corollary~\ref{corDefHkf}, we may define
$H_k(\F): H_k(\C) \to H_k(\tilde{\C})$ by
$H_k(\F)([c]) = [f_k(c)]$ for all $c \in Z_k(\C)$.

The next result is traditionally called a lemma but it really should
be a theorem because it is used in very many proofs in algebraic
topology.

\begin{lemma}[Zig-Zag Lemma]  \label{lemZigZag}
Suppose that $\F = \{f_k\}_{k\in\ZZ}$ is a    \index{Zig-Zag Lemma}
chain map between the chain
complexes $\displaystyle \A = \{(A_k,\partial_k^{\A})\}_{k\in \ZZ}$ and
$\displaystyle \BB = \{(B_k,\partial_k^{\BB})\}_{k\in \ZZ}$, and
$\GG = \{g_k\}_{k\in\ZZ}$  is a chain map between the chain complexes
$\displaystyle \BB = \{(B_k,\partial_k^{\BB})\}_{k\in \ZZ}$ and
$\displaystyle \C = \{(C_k,\partial_k^{\C})\}_{k\in \ZZ}$.
If
\[
\xymatrix{
0 \ar[r] & A_k \ar[r]^{f_k} & B_k \ar[r]^{g_k} & C_k \ar[r] & 0
}
\]
is an exact sequence for all $k$, then there exists a long exact
sequence
\[
\xymatrix@C+2ex{
\ar[r]^-{\tilde{\partial}_{\,k+1}} & H_k(\A) \ar[r]^{H_k(\F)}
& H_k(\BB) \ar[r]^{H_k(\GG)} & H_k(\C) \ar[r]^-{\tilde{\partial}_{\,k}}
& H_{k-1}(\A) \ar[r]^-{H_{k-1}(\F)} &
}
\]
where $\tilde{\partial}_{\,k}$ is induced by
$\displaystyle \partial_k^{\BB}$.
\end{lemma}

\begin{proof}
The proof is based on the following commutative diagram.
\[
\xymatrix{
0 \ar[r] & A_{k+1} \ar[r]^{f_{k+1}}\ar[d]^{\partial_{k+1}^{\A}}
& B_{k+1} \ar[r]^{g_{k+1}}\ar[d]^{\partial_{k+1}^{\BB}}
& C_{k+1} \ar[r]\ar[d]^{\partial_{k+1}^{\C}} & 0 \\
0 \ar[r] & A_k \ar[r]^{f_k}\ar[d]^{\partial_k^{\A}}
& B_k \ar[r]^{g_k}\ar[d]^{\partial_k^{\BB}}
& C_k \ar[r]\ar[d]^{\partial_k^{\C}} & 0 \\
0 \ar[r] & A_{k-1} \ar[r]^{f_{k-1}} & B_{k-1} \ar[r]^{g_{k-1}}
& C_{k-1} \ar[r] & 0
}
\]

\stage{i} Our first task is to define $\tilde{\partial}_k$ on $H_k(\C)$.
Given $[c] \in H_k(\C)$, there exists $b \in B_k$ such that
$g_k(b) = c$ because $g_k$ is onto.  We have that
$\displaystyle \partial_k^{\BB}(b) \in B_{k-1} \cap \IMG(f_{k-1})$ because
$\displaystyle \partial_k^{\BB}(b) \in \KE(g_{k-1}) = \IMG(f_{k-1})$ since
$\displaystyle g_{k-1}(\partial_k^{\BB}(b)) = \partial_k^{\C}(g_k(b))  
= \partial_k^{\C}(c) = 0$ for $[c] \in H_k(\C)$.  There exists
$a \in A_{k-1}$ such that $\displaystyle f_{k-1}(a) = \partial_k^{\BB}(b)$.  In
fact, $a$ is unique because $f_{k-1}$ is one-to-one.

We set $\tilde{\partial}_k([c]) = [a] \in H_{k-1}(\A)$.

We have to show that $\tilde{\partial}_k([c])$ is well defined;
namely, that it is independent of the representative of the
equivalence class $[c] \in H_k(\C)$ that we choose to find $a \in A_{k-1}$.
Suppose that $\tilde{c} \in Z_k(\C)$ is such that $[\tilde{c}] = [c]$.
As before, there exists $\tilde{b} \in B_k$ such that
$g_k(\tilde{b}) = \tilde{c}$, and
$\tilde{a} \in A_{k-1}$ such that $\displaystyle f_{k-1}(\tilde{a})
= \partial_k^{\BB}(\tilde{b})$.

Since $[c]=[\tilde{c}]$, we have that
$\displaystyle c - \tilde{c} = \partial_{k+1}^{\C}(\breve{c})$ for some
$\breve{c} \in C_{k+1}$.  Choose $\breve{b} \in B_{k+1}$ such that
$g_{k+1}(\breve{b}) = \breve{c}$.  This is possible because $g_{k+1}$
maps $B_{k+1}$ onto $C_{k+1}$.   Then
\[
g_k\big( b - \tilde{b} - \partial_{k+1}^{\BB}(\breve{b})\big)
= g_k(b) - g_k(\tilde{b}) - \partial_{k+1}^{\C}(g_{k+1}(\breve{b}))\big)
= c - \tilde{c} - \partial_{k+1}^{\C}(\breve{c}) = 0 \ .
\]
Since $\displaystyle
b - \tilde{b} - \partial_{k+1}^{\BB}(\breve{b}) \in \KE(g_k) = \IMG(f_k)$,
there exists $\breve{a} \in A_k$ such that
$\displaystyle f_k(\breve{a}) = b - \tilde{b} - \partial_{k+1}^{\BB}(\breve{b})$.
Moreover, since
\begin{align*}
f_{k-1}( \partial_k^{\A}(\breve{a})) 
&= \partial_k^{\BB}(f_k(\breve{a}))
= \partial_k^{\BB}\big( b - \tilde{b} -\partial_{k+1}^{\BB}(\breve{b}) \big)
= \partial_k^{\BB}(b) - \partial_k^{\BB}(\tilde{b}) \\
&=  f_{k-1}(a) - f_{k-1}(\tilde{a}) = f_{k-1}(a - \tilde{a})
\end{align*}
and $f_{k-1}$ is one-to-one, we get that
$\displaystyle a - \tilde{a} = \partial_k^{\A}(\breve{a})$ with
$\breve{a} \in A_k$.  Therefore $[a]=[\tilde{a}]$.

To prove that $\tilde{\partial}_k$ is an homomorphism, suppose that
$[c], [\tilde{c}] \in H_k(\C)$.  We have by definition of
$\tilde{\partial}_k$ that $\tilde{\partial}_k([c]) = [a]$ and
$\tilde{\partial}_k([\tilde{c}]) = [\tilde{a}]$ where
$\displaystyle f_{k-1}(a) = \partial_k^{\BB}(b)$
with $g_k(b) = c$ for some $b \in B_k$ and
$\displaystyle f_{k-1}(\tilde{a}) = \partial_k^{\BB}(\tilde{b})$ 
with $g_k(\tilde{b}) = \tilde{c}$ for some $\tilde{b} \in B_k$.
Since
$\displaystyle f_{k-1}(a + \tilde{a}) = \partial_k^{\BB}(b + \tilde{b})$ 
with $g_k(b+ \breve{b}) = c + \breve{c}$, we get
\[
\tilde{\partial}_k([c]+[\tilde{c}])
= \tilde{\partial}_k([c+\tilde{c}]) = [ a + \tilde{a}]
= [a] + [\tilde{a}] =
\tilde{\partial}_k([c]) + \tilde{\partial}_k([\tilde{c}]) \ .
\]
Similarly, $\tilde{\partial}_k(r[c]) = r \tilde{\partial}_k([c])$
for all $r \in R$.

\stage{ii} We prove that $\IMG(H_k(\F)) = \KE(H_k(\GG))$.
Since $g_k \circ f_k = 0$ on $A_k$, we get that
$\big(H_k(\GG) \circ H_k(\F)\big)([c])
= [(g_k \circ f_k)(c)] = [0]$ for all $[c] \in H_k(\A)$.  Thus
$\IMG(H_k(\F)) \subset \KE(H_k(\GG))$.

Suppose that $[b] \in \KE(H_k(\GG))$.  We therefore have that
$H_k(\GG)([b]) = [g_k(b)] = [0]$.  Thus
$\displaystyle g_k(b) = \partial_{k+1}^{\C}(c)$ for some $c \in C_{k+1}$.
Since $g_{k+1}$ is onto, there exists $\tilde{b} \in B_{k+1}$ such
that $g_{k+1}(\tilde{b}) = c$.  We get
\[
g_k(b - \partial_{k+1}^{\BB}(\tilde{b}))
= g_k(b) - \partial_{k+1}^{\C}(g_{k+1}(\tilde{b}))
= g_k(b) - \partial_{k+1}^{\C}(c) = 0 \ .
\]
Since $\IMG(f_k) = \KE(g_k)$, there exists $a \in A_k$ such that
$f_k(a) = b - \partial_{k+1}^{\BB}(\tilde{b})$.

We have that $a \in Z_k(\A)$ because $b \in Z_k(\BB)$ implies that
\[
f_{k-1}(\partial_k^{\A}(a)) = \partial_k^{\BB}(f_k(a))
= \partial_k^{\BB}\big( b - \partial_{k+1}^{\BB}(\tilde{b}) \big)
= \partial_k^{\BB}(b) = 0 
\]
and $f_{k-1}$ being one-to-one implies that
$\displaystyle \partial_k^{\A}(a) = 0$.  Thus $[a] \in H_k(\A)$.
Moreover,\\
$\displaystyle H_k(\F)([a]) = [f_k(a)]
= [b - \partial_{k+1}^{\BB}(\tilde{b})] = [b]$.
Thus $[b] \in \IMG(H_k(\F)$.  This proves that
$\KE(H_k(\GG)) \subset \IMG(H_k(\F))$.

\stage{iii} We prove that $\IMG(H_k(\GG)) = \KE(\tilde{\partial}_k)$.
Given $[c] \in H_k(\C)$, we have that $\tilde{\partial}_k([c]) = [a]$ where
$\displaystyle f_{k-1}(a) = \partial_k^{\BB}(b)$ with $g_k(b) = c$ for
some $b \in B_k$.

If $[c] \in \IMG(H_k(\GG))$, then $[c] = H_k(\GG)([b]) = [g_k(b)]$
for some $[b] \in H_k(\BB)$.  In particular, $b \in Z_k(\BB)$.
Thus $\displaystyle f_{k-1}(a) = \partial_k^{\BB}(b) = 0$ implies that
$a=0$ because $f_{k-1}$ is one-to-one.  Thus $\tilde{\partial}_k([c]) = [0]$;
namely, $[c] \in \KE(\tilde{\partial}_k)$.  We have proved that
$\IMG(H_k(\GG) \subset \KE(\tilde{\partial}_k)$.

Suppose that $[c] \in \KE(\tilde{\partial}_k)$.
Then $\displaystyle a = \partial_k^{\A}(\tilde{a})$ for some
$\tilde{a} \in A_k$.  We have that $b - f_k(\tilde{a}) \in Z_k(\BB)$ because
\[
\partial_k^{\BB}(b - f_k(\tilde{a}))
= \partial_k^{\BB}(b) - f_{k-1}(\partial_k^{\A}(\tilde{a}))
= \partial_k^{\BB}(b) - f_{k-1}(a) = 0
\]
and $[c] = H_k(\GG)\big([b-f_k(\tilde{a})]\big)$ because
\[
g_k\big(b - f_k(\tilde{a})\big)
= g_k(b) - g_k(f_k(\tilde{a})) = g_k(b) - 0 = c \ .
\]
Thus $[c] \in \IMG(H_k(\GG))$.  This proves that
$\KE(\tilde{\partial}_k) \subset \IMG(H_k(\GG)$.

\stage{iv} We prove that $\IMG(\tilde{\partial}_k) = \KE(H_{k-1}(\F))$.
If $[a] \in \IMG(\tilde{\partial}_k)$, then
$\displaystyle f_{k-1}(a) = \partial_k^{\BB}(b)$ for some $b \in B_k$.
Hence $\displaystyle 
H_{k-1}(\F)([a]) = [f_{k-1}(a)] = [ \partial_k^{\BB}(b) ] = [0]$.
Thus $[a] \in \KE(H_{k-1}(\F))$.  Hence
$\IMG(\tilde{\partial}_k) \subset \KE(H_{k-1}(\F))$.

Suppose that $[a] \in \KE(H_{k-1}(\F))$.  Since $H_{k-1}(\F)([a])
= [f_{k-1}(a)] = [0]$, we get that
$\displaystyle f_{k-1}(a) = \partial_k^{\BB}(b)$
for some $b \in B_k$.  Let $c = g_k(b)$.  We have that $c \in Z_k(C)$
because
\[
\partial_k^{\C}(c) = \partial_k^{\C}(g_k(b)) = 
g_{k-1}\big(\partial_k^{\BB}(b)\big) = g_{k-1}(f_{k-1}(a)) = 0 \ .
\]
We also have by definition of $\tilde{\partial}_k$ that
$\tilde{\partial}_k([c]) = [a]$.  Thus $[a] \in \IMG(\tilde{\partial}_k)$.
This proves that $\KE(H_{k-1}(\F)) \subset \IMG(\tilde{\partial}_k)$.
\end{proof}

\begin{defn} \label{defnChHom}
Let $\F = \{f_k\}_{k\in \ZZ}$ and $\GG = \{g_k\}_{k\in \ZZ}$ be two
chain maps between the chain complexes
$\C = \{(C_k,\partial_k)\}_{k\in \ZZ}$ and
$\tilde{\C} = \{ (\tilde{C}_k,\tilde{\partial}_k)\}_{k\in \ZZ}$.
A {\bfseries chain homotopy}\index{Chain Homotopy} between $\F$ and
$\GG$ is a sequence
$\D = \{D_k\}_{k\in \ZZ}$ where the maps $D_k:C_k \to \tilde{C}_{k+1}$
are homomorphism such that 
$\tilde{\partial}_{k+1} \circ D_k + D_{k-1} \circ \partial_k = f_k - g_k$
for $k \in \ZZ$.
\end{defn}

A commutative diagram representing a chain homotopy is given below.
\[
\xymatrix@R+2em@C+2em{
\quad \ar[r]^-{\partial_{k+2}}
&  C_{k+1}  \ar[r]^{\partial_{k+1}} \ar@/^/[d]^{f_{k+1}} \ar@/_/[d]_{g_{k+1}}
\ar[ld]^(0.6){D_{k+1}}
&  C_k  \ar[r]^{\partial_k} \ar@/^/[d]^{f_k} \ar@/_/[d]_{g_k}  \ar[ld]^(0.6){D_k}
&  C_{k-1}  \ar[r]^-{\partial_{k-1}} \ar@/^/[d]^{f_{k-1}} \ar@/_/[d]_{g_{k-1}}
\ar[ld]^(0.6){D_{k-1}}
& \rule{0em}{1em} \rule{1.3em}{0em} \ar[ld]^(0.6){D_{k-2}}\\
\rule{0em}{1em} \rule{1.3em}{0em}
\ar[r]_-{\tilde{\partial}_{k+2}} &  \tilde{C}_{k+1}
\ar[r]_{\tilde{\partial}_{k+1}} &  \tilde{C}_k
\ar[r]_{\tilde{\partial}_k} &  \tilde{C}_{k-1}
\ar[r]_-{\tilde{\partial}_{k-1}} &
}
\]

\begin{prop} \label{propCHiFeG}
If we have a chain homotopy as in Definition~\ref{defnChHom}, then
$H_k(\F) = H_k(\GG)$ for all $k$.
\end{prop}

\begin{proof}
Given $[c] \in H_k(C)$.  Then $\partial_k(c) = 0$.  It follows from
$\tilde{\partial}_{k+1}(D_k(c)) + D_{k-1}(\partial_k(c)) = f_k(c) - g_k(c)$
that $f_k(c) - g_k(c) = \tilde{\partial}_{k+1}(D_k(c))$ with
$D_k(c) \in \tilde{C}_{k+1}$.  Thus $H_k(\F)([c]) = [f_k(c)]= [g_k(c)]
= H_k(\GG)([c])$.
\end{proof}

\begin{defn}
Let $\displaystyle \C = \{ (C_k, \partial_k) \}_{k \in \ZZ}$ be a chain complex
A surjective linear operator $\epsilon : C_0 \to R$ such that
$\epsilon \circ \partial_1 = 0$ is called an
{\bfseries augmentation}\index{Augmentation}
of the chain complex $\C$.
\end{defn}

Suppose that $\epsilon$ is an augmentation of the
$\displaystyle \C = \{ (C_k, \partial_k) \}_{k \in \ZZ}$.  If we set
\[
C^\sharp_k =
\begin{cases}
C_k & \quad \text{if} \ k \neq -1 \\
R & \quad \text{if} \ k = -1
\end{cases}
\qquad \text{and} \qquad
\partial^\sharp_k =
\begin{cases}
\partial_k & \quad \text{if} \ k \neq 0 \\
\epsilon & \quad \text{if} \ k = 0  
\end{cases}
\]
then we get a new chain complex
$\displaystyle \C^\sharp = \{ (C_k^\sharp,\partial_k^\sharp) \}_{k\in \ZZ}$.

\begin{defn} \label{defnRedCC}
The chain complex
$\displaystyle \C^\sharp = \{ (C_k^\sharp,\partial_k^\sharp) \}_{k\in \ZZ}$
defined above is called the
{\bfseries reduced chain complex}\index{Reduced Chain Complex}
of the chain complex $\C$ with augmentation $\epsilon$ \footnotemark.
\end{defn}

\footnotetext{We have not found any reason why an ``augmentation'' of
a chain complex leads to a ``reduced'' chain complex.  It is perhaps
due to Proposition~\ref{propHkEquHskpR}.  The choice of terminology in
mathematics is not always obvious or even logic.}

Suppose that
$\displaystyle \tilde{\C} = \{ (\tilde{C}_k,\tilde{\partial}_k) \}_{k\in \ZZ}$
is another chain complex and
$\displaystyle \tilde{\C}^\sharp
= \{ (\tilde{C}_k^\sharp,\tilde{\partial}_k^\sharp) \}_{k\in \ZZ}$ is
a reduced chain complex of $\tilde{\C}$ with augmentation $\tilde{\epsilon}$.
Moreover, suppose that $\F = \{f_k\}_{k\in \ZZ}$ is a chain map
between $\C$ and $\tilde{\C}$.

\begin{defn}
We say that the chain map $\F$ defined above is
{\bfseries augmentation-preserving}\index{Augmentation-Preserving} 
if $\tilde{\epsilon} \circ f_0 = \epsilon$.
\end{defn}

If we set $f_{-1} = \Id_R$, then $\F = \{f_k\}_{k\in \ZZ}$ becomes a chain map
between $\displaystyle \C^\sharp$ and $\displaystyle \tilde{\C}^\sharp$.

\begin{prop} \label{propCsCtfauHsH}
Let $\displaystyle \C^\sharp = \{ (C_k^\sharp, \partial_k^\sharp) \}_{k \in \ZZ}$
be a reduced chain complex of the chain complex
$\C = \{ (C_k, \partial_k) \}_{k \in \ZZ}$ with augmentation $\epsilon$, and
$\displaystyle \tilde{\C}^\sharp
= \{ (\tilde{C}_k^\sharp,\tilde{\partial}_k^\sharp) \}_{k\in \ZZ}$ be a
reduced chain complex of the chain complex
$\displaystyle \tilde{\C} = \{ (\tilde{C}_k,\tilde{\partial}_k) \}_{k\in \ZZ}$
with augmentation $\tilde{\epsilon}$.
Suppose that $\F = \{ f_k\}_{k\in \ZZ}$ is an augmentation-preserving
chain map between $\displaystyle \C$ and $\displaystyle \tilde{\C}$.
Then $\displaystyle H_k(\F): H_k(\C^\sharp) \to H_k(\tilde{\C}^\sharp)$
is an isomorphism if and only if
$H_k(\F) : H_k(\C) \to H_k(\tilde{\C})$ is an isomorphism.
\end{prop}

\begin{proof}
The result is obvious for $k \neq 0$ because
$\displaystyle H_k(\C^\sharp) = H_k(\C)$ and 
$\displaystyle H_k(\tilde{\C}^\sharp) = H_k(\tilde{\C})$ for $k\neq 0$.

For $k=0$, we have the following commutative diagram
\begin{equation} \label{CsCtfauHsHEq1}
\xymatrix{ 0 \ar[r] & \KE(\epsilon) \ar[r]^{\iota} \ar[d]^{f_0}
& C_0 \ar@/^/[r]^{\epsilon} \ar[d]^{f_0}
& R \ar[r] \ar@/^/[l]^{g} \ar[d]^{\Id_R} & 0 \\
0 \ar[r] & \KE(\tilde{\epsilon}) \ar[r]^{\iota}
& \tilde{C}_0 \ar@/^/[r]^{\tilde{\epsilon}}
& R \ar[r] \ar@/^/[l]^{\tilde{g}} & 0
}
\end{equation}
where the two horizontal sequence are exact.  The map
$\iota$ is the inclusion map and $g:R \to C_0$ is a map such that
$\epsilon \circ g = \Id_R$.  A map $g$ exists because $\epsilon$ is onto.
Let $\tilde{g} = f_0 \circ g \circ \Id_R$.

It follows from Proposition~\ref{propSplit2Cond} that both horizontal
sequence are split.  Hence, we get from Proposition~\ref{propSplitOplus} that
$C_0 = \KE(\epsilon) \oplus \IMG(g)$ and
$\tilde{C}_0 = \KE(\tilde{\epsilon}) \oplus \IMG(\tilde{g})$.

Since $C_0 = \KE(\partial_0)$ because we assume that $C_{-1} = \{0\}$,
we have
\[
H_0(\C) = C_0/\IMG(\partial_1) = \big(\KE(\epsilon) \oplus
\IMG(g)\big) / \IMG(\partial_1)
\cong \big(\KE(\epsilon) / \IMG(\partial_1)\big) \oplus \IMG(g)
\]
because $\IMG(\partial_1) \subset \KE(\epsilon)$ and
$\KE(\epsilon) \cap \IMG(g) = \{0\}$.  More precisely, we have that
$\IMG(g) / \IMG(\partial_1) \cong \IMG(g)$.

Similarly, since $\tilde{C}_0 = \KE(\tilde{\partial}_0)$ because we assume
that $\tilde{C}_{-1} = \{0\}$, we have
\[
H_0(\tilde{\C}) = \tilde{C}_0/\IMG(\tilde{\partial}_1)
= \big(\KE(\tilde{\epsilon}) \oplus \IMG(\tilde{g})\big)
/ \IMG(\tilde{\partial}_1)
\cong \big(\KE(\tilde{\epsilon}) / \IMG(\tilde{\partial}_1)\big)
\oplus \IMG(\tilde{g})
\]
because $\IMG(\tilde{\partial}_1) \subset \KE(\tilde{\epsilon})$ and
$\KE(\tilde{\epsilon}) \cap \IMG(\tilde{g}) = \{0\}$.  Again,
we have that
$\IMG(\tilde{g}) / \IMG(\tilde{\partial}_1) \cong \IMG(\tilde{g})$.

Using the notation of the reduced chain complexes, we get
\begin{equation} \label{HFHCHCtEq1}
H_0(\C) \cong \big(\KE(\partial_0^\sharp) / \IMG(\partial_1^\sharp)\big)
\oplus \IMG(g) = H_0(\C^\sharp) \oplus \IMG(g)
\end{equation}
and
\begin{equation} \label{HFHCHCtEq2}
H_0(\tilde{\C}) \cong \big(\KE(\tilde{\partial}_0^\sharp) /
\IMG(\tilde{\partial}_1^\sharp)\big) \oplus \IMG(\tilde{g})
\cong H_0(\tilde{\C}^\sharp) \oplus \IMG(\tilde{g}) \ .
\end{equation}
Since $\IMG(g) \cong R$ and $\IMG(\tilde{g}) \cong R$ because the
two horizontal sequences in (\ref{CsCtfauHsHEq1}) are split, we
get that $\displaystyle H_0(\C^\sharp) \cong H_0(\tilde{\C}^\sharp)$ if
and only if $H_0(\C) \cong H_0(\tilde{\C})$.

We have that
$f_0 : \IMG(g) \to \IMG(\tilde{g})$ is an isomorphism.
Suppose that $f_0(a) = f_0(b)$ with $a,b \in \IMG(g)$.  Then
$\epsilon(a) = \tilde{\epsilon}(f_0(a)) = \tilde{\epsilon}(f_0(b)) = 
\epsilon(b)$.  Thus $a-b \in \KE(\epsilon) \cap \IMG(g) = \{0\}$.
Hence $a = b$.  This proves that $f_0: \IMG(g) \to \IMG(\tilde{g})$ is
one-to-one.  To prove that it is onto, consider $b \in \IMG(\tilde{g})$.
Therefore there exists $r \in R$ such that
$b = \tilde{g}(r) = f_0(g(r))$ and $g(r) \in \IMG(g)$.

It then follows from (\ref{HFHCHCtEq1}) and
(\ref{HFHCHCtEq2}) that $H_0(\F):H_0(\C) \to H_0(\tilde{\C})$ is an
isomorphism if and only
$\displaystyle H_0(\F):H_0(\C^\sharp) \to H_0(\tilde{\C}^\sharp)$
is an isomorphism because $H_0(\F): \IMG(g) / \IMG(\partial_1) \cong
\IMG(g) \to \IMG(\tilde{g}) / \IMG(\tilde{\partial}_1) \cong
\IMG(\tilde{g})$ is an isomorphism.
\end{proof}

\begin{defn}
A chain complex $\{(C_k,\partial_k)\}_{k\in \ZZ}$ is
{\bfseries acyclic}\index{Acyclic} if $H_k(\C) = 0$ for all $k$.
\end{defn}

\begin{defn}
A topological space $X$ is {\bfseries aspherical}\index{Aspherical Space}
if every continuous map $\displaystyle f: S^k\to X$ can be extended to
a continuous map $\displaystyle F: B_{k+1} \to X$ where
$\displaystyle B_{k+1} = \overline{B_1(\VEC{0})} \subset \RR^{k+1}$ is the
closed unit ball centred at the origin.
\end{defn}

\begin{rmkList} \label{rmkSasph}
\begin{enumerate}
\item Convex subsets $\displaystyle X \subset \RR^n$ are aspherical.
Suppose that $\displaystyle f: S^k \to X$ is a continuous map.
Choose $y \in X$.  Since
$\displaystyle \overline{B_1(\VEC{0})} \cong \big(S^k \times [0,1]\big) /
\big(S^k \times \{0\}\big)$, we define
$F: \overline{B_1(\VEC{0})} \to X$ as it follows.
$F([x,t]) = (1-t)y +tf(x)$ for all $[x,t] \in \overline{B_1(\VEC{0})}$.
Note that $F([x,t] \in X$ for all $x\in X$ and $t \in [0,1]$ because
$X$ is convex.  We also have that $F([x,1]) = f(x)$ for all
$\displaystyle x\in S^k$.  For the readers who may be wondering why we
had to quotient $\displaystyle S^k \times [0,1]$ by
$\displaystyle S^k \times \{0\}$, they should remember that
we always have this singularity at the origin when using polar
coordinates.  The quotient by $\displaystyle S^k \times \{0\}$
eliminates this problem.
\item Since $\Delta_k \times [0,1]$ is convex, it is aspherical.
\end{enumerate}
\end{rmkList}

\begin{theorem}
Suppose that $X$ is an aspherical space and that $R$ is an integral
domain.  Then a reduced chain complex
$\displaystyle \C^\sharp = \big\{ \big( S_k^\sharp(X;R),
\partial_k^\sharp\big)\big\}_{k\in \ZZ}$ of the chain complex
$\displaystyle \C = \{ ( S_k(X;R), \partial_k)\}_{k\in \ZZ}$
with augmentation $\epsilon$ is acyclic.
\end{theorem}

\begin{proof}
The proof is in two steps.

\stage{I}
Define the linear map $\eta: R \to S_0(X;R)$ by $\eta(s) = s\,e_x$ for
some $x \in X$ such that $\epsilon(e_x) = 1$.  Recall that
$e_x:\Delta_0 \to X$ is the singular $0$-simplex defined by
$e_x(\VEC{e}_0) = x$.  Hence, $\eta$ is a right inverse of $\epsilon$;
namely, $\epsilon \circ \eta = \Id_R$.

Let $\F = \{ f_k\}_{k\in \ZZ}$ be the chain map between $\C$ and
itself defined by $f_0 = \eta \circ \epsilon$ and $f_k = 0$ otherwise.
Moreover, Let $\GG = \{ g_k\}_{k\in \ZZ}$ be another chain map between $\C$ and
itself defined by $g_k = \Id_{S_k(X;R)}$ for $k \geq 0$ and $g_k = 0$ otherwise.

We prove that if there is a chain homotopy $\D = \{D_k\}_{k\in \ZZ}$
between $\F$ and $\GG$, then $\displaystyle \C^\sharp$ is acyclic.
We have that $D_k:S_k(X;R) \to S_{k+1}(X;R)$ are homomorphism such that 
\begin{equation} \label{pHkDI0Eq1}
\partial_{k+1} \circ D_k + D_{k-1} \circ \partial_k = g_k - f_k
\end{equation}
for $k \in \ZZ$.

Suppose that $\sigma \in Z_k(S;R)$.  It follows from (\ref{pHkDI0Eq1}) that
\[
\partial_{k+1}(D_k(\sigma))
= \begin{cases}
\sigma & \quad \text{if} \ k \neq 0 \\
\sigma - \eta(\epsilon(\sigma)) & \quad \text{if} \ k = 0
\end{cases}
\]
We have that $\sigma \in B_k(S;R)$ for $k>0$.  Thus $[\sigma] = 0$.
Since $\sigma \in Z_k(S;R)$ is arbitrary, we get that
$\displaystyle H_k^\sharp(S;R) = H_k(S;R) = 0$ for $k>0$.
If $k=0$, then we have
\begin{equation} \label{pHkDI0Eq2}
\sigma - \eta(\epsilon(\sigma))
= \partial_1(D_0(\sigma)) \ .
\end{equation}
By definition of the augmentation $\epsilon$, we have that
$\displaystyle \KE(\epsilon) = Z^\sharp_0(S;R)$.  We need to prove
that $\displaystyle \KE(\epsilon) \subset \IMG(\partial_1) = B_0^\sharp(S;R)$
to obtain that $\displaystyle H_0^\sharp(S;R) = 0$.  Consider
$\sigma \in \KE(\epsilon)$. 
It follows from (\ref{pHkDI0Eq2}) that
$\sigma = \partial_1(D_0(\sigma))$.  Thus
$\displaystyle \sigma \in \IMG(\partial_1) = B_0^\sharp(S;R)$.
Hence $\displaystyle [\sigma] = 0 \in H_0^\sharp(S;R)$.
Since $\sigma \in \KE(\epsilon)$ is arbitrary, we get that
$\displaystyle H_0^\sharp(S;R) = 0$.

\stage{II}  We prove that there is a chain homotopy
$\D = \{D_k\}_{k\in \ZZ}$ between the chain maps $\F$ and
$\GG$ defined in (i).

Let $\displaystyle \dot{\Delta}_{k+1} = \bigcup_{i=0}^{k+1}
F_{k+1}^i(\Delta_k)$.  It is the topological boundary of $\Delta_{k+1}$. 

Given a singular $k$-simplex, we proceed by induction
to define\\
$D_k :S_k(X;R) \to S_{k+1}(X;R)$.  Obviously, we may set
$D_k = 0$ for $k<0$.

\stage{II.i} Given a singular $0$-form $\sigma$ in $X$, we define
$\beta_\sigma:\dot{\Delta}_1 \to X$ such that
$\displaystyle \beta_\sigma \circ F^0_1 = \sigma$ and
$\displaystyle \beta_\sigma \circ F^1_1 = (\eta \circ \epsilon)(\sigma)$.
Since $\dot{\Delta}_1 = \{\VEC{e}_0,\VEC{e}_1\}$, two isolated points,
we have that $\beta_\sigma$ is obviously continuous on
$\dot{\Delta}_0$.  Thus, the map $\beta_\sigma$ can be extended to
$\Delta_1$ because $X$ is aspherical to give $D_0(\sigma):\Delta_1 \to X$.
Note that $\displaystyle B_k$ is homeomorphic to $\Delta_k$ for all $k\geq 0$.
Now that $D_0$ is defined for arbitrary singular $0$-forms in $X$, we
may extend $D_0$ by linearity to $S_0(X;R)$.

\stage{II.ii} Given a singular $1$-simplex $\sigma$ in $X$, we define
$\beta_\sigma:\dot{\Delta}_2 \to X$ such that
$\displaystyle \beta_\sigma\circ F^0_2 = \sigma$ and
$\displaystyle \beta_\sigma \circ F^i_2 =
D_0(\sigma\circ F^{i-1}_1)$ for $i =1,2$.
To prove that $\beta_\sigma$ is continuous on $\dot{\Delta}_2$, we
have to prove that $\beta_\sigma$ is continuous on the intersection of
the distinct faces of $\Delta_2$.  Let
$\displaystyle K_{i,j} = F^i_2(\Delta_1) \cap F^j_2(\Delta_1)$ for
$0 \leq j < i \leq 2$.

We have proved in the context of the proof of
Proposition~\ref{propdkmqdk0} that
$\displaystyle F_2^i \circ F_1^j = F_2^j \circ F_1^{i-1}$
for $0 \leq j < i \leq 2$.  Thus
$\displaystyle K_{i,j} = (F^i_2\circ F^j_1)(\Delta_0)
= (F_2^j \circ F_1^{i-1})(\Delta_0)$ for $0 \leq j < i \leq 2$.
Hence
$\displaystyle K_{0,1} = \{\VEC{e}_2\} = \{(F^1_2\circ F^0_1)(\VEC{e}_0) \}$,
$\displaystyle K_{0,2} = \{\VEC{e}_1\} = \{ (F^2_2\circ F^0_1)(\VEC{e}_0) \}$ and
$\displaystyle K_{1,2} = \{\VEC{e}_0\} = \{ (F_2^2 \circ F_1^1)(\VEC{e}_0) \}$.
Since the intersection of the faces is only one point, we obviously
have that $\beta_\sigma$ is continuous on $\dot{\Delta}_1$.
Therefore, the map $\beta_\sigma$ can be extended to $\Delta_2$ because $X$ is
aspherical to give $D_1(\sigma):\Delta_2 \to X$.
Now that $D_1$ is defined for arbitrary singular $1$-forms in $X$, we
may extend $D_1$ by linearity to $S_1(X;R)$.

\stage{II.iii} We assume that $D_{k-1} : S_{k-1}(X;R) \to S_k(X;R)$ has
been defined.  Given a singular $k$-simplex $\sigma$ in $X$, 
we define $\beta_\sigma:\dot{\Delta}_{k+1} \to X$ such that
$\displaystyle \beta_\sigma\circ F^0_{k+1} = \sigma$ and
$\displaystyle \beta_\sigma \circ F^i_{k+1} =
D_{k-1}(\sigma\circ F^{i-1}_k)$ for $1 \leq i \leq k+1$.
To prove that $\beta_\sigma$ is continuous on $\dot{\Delta}_{k+1}$, we
have to prove that $\beta_\sigma$ is continuous on the intersection of
the distinct faces of $\Delta_{k+1}$.

Let $\displaystyle K_{i,j} = F^i_{k+1}(\Delta_k) \cap F^j_{k+1}(\Delta_k)$ for
$0 \leq j < i \leq k+1$.  As we mentioned before, we have proved in
the context of the proof of Proposition~\ref{propdkmqdk0} that
$\displaystyle F_{k+1}^i \circ F_k^j = F_{k+1}^j \circ F_k^{i-1}$
for $0 \leq j < i \leq k+1$.  Thus
$\displaystyle K_{i,j} = (F^i_{k+1}\circ F^j_k)(\Delta_{k-1})
= (F_{k+1}^j \circ F_k^{i-1})(\Delta_{k-1})$ for
$0 \leq j < i \leq k+1$.  Since $i > j \geq 0$, we get from the
definition of $D_{k-1}$ that
\begin{align}
\beta_\sigma \circ F^i_{k+1} \circ F^j_k
&= (\beta_\sigma \circ F^i_{k+1}) \circ F^j_k
= D_{k-1}(\sigma \circ F_k^{i-1}) \circ F^j_k \nonumber \\
&= \begin{cases}
\sigma \circ F_k^{i-1} & \quad \text{if}\ j = 0 \\
D_{k-2}(\sigma \circ F_k^{i-1} \circ F^{j-1}_{k-1}) & \quad \text{if}
\ 0 < j < i
\end{cases}  \label{pHkDI0Eq3}
\end{align}
To get the last equality, wee use the fact that
$\displaystyle D_{k-1}(\sigma \circ F_k^{i-1})\Big|_{\dot{\Delta}_k}
= \beta_{\sigma\circ F_k^{i-1}}:\dot{\Delta}_k \to X$ for the
definition of $D_{k-1}$.

Moreover, we get from the definition of $\beta_\sigma$ and $D_{k-1}$ that
\begin{align}
\beta_\sigma \circ F^j_{k+1} \circ F^{i-1}_k
&= (\beta_\sigma \circ F^j_{k+1}) \circ F^{i-1}_k
= \begin{cases}
\sigma \circ F_k^{i-1} & \quad \text{if}\ j = 0 \\
D_{k-1}(\sigma \circ F_k^{j-1}) \circ F^{i-1}_k & \quad \text{if}
\ 0 < j < i
\end{cases} \nonumber \\
&= \begin{cases}
\sigma \circ F_k^{i-1} & \quad \text{if}\ j = 0 \\
D_{k-2}(\sigma \circ F_k^{j-1} \circ F^{i-2}_{k-1}) & \quad \text{if}
\ 0 < j < i
\end{cases}  \label{pHkDI0Eq4}
\end{align}
To get the last equality, we again use the fact that
$\displaystyle D_{k-1}(\sigma \circ F_k^{i-1})\Big|_{\dot{\Delta}_k}
= \beta_{\sigma\circ F_k^{i-1}}:\dot{\Delta}_k \to X$ for the
definition of $D_{k-1}$.

Since $\displaystyle F_k^{i-1} \circ F_{k-1}^{j-1}
= F_k^{j-1} \circ F_{k-1}^{i-2}$
for $1 \leq j < i \leq k+1$, we get from (\ref{pHkDI0Eq3}) and
(\ref{pHkDI0Eq4}) that
$\displaystyle \beta_\sigma \circ F^i_{k+1} \circ F^j_k
= \beta_\sigma \circ F^j_{k+1} \circ F^{i-1}_k$; namely,
$\beta_\sigma$ on the $\displaystyle j^{th}$ face of the 
$\displaystyle i^{th}$ face of $\dot{\Delta}_{k+1}$ is equal to
$\beta_\sigma$ on the $\displaystyle (i-1)^{th}$ face of the 
$\displaystyle j^{th}$ face of $\dot{\Delta}_{k+1}$.
Thus $\beta_\sigma: \dot{\Delta}_k \to X$ is continuous.
Therefore, the map $\beta_\sigma$ can be extended to $\Delta_{k+1}$
because $X$ is aspherical to give $D_k(\sigma):\Delta_{k+1} \to X$.
Now that $D_k$ is defined for arbitrary singular $k$-forms in $X$, we
may extend $D_k$ by linearity to $S_k(X;R)$.

\stage{II.iv} It is left to prove that $\D = \{D_k\}_{k\in \ZZ}$ is a
chain homotopy between the chain maps $\F$ and $\GG$ defined in (i);
namely, that
\begin{equation} \label{pHkDI0Eq5}
\partial_{k+1} \circ D_k + D_{k-1} \circ \partial_k = g_k - f_k
\end{equation}
for $k \in \ZZ$.  The equation is trivially true for $k<0$.
For $k=0$, we have from the definition of $D_0$ that
\[
\partial_1(D_0(\sigma)) = D_0(\sigma) \circ F^0_1 -
D_0(\sigma) \circ F^1_1 = \sigma - (\eta \circ \epsilon)(\sigma)
= g_0(\sigma) - f_0(\sigma)
\]
for all singular $0$-simplices in $X$.  By linearity,
(\ref{pHkDI0Eq5}) is true on $S_0(X;R)$.

For $k>0$, we have
\begin{align*}
&\partial_{k+1} (D_k(\sigma)) + D_{k-1}(\partial_k(\sigma))
= \sum_{i=0}^{k+1} (-1)^i D_k(\sigma) \circ F^i_{k+1}
+ D_{k-1}\left( \sum_{i=0}^k (-1)^i \sigma \circ F^i_k \right) \\
&\qquad = (D_k(\sigma)) \circ F^0_{k+1}
+ \sum_{i=1}^{k+1} (-1)^i D_k(\sigma) \circ F^i_{k+1}
+ \sum_{i=0}^k (-1)^i  D_{k-1}(\sigma \circ F^i_k ) \\
&\qquad = \underbrace{D_k(\sigma) \circ F^0_{k+1}}_{=\sigma}
+ \sum_{i=0}^k (-1)^{i+1}
\underbrace{D_k(\sigma) \circ F^{i+1}_{k+1}}_{= D_{k-1}(\sigma\circ F_k^i)}
+ \sum_{i=0}^k (-1)^i  D_{k-1}(\sigma \circ F^i_k ) \\
&\qquad = \sigma = g_k(\sigma) - f_k(\sigma)
\end{align*}
for all singular $k$-simplices in $X$.
By linearity, (\ref{pHkDI0Eq5}) is true on $S_k(X;R)$ for $k>0$.
\end{proof}

\begin{prop} \label{propHkDI0}
Let $R$ be an integral domain.  Then
$H_k(\Delta_k \times [0,1];R) = 0$ for $k>0$.  
\end{prop}

\begin{proof}
According to the Remark~\ref{rmkSasph}, $\Delta_k \times [0,1]$ is aspherical.  
It follows from the previous theorem that the reduced chain complex
$\displaystyle \C^\sharp = \big\{ \big( S_k^\sharp(\Delta_k \times [0,1];R),
\partial_k^\sharp\big)\big\}_{k\in \ZZ}$ of the chain complex
$\displaystyle \C = \{ ( S_k(\Delta_k \times [0,1];R), \partial_k)\}_{k\in \ZZ}$
with augmentation $\epsilon$ is acyclic.  Therefore
$\displaystyle H_k(\Delta_k \times [0,1];R)
= H_k^\sharp(\Delta_k \times [0,1];R) = 0$ for $k>0$.  
\end{proof}

The proof of the next theorem is based on the previous proposition.

\begin{theorem} \label{thmExistCH}
Let $X$ and $Y$ be two topological spaces, and $f,g:X \to Y$ be two
homotopic functions.  Consider chain maps
$\F = \{f_k\}_{k\in \ZZ}$ and $\GG = \{g_k\}_{k\in \ZZ}$
between the chain complexes
$\C = \{(S_k(X;R),\partial_k)\}_{k\in \ZZ}$ and
$\tilde{\C} = \{ (S_k(Y;R) ,\partial_k)\}_{k\in \ZZ}$ defined by
$f_k = S_k(f)$ and $g_k = S_k(g)$ for all $k$.  There exists a chain homotopy
$\D = \{ D_k\}_{k\in Z}$ between $\F$ and $\GG$
\end{theorem}

\begin{proof}
\stage{i} Let $\iota_y: X \to X \times [0,1]$ be the functions defined by
$\iota_y(x) = (x,y)$ for $x \in X$ and $y \in [0,1]$.  Moreover, let
$\SS_0 = \{S_k(\iota_0)\}_{k\in \ZZ}$ and $\SS_1 = \{S_k(\iota_1)\}_{k\in \ZZ}$
be chain maps between the chain complexes
$\C = \{(S_k(X;R),\partial_k)\}_{k\in \ZZ}$ and
$\breve{\C} = \{ (S_k(X\times [0,1];R) ,\partial_k)\}_{k\in \ZZ}$.  We prove that
there exists a chain homotopy $\tilde{\D} = \{ \tilde{D}_k\}_{k\in Z}$ between
$\SS_0$ and $\SS_1$.   To be more precise, we prove by induction the
following statement.

\noindent{$\mathbf{P_k}$}:
There exists a homomorphism $\tilde{D}_k:S_k(X;R) \to S_{k+1}(X\times [0,1];R)$
such that:
\begin{enumerate}
\item $\partial_{k+1} \circ \tilde{D}_k + \tilde{D}_{k-1} \circ \partial_k
= S_k(\iota_1) - S_k(\iota_0)$.
\item $\tilde{D}_k \circ S_k(h) = S_{k+1}(h \times \Id_{[0,1]})
\circ \tilde{D}_k$ for all continuous map $h:X \to Y$
\footnote{This property is called
{\bfseries naturality}\index{Naturality} and is really important in
category theory.  In particular, we can show that the definition of
$\tilde{D}_k$ is independent of the topological space homeomorphic to
$X$.  We leave this subject to those in pure algebra and logic.}.
\item $\partial_{k+1} \circ \big( S_{k+1}(\iota_1) - S_{k+1}(\iota_0)
  - \tilde{D}_k \circ \partial_{k+1} \big) = 0$.
\end{enumerate}
Note that $h \times \Id_{[0,1]}:X\times [0,1] \to \tilde{X} \times [0,1]$ is
defined by $(h \times \Id_{[0,1]})(x,t) = (h(x),t)$ for all
$(x,t) \in X \times [0,1]$.

We may set $\tilde{D}_k = 0$ for $k<0$.  Hence, we only have to consider
$k \geq 0$.

If $\sigma$ is a singular $k$-simplex in $X$, then we also
note that the singular $k$-simplex $S_k(\iota_y)(\sigma)$ in
$X \times [0,1]$ is defined by
$(S_k(\iota_y)(\sigma))(\VEC{x}) = (\sigma, e_y)(\VEC{x})
= (\sigma(\VEC{x}),y)$
for all $\VEC{x} \in \Delta_k$, where the singular $k$-simplex
$e_y :\Delta_k \to [0,1]$ is defined by
$e_y(\VEC{x}) = y$ for all $\VEC{x} \in \Delta_k$.

We proof that $P_0$ is true.
Given a singular $0$-simplex $\sigma$ in $X$, we define
$\tilde{D}_0(\sigma) \in S_1(X\times [0,1];R)$ as
$\tilde{D}_0(\sigma)(\VEC{x})
= \big(\rho_{\sigma}(\VEC{x}),\rho_I(\VEC{x})\big)$
for all $\VEC{x} \in \Delta_1$, where $\rho_{\sigma}:\Delta_1 \to X$ and
$\rho_I:\Delta_1 \to [0,1]$ are defined by
$\rho_{\sigma}(\VEC{x}) = \sigma(\VEC{e}_0)$ and
$\rho_I(\VEC{x}) = \VEC{x} \cdot \VEC{e}_1$ for all $\VEC{x} \in \Delta_1$.
We extend $\tilde{D}_0$ to $S_1(X\times [0,1];R)$ by linearity.
Moreover, because of the linearity, it is enough to verify the three
properties of $P_0$ for arbitrary singular simplices in $X$.

Suppose that $\sigma$ is an arbitrary singular $0$-simplex in $X$.
Since $\tilde{D}_{-1} = 0$, we have
\[
(\partial_1 \circ \tilde{D}_0)(\sigma)
+ (\tilde{D}_{-1} \circ \partial_0)(\sigma)
= (\partial_1 \circ \tilde{D}_0)(\sigma)
= S_0(\iota_1)(\sigma) - S_0(\iota_0)(\sigma)
\]
because
\begin{align*}
\big(\partial_1(\tilde{D}_0(\sigma))\big)(\VEC{e}_0)
&= (\tilde{D}_0(\sigma) \circ F^0_1)(\VEC{e}_0) -
(\tilde{D}_0(\sigma) \circ F^1_1)(\VEC{e}_0)
= \tilde{D}_0(\sigma) (\VEC{e}_1) - \tilde{D}_0(\sigma)(\VEC{e}_0) \\
&= (\sigma(\VEC{e}_0),1) - (\sigma(\VEC{e}_0),0)
= S_0(\iota_1)(\sigma)(\VEC{x}_0) - S_0(\iota_0)(\sigma)(\VEC{x}_0) \ .
\end{align*}
So (1) of $P_0$ is satisfied.

Since
\begin{align*}
\tilde{D}_0(S_0(h)(\sigma))
&= \tilde{D}_0(h \circ \sigma) = (\rho_{h\circ \sigma}, \rho_I)
= (h\circ \rho_{\sigma} ,\rho_I)
= (h \times \Id_{[0,1]})(\rho_{\sigma}, \rho_I) \\
&= S_1(h \times \Id_{[0,1]})(\tilde{D}_0(\sigma)) \ ,
\end{align*}
we get that (2) of $P_0$ is satisfied.

Suppose that $\sigma$ is an arbitrary singular $1$-simplex in $X$.
To prove (3) of $P_0$, namely
$\partial_1 \circ \big( S_1(\iota_1) - S_1(\iota_0)
- \tilde{D}_0 \circ \partial_1 \big)\big) = 0$, we note that
\begin{align*}
&\big(\partial_1(\tilde{D}_0(\partial_1(\sigma)))\big)(\VEC{e}_0)
= \big( \big(\tilde{D}_0 (\partial_1(\sigma))\big) \circ F^0_1 -
\big(\tilde{D}_0 (\partial_1(\sigma))\big) \circ F^1_1\big)(\VEC{e}_0) \\
&\qquad = \big(\tilde{D}_0 (\partial_1(\sigma))\big) (\VEC{e}_1) -
\big(\tilde{D}_0 (\partial_1(\sigma))\big) (\VEC{e}_0)
= \big( \partial_1(\sigma)(\VEC{e}_0),1 \big)
-\big( \partial_1(\sigma)(\VEC{e}_0),0 \big) \\
&\qquad
= \big( (\sigma\circ F^0_1)(\VEC{e}_0),1 \big)
- \big( (\sigma\circ F^1_1)(\VEC{e}_0),1 \big)
- \big( (\sigma\circ F^0_1)(\VEC{e}_0),0 \big)
+ \big( (\sigma\circ F^1_1)(\VEC{e}_0),0 \big) \\
&\qquad = (\sigma(\VEC{e}_1),1) - (\sigma(\VEC{e}_0),1)
- (\sigma(\VEC{e}_1),0) + (\sigma(\VEC{e}_0),0) \\
&\qquad= S_1(\iota_1)(\sigma)(\VEC{e}_1)  - S_1(\iota_1)(\sigma)(\VEC{e}_0)
-S_1(\iota_0)(\sigma)(\VEC{e}_1) + S_1(\iota_0)(\sigma)(\VEC{e}_0) \\
&\qquad= \big(S_1(\iota_1)(\sigma) \circ F^0_1
- S_1(\iota_1)(\sigma) \circ F^1_1\big)(\VEC{e}_0)
- \big( S_1(\iota_0)(\sigma) \circ F^0_1
- S_1(\iota_0)(\sigma) \circ F^1_1\big)(\VEC{e}_0) \\
&\qquad = \partial_1 (S_1(\iota_1)(\sigma)) (\VEC{e}_0)
- \partial_1 (S_1(\iota_0)(\sigma)) (\VEC{e}_0)
\end{align*}

We assume that $P_{k-1}$ is true and prove that $P_k$ is true.

It follows from (3) of $P_{k-1}$ with $X$ replaced by $\Delta_k$ that
\[
\partial_k \circ \big( (S_k(\iota_1) - S_k(\iota_0)
- \tilde{D}_{k-1} \circ \partial_k )(\delta)\big) = 0
\]
where $\delta:\Delta_k \to \Delta_k$ is defined by $\delta(\VEC{x}) = \VEC{x}$
for all $\VEC{x} \in \Delta_k$.  Note that everything that we
have said so far is true for any topological space $X$.  This is the
reason why we may use $X = \Delta_k$.

Thus $\big( S_k(\iota_1) - S_k(\iota_0)
- \tilde{D}_{k-1} \circ \partial_k \big)(\delta)
\in Z_k(\Delta_k \times [0,1];R)$.  Since
$H_k(\Delta_k \times [0,1];R) = 0$ for $k>0$ according to
Proposition~\ref{propHkDI0},
there exists $\tilde{\delta} \in S_{k+1}(\Delta_k \times [0,1];R)$ such
that
\begin{equation} \label{singCHEq1}
\partial_{k+1} \tilde{\delta} = \big( S_k(\iota_1) - S_k(\iota_0)
- \tilde{D}_{k-1} \circ \partial_k \big)(\delta) \ .
\end{equation}
We set $\tilde{D}_{k}(\delta) = \tilde{\delta}$.

Given a singular $k$-simplex $\sigma$ in $X$, we set
$\tilde{D}_k(\sigma) = S_{k+1}(\sigma \times \Id_{[0,1]})(\tilde{\delta}) \in
S_{k+1}(X \times [0,1];R)$.  Recall that
$(\sigma \times \Id_{[0,1]}):\Delta_k \times [0,1] \to \Delta_k \times [0,1]$
is defined by $(\sigma \times \Id_{[0,1]})(\VEC{x},t)
= (\sigma(\VEC{x}),t)$ for all $(\VEC{x},t) \in \Delta_k \times [0,1]$.
We extend $\tilde{D}_k$ to $S_k(X;R)$ by linearity.

As for $P_0$, because of the linearity, it is enough to verify the
three properties of $P_k$ for arbitrary singular simplices in $X$.

Suppose that $\sigma$ is an arbitrary singular $k$-simplex in $X$.
(2) of $P_k$ follows from
\begin{align*}
\tilde{D}_k\big(S_k(h)(\sigma)\big) &= \tilde{D}_k( h \circ \sigma)
= S_{k+1}\big((h\circ \sigma) \times \Id_{[0,1]}\big)(\tilde{\delta})
= S_{k+1}(h\times \Id)\big(S_{k+1}(\sigma \times
\Id)(\tilde{\delta})\big) \\
&= S_{k+1}(h\times \Id_{[0,1]})( \tilde{D}_k(\sigma)) \ .
\end{align*}

To prove (1) of $P_k$, we get from (\ref{singCHEq1}) that
\begin{equation} \label{singCHEq2}
\partial_{k+1} (\tilde{D}_k(\delta)) + \tilde{D}_{k-1}(\partial_k(\delta))
= \partial_{k+1} (\tilde{\delta}) + \tilde{D}_{k-1}(\partial_k(\delta))
= S_k(\iota_1)(\delta) - S(\iota_0)(\delta)
\end{equation}
and note that $\sigma = S_k(\sigma)(\delta)$.  Hence
\begin{align*}
&\partial_{k+1}(\tilde{D}_k(\sigma)) + \tilde{D}_{k-1}(\partial_k(\sigma))
= \partial_{k+1}\big(S_{k+1}(\sigma \times
\Id_{[0,1]})(\tilde{D}_k(\delta))\big) +
\tilde{D}_{k-1}(\partial_k(S_k(\sigma)(\delta))) \\
&= S_k(\sigma \times \Id_{[0,1]})(\partial_{k+1}(\tilde{D}_k(\delta))) +
\tilde{D}_{k-1}(S_{k-1}(\sigma)(\partial_k(\delta))) \\
&= S_k(\sigma \times \Id_{[0,1]})(\partial_{k+1}(\tilde{D}_k(\delta))) +
S_k(\sigma \times \Id_{[0,1]})(\tilde{D}_{k-1}(\partial_k(\delta))) \\
&= S_k(\sigma \times \Id_{[0,1]})\big( \partial_{k+1}(\tilde{D}_k(\delta))
+ \tilde{D}_{k-1}(\partial_k(\delta))\big)
= S_k(\sigma \times \Id_{[0,1]})\big( S_k(\iota_1)(\delta) -
S(\iota_0)(\delta)\big) \\
&= S_k(\sigma \times \Id_{[0,1]})\big( (\delta,e_1) - (\delta,e_0) \big)
= (\sigma,e_1) - (\sigma,e_0)
= S_k(\iota_1)(\sigma) - S_k(\iota_0)(\sigma)
\end{align*}
where the second equality comes from (2) of
Proposition~\ref{propSkPropr}, the third equality comes from (2) for
$P_{k-1}$ and the fifth equality comes from (\ref{singCHEq2}).

Finally, (3) of $(P_k)$ follows from (1) because
\begin{align*}
\partial_{k+1}\circ \big( S_{k+1}(\iota_1) - S_{k+1}(\iota_0) \big)
&= \big( S_k(\iota_1) - S_k(\iota_0) \big) \circ \partial_{k+1}
= \big( \partial_{k+1} \circ \tilde{D}_k + \tilde{D}_{k+1} \circ
\partial_k \big) \circ \partial_{k+1}\\
&= \partial_{k+1} \circ \tilde{D}_k \circ \partial_{k+1}
\end{align*}
where again we have use (2) of Proposition~\ref{propSkPropr} to obtain
the first equality.  This completes the proof by induction.

\stage{ii} Since $f \sim g$, there exists a continuous function
$H:X \times [0,1] \to X$ such that $H(x,0) = f(x)$ and $H(x,1) = g(x)$
for all $x \in X$.  Thus $f = H\circ \iota_0$ and $g = H\circ \iota_1$.

It follows from (i) that
$\partial_{k+1} \circ \tilde{D}_k+ \tilde{D}_{k-1} \circ \partial_k
= S_k(i_1) - S_k(i_0)$
for all $k$.  Hence,
\begin{align*}
& \partial_{k+1} \circ (S_{k+1}(H) \circ \tilde{D}_k) +
(S_k(H) \circ \tilde{D}_{k-1}) \circ \partial_k
= S_k(H) \circ ( \partial_{k+1} \circ \tilde{D}_k) +
S_k(H)\circ (\tilde{D}_{k-1} \circ\partial_k) \\
&\qquad
= S_k(H) \big(  \partial_{k+1} \circ \tilde{D}_k
+ \tilde{D}_{k-1} \circ\partial_k\big)
= S_k(H)\circ (S_k(i_1) - S_k(i_0)) \\
&\qquad = S_k(H \circ i_1) - S_k(H\circ i_0)
= S_k(g) - S_k(f)
\end{align*}
where the first and fourth equality come from Proposition~\ref{propSkPropr},
and the third equality comes from (i).
This complete the proof of the theorem if we take $D_k = S_{k+1}(H)
\circ \tilde{D}_k$ for all $k$.
\end{proof}

The next corollary is a consequence of Theorem~\ref{thmExistCH} and
Proposition~\ref{propCHiFeG}.

\begin{cor} \label{corXCHom}
Suppose that $X$ and $Y$ are two topological spaces.  If $f,h:X \to Y$ are
homotopic maps, then $H_k(f) = H_k(g)$ for all $k$.
\end{cor}

\begin{cor} \label{corContrHk}
If $X$ is a contractible topological space and $R$ is an integral
domain, then $H_k(X;R) = 0$ for $k>0$ and $H_0(X;R) \cong R$.
\end{cor}

\begin{proof}
It follows from the definition of contractible spaces that there
exists $x_0 \in X$ such that the identity function $\Id_X:X \to X$ is
homotopic to the constant function $f:X\to X$ defined by $f(x) = x_0$
for all $x\in X$.  From the previous corollary, we have that
$H_k(\Id_X) = H_k(f)$.  It follows from Example~\ref{eggHk1} that
\[
H_k(X;R) = H_k(\Id_X)(H_k(X;R)) = H_k(f)(H_k(X;R))
\cong \begin{cases}
0 & \quad \text{if} \ k > 0 \\
R & \quad \text{if} \ k = 0
\end{cases}
\]
because $H_k(f)(H_k(X;R)) = H_k(\{x_0\};R)$.
\end{proof}

\begin{defn}
Let $X$ and $Y$ be two topological spaces.  We say that a continuous
map $f:X \to Y$ is a {\bfseries homotopic
equivalence}\index{Homotopic Equivalence} if there exists a continuous
map $h:Y \to X$ such that $h \circ f \sim \Id_X$ and
$f \circ h \sim \Id_Y$.  In that situation, we say that $X$ and $Y$ are
{\bfseries homotopic equivalent}\index{Homotopic Equivalent}.
\end{defn}

\begin{prop}
Suppose that $X$ and $Y$ are two homotopic equivalent topological
spaces.  Then $H_k(X;R) \cong H_k(Y;R)$.
\end{prop}

\begin{proof}
The previous proposition follows from Corollaries~\ref{corCompHgHf} and
\ref{corXCHom}.

Suppose that $f:X \to Y$ and $h:Y \to X$ are two continuous maps
such that $h \circ f \sim \Id_X$ and $f \circ h \sim \Id_Y$.  Then
$H_k(h) \circ H_k(f) = H_k(h \circ f) = H_k(\Id_X) = \Id_{H_k(X;R)}$
and
$H_k(f) \circ H_k(h) = H_k(f \circ h) = H_k(\Id_Y) = \Id_{H_k(Y;R)}$.
Thus $H_k(h)$ is the inverse of $H_k(f)$.  Hence
$H_k(f)$ is an isomorphism between $H_k(X;R)$ and $H_k(Y;R)$.
\end{proof}

\subsection{Reduced Homology}

Let $X$ be a topological space and $R$ be an integral domain.
This section presents an example of an augmentation for the
chain complex $\displaystyle \C = \{ (S_k(X;R), \partial_k) \}_{k \in \ZZ}$.
This example plays a crucial role later.

We consider the reduced chain complex
$\displaystyle \C^\sharp = \{ (S_k^\sharp(X;R),\partial_k^\sharp) \}_{k\in \ZZ}$
of the chain complex $\C$ with augmentation
\[
\partial_0^\sharp \left( \sum_{j\in J} a_j \sigma_j \right)  
= \sum_{j\in J} a_j \ ,
\]
where $J \subset \NN$ is a finite set, $a_j \in R$ and $\sigma_j$ is a
singular $0$-simplex for $j \in J$.  According to the definition of a
reduced chain complex, Definition~\ref{defnRedCC}, we have
\[
S_k^\sharp(X;R) = \begin{cases}
R & \quad \text{if} \ k = -1 \\
S_k(X;R) & \quad \text{if} \ k \neq -1
\end{cases}
\]
and $\displaystyle \partial_k^\sharp = \partial_k$ for $k \neq 0$.
We have already seen this augmentation in the proof of
Proposition~\ref{propH0XR} where $\displaystyle \partial_0^\sharp$ was
denoted $\epsilon$.  We also proved that
$\displaystyle \partial_0^\sharp \circ \partial_1^\sharp= 0$.
We can then proceed as we did before to define
$\displaystyle Z_k^\sharp(X;R)$, 
$\displaystyle B_k^\sharp(X;R)$ and $\displaystyle H_k^\sharp(X;R)$.

\begin{defn}
Let $X$ be a topological space $X$ and $R$ be an integral domain.
The $R$-module $\displaystyle H_k^\sharp(X;R)$ is called the
{\bfseries reduced (singular) $\mathbf{k^{th}}$ homology
module}\index{Homology Module!Reduced (Singular) $k^{th}$ Homology Module}
of $X$.
\end{defn}

In fact $\displaystyle H_k^\sharp(X;R) \cong H_k(X;R)$ for $k\neq 0$.
Only $\displaystyle H_0^\sharp(X;R)$ and $\displaystyle H_0(X;R)$ may
be different.  For instance, we saw that in Example~\ref{eggHk1} that
$H_0(X,R) = R$ if $X = \{x\}$.  However, we have that
$\displaystyle H_0^\sharp(X,R) = 0$ if $X = \{x\}$.  In general, we have the
following result.

\begin{prop} \label{propHkEquHskpR}
Let $X$ be a topological space $X$ and $R$ be an integral domain.
Then $\displaystyle H_0(X;R) \cong H_0^\sharp(X;R) \oplus R$.
\end{prop}

\begin{proof}
We first note that $Z_0(X;R) = S_0(X;R)$.
Moreover, $\displaystyle B_0(X;R) = B_0^\sharp(X;R)$ because
$\partial_1^\sharp = \partial_1$.
Since $\displaystyle \partial_0^\sharp \circ \partial_1 = 0$, we also
have that $\displaystyle B_0(X;R) \subset \KE(\partial_0^\sharp)
= Z_0^\sharp(X;R)$.

Given $b \in S_0(X;R)$ such that
$\displaystyle \partial_0^\sharp(b) = 1$, we prove that
\[
Z_0(X;R) = Z_0^\sharp(X;R) \oplus \{ r b : r \in R\} \ .
\]
Suppose that
$\displaystyle c \in Z_0^\sharp(X;R) \cap \{ r b : r \in R\}$.
Then $c = q b$ for some $q \in R$.  But
$\displaystyle q = q \partial_0^\sharp(b) = \partial_0^\sharp(q b)
= \partial_0^\sharp(c) = 0$.
Thus $c = 0$ and so
$\displaystyle Z_0^\sharp(X;R) \cap \{ r b : r \in R\} = \{0\}$.
Given $c \in S_0(S;R)$, suppose that $\displaystyle \partial_0^\sharp(c) = q$.
Then $c = (c - q b) + qb$ with
$\displaystyle c - q b \in \KE(\partial_0^\sharp) = Z_0^\sharp(X;R)$ and
$q b \in \{ r b : r \in R\}$.  Thus
$\displaystyle S_0(X;R) = Z_0(X;R) = Z_0^\sharp(X;R) + \{ r b : r \in R\}$.  

Hence
\begin{align*}
H_0(X;R) &= Z_0(X;R) / B_0(X;R)
= \big( Z_0^\sharp(X;R) \oplus \{ r b : r \in R\} \big)/  B_0^\sharp(X;R) \\
&\cong \big( Z_0^\sharp(X;R) / B_0^\sharp(X;R) \big) \oplus \{ r b : r \in R\}
\cong H_0^\sharp(X;R) \oplus R \ .
\end{align*}
Note that $\displaystyle b \not\in B_0^\sharp(X;R)$ because
$\displaystyle \partial_0^\sharp(b) \neq 0$.
\end{proof}

In general, the results about singular homology and reduced
singular homology differ only at the level of $S_0(X;R)$.

\subsection{Relation Between $\pi_1(X,x_0)$ and $H_1(X;R)$.}
\label{subsectP1EquH1}

We take a short pause in our study of singular homology to present the
relation between the concept of fundamental groups introduced in
Section~\ref{sectFundGr} and the singular
$\displaystyle 1^{st}$ homology module.

\begin{theorem}
Let $X$ be a topological space and $x_0 \in X$.  There exists a
natural group homomorphism $\Chi$ from $\pi_1(X,x_0)$ to $H_1(X;\ZZ)$.
Moreover, if $X$ is path connected, then $\Chi$ is onto and the kernel
of $\Chi$, denoted $\KE(\Chi)$, is the commutator subgroup of
$\pi_1(X,x_0)$.
\end{theorem}

\begin{proof}
The map $\Chi : \pi_1(X,x_0) \to H_1(X;\ZZ)$ is define by
$\Chi([\alpha]) = [\breve{\alpha}]$ where
$\breve{\alpha}$ is the singular $1$-simplex defined by
$\breve{\alpha}\big((1-t)\VEC{e}_0 + t\VEC{e}_1\big)
= \alpha(t)$ for $0\leq t \leq 1$.

\stage{i} We first show that $\Chi$ is well defined.
Let $\alpha_0$ and $\alpha_1$ be two loops at $x_0$
such that $\alpha_0 \dotsim \alpha_1$.  Therefore, according to the
definition of homotopic paths, Definition~\ref{defnHomPaths}, 
there exists a continuous function $H:[0,1]\times[0,1] \to X$ such
that $H(t,0) = \alpha_0(t)$ and $H(t,1) = \alpha_1(t)$
for $0 \leq t \leq 1$ and $H(0,s) = H(1,s) = x_0$ for $0 \leq s \leq 1$.

In the context of this proof, we will associate to each path
$\alpha:[0,1]\to X$ the singular $1$-simplex
$\breve{\alpha}:\Delta_1 \to X$ defined by
$\breve{\alpha}\big( (1-t)\VEC{e}_0 +t\VEC{e}_1\big) = \alpha(t)$
for $0 \leq t \leq 1$, and vice-versa.  In particular, singular
simplices and singular chains are identified with the symbol $\breve{}$\ . 

We define a singular $2$-simplex $\beta:\Delta_2 \to X$ in $X$ as
\[
\breve{\beta}( t(s\VEC{e}_2 + (1-s)\VEC{x}_1) + (1-t) \VEC{e}_0) = H(t,s)
\]
for $t,s \in [0,1]$.  The only issue with this definition may be at
$\VEC{e}_0$ because $\VEC{e}_0 = 0 (s\VEC{e}_2 + (1-s)\VEC{x}_1) + \VEC{e}_0$
for $0\leq s \leq 1$.  However, this is not an issue because
$H(0,s) = x_0$ for $0 \leq s \leq 1$.  We also have that $\breve{\beta}$
is continuous.

We have
\begin{equation} \label{thpi1equH1Eq1}
\partial_2 \breve{\beta} = \beta \circ F^0_2
- \breve{\beta} \circ F^1_2 + \breve{\beta} \circ F^2_2
\end{equation}
where
\begin{align*}
\breve{\beta}(F^2_2((1-t)\VEC{e}_0 + t\VEC{e}_1)
&= \breve{\beta}((1-t)\VEC{e}_0 + t\VEC{e}_1) = \breve{\beta}( t(0 \VEC{e}_2 +
\VEC{e}_1) + (1-t) \VEC{e}_0) = H(t,0) \\
& = \alpha_0(t) = \breve{\alpha}_0\big( (1-t)\VEC{e}_0 +t\VEC{e}_1\big)
\end{align*}
for $0\leq t \leq 1$,
\begin{align*}
\breve{\beta}(F^1_2((1-t)\VEC{e}_0 + t\VEC{e}_1)
&= \breve{\beta}((1-t)\VEC{e}_0 + t\VEC{e}_2) = \breve{\beta}( t( \VEC{e}_2 +
0\VEC{e}_1) + (1-t) \VEC{e}_0) = H(t,1) \\
& = \alpha_1(t) = \breve{\alpha}_1\big((1-t)\VEC{e}_0 + t\VEC{e}_1\big)
\end{align*}
for $0\leq t \leq 1$, and
\begin{align*}
\breve{\beta} \circ F^0_2((1-s) \VEC{e}_0 + s \VEC{e}_1) 
&= \breve{\beta}((1-s)\VEC{e}_1+s\,\VEC{e}_2)
= \breve{\beta}( (s \VEC{e}_2 + (1-s) \VEC{e}_1) + 0\,\VEC{e}_0) = H(0,s) \\
&= x_0 = \breve{e}_{x_0}\big((1-s)\VEC{e}_0 + s\VEC{e}_1\big)
\end{align*}
for $0\leq s \leq 1$.  In the previous expression,
$\breve{e}_{x_0}:\Delta_1 \to X$ is defined by
$\breve{e}_{x_0}(\VEC{x}) = x_0$ for $\VEC{x} \in \Delta_1$.
Since $\displaystyle \breve{\beta} \circ F^0_2 = \breve{e}_{x_0} =
\partial_2(\breve{\gamma})$ where the singular $2$-simplex
$\breve{\gamma}$ in $X$ is defined by $\breve{\gamma}(\VEC{x}) = x_0$
for all $\VEC{x} \in \Delta_2$,
we get from (\ref{thpi1equH1Eq1}) that
$\displaystyle \breve{\alpha}_0 - \breve{\alpha}_1 = \breve{\beta} \circ F^2_2 
- \breve{\beta} \circ F^1_2 \in B_1(X;\ZZ)$.  Thus
$\Chi([\alpha_0]) = [\breve{\alpha}_0] = [\breve{\alpha}_1]
= \Chi([\alpha_1])$ in $H_1(X;\ZZ)$.

\stage{ii} We show that $\Chi$ is an homomorphism.
Let $\alpha_0$ and $\alpha_1$ be two loops at $x_0$.
We need to show that $\Chi([\alpha_0]\,[\alpha_1])
= \Chi([\alpha_0]) + \Chi([\alpha_1])$.
Since $[\alpha_0]\,[\alpha_1] = [\alpha_0 \alpha_1]$, we need to show
that $[\breve{\alpha_0 \alpha_1}]
= [\breve{\alpha}_0] + [\breve{\alpha}_1]$ in $H_1(X;\ZZ)$.

Let $\VEC{y}_s = (1-s)\VEC{e}_1 + (s/2)\VEC{e}_2$ for $0 \leq s \leq 1$
and $\breve{\beta}$ be the singular $2$-simplex in $X$ defined by
\[
\breve{\beta}(\VEC{x})
= \begin{cases}
\displaystyle \breve{\alpha}_0\big( (1 - 2t(2-s)^{-1})\VEC{e}_0 +
2t(2-s)^{-1} \VEC{e}_1\big)
 & \displaystyle \ \ \text{if} \ \VEC{x} =
\big( 1 - 2t(2-s)^{-1}\big)\VEC{e}_0 \\
&\qquad + 2t(2-s)^{-1} \VEC{y}_s \ \text{with} \\
&\displaystyle \quad 
0 \leq s \leq 1 \ \text{and} \ 0 \leq t \leq (2-s)/2 \\[0.8em]
\displaystyle \breve{\alpha}_1\big( 2(1-t)(2-s)^{-1} \VEC{e}_0
& \displaystyle \ \ \text{if} \ \VEC{x} =
\big( 1 - 2(1-t)(2-s)^{-1}\big)\VEC{e}_2\big) \\
\quad + (1 - 2(1-t)(2-s)^{-1})\VEC{e}_1\big)
&\qquad + 2(1-t)(2-s)^{-1} \VEC{y}_s \ \text{with} \\
&\displaystyle \quad 0 \leq s \leq 1 \ \text{and} \ s/2 < t \leq 1
\end{cases}
\]
Note that $\breve{\beta}$ is well defined because
$\breve{\beta}(\VEC{x}) =
\breve{\alpha}_0(\VEC{e}_1) = \breve{\alpha}_1(\VEC{e}_0) = x_0$ for
$\displaystyle \VEC{x} = \VEC{y}_s$ with $0 \leq s \leq 1$.  The
following figure illustrates the singular $2$-simplex $\breve{\beta}$.
\pdfbox{alg_top/singhm1}

We have
\begin{equation} \label{thpi1equH1Eq2}
\partial \breve{\beta} = \breve{\beta} \circ F^0_2 - \breve{\beta} \circ F^1_2
+ \breve{\beta} \circ F^2_2
\end{equation}
where
\[
(\breve{\beta} \circ F^2_2)((1-t)\VEC{e}_0 + t\VEC{e}_1)
= \breve{\beta}((1-t)\VEC{e}_0 + t\VEC{e}_1)
= \breve{\beta}((1-t)\VEC{e}_0 + t\VEC{y}_0)
= \breve{\alpha}_0((1 - t)\VEC{e}_0 + t \VEC{e}_1)
\]
for $0 \leq t \leq 1$,
\[
(\breve{\beta} \circ F^0_2)((1-t)\VEC{e}_0 + t\VEC{e}_1)
= \breve{\beta}((1-t)\VEC{e}_1 + t\VEC{e}_2)
= \breve{\beta}((1-t)\VEC{y}_0 + t\VEC{e}_2)
= \breve{\alpha}_1((1 - t)\VEC{e}_0 + t \VEC{e}_1)
\]
for $0 \leq t \leq 1$, and
\begin{align*}
&(\breve{\beta} \circ F^1_2)((1-t)\VEC{e}_0 + t\VEC{e}_1)
= \breve{\beta}((1-t)\VEC{e}_0 + t\VEC{e}_2) \\
&\qquad = \begin{cases}
\breve{\beta}((1 - 2t)\VEC{e}_0 + 2t \VEC{y}_1) & \quad \text{if} \ 0
\leq t \leq 1/2 \\
\breve{\beta}(1-2(1-t))\VEC{e}_2 + 2(1-t) \VEC{y}_1) & \quad \text{if} \ 1/2
\leq t \leq 1 \\
\end{cases} \\
&\qquad = \begin{cases}
\breve{\alpha}_0((1 - 2t)\VEC{e}_0 + 2t \VEC{e}_1) & \quad \text{if} \ 0
\leq t \leq 1/2 \\
\breve{\alpha}_1(2(1-t)\VEC{e}_0 + (1-2(1-t)) \VEC{e}_1) & \quad \text{if} \ 1/2
\leq t \leq 1 \\
\end{cases} \\
&\qquad = \breve{\alpha_0 \alpha_1}((1-t)\VEC{e}_0 + t \VEC{e}_1))
\end{align*}
for $0 \leq t \leq 1$.  It then follows from (\ref{thpi1equH1Eq2})
that
$\partial \breve{\beta} = \breve{\alpha}_1 - \breve{\alpha_0\alpha_1}
+ \breve{\alpha}_0$.  Thus $\breve{\alpha}_1 - \breve{\alpha_0\alpha_1}
+ \breve{\alpha}_0 \in B_1(X;\ZZ)$.  Hence
$\Chi([\alpha_0\alpha_1]) = [\breve{\alpha_0 \alpha_1}]
= [\breve{\alpha}_0] + [\breve{\alpha}_1]
= \Chi([\alpha_0]) + \Chi([\alpha_1])$ in $H_1(X;\ZZ)$.

\stage{iii} We assume that $X$ is path-connected and show that $\Chi$
is onto.  Suppose that $\breve{c} \in Z_1(X;\ZZ)$.  We can write $\breve{c}$ as
$\displaystyle \breve{c} = \sum_{j=1}^J a_j \breve{\alpha}_j$ where
$a_j\in \ZZ$ and $\breve{\alpha}_j$ is singular $1$-simplex for
$1 \leq j \leq J$.  Hence
\begin{equation} \label{thpi1equH1Eq3}
\partial_1(\breve{c}) =
\sum_{j=1}^J a_j \big( \breve{\alpha}_j \circ F^0_1
- \breve{\alpha}_j\circ F^1_1 \big) = 0 \ .
\end{equation}
Since $X$ is path-connected, we can choose for each $\breve{\alpha}_j$ in
the summation above, a path $\eta_{j,1}$ in $X$ from $x_0$ to
$\displaystyle (\breve{\alpha}_j \circ F^0_1)(\VEC{e}_0) =
\breve{\alpha}_j(\VEC{e}_1)$ and a path $\eta_{j,0}$ in $X$ from $x_0$ to
$\displaystyle (\breve{\alpha}_j \circ F^1_1)(\VEC{e}_0) =
\breve{\alpha}_j(\VEC{e}_0)$ that satisfy the following conditions:
\begin{enumerate}
\item If $\displaystyle (\breve{\alpha}_{j_1} \circ F^0_1)(\VEC{e}_0) =
(\breve{\alpha}_{j_2} \circ F^0_1)(\VEC{e}_0)$ for some $j_1 \neq j_2$,
then $\eta_{j_1,1} = \eta_{j_2,1}$.
\item If $\displaystyle (\breve{\alpha}_{j_1} \circ F^0_1)(\VEC{e}_0) =
(\breve{\alpha}_{j_2} \circ F^1_1)(\VEC{e}_0)$ for some $j_1 \neq j_2$,
then $\eta_{j_1,1} = \eta_{j_2,0}$.
\item If $\displaystyle (\breve{\alpha}_{j_1} \circ F^1_1)(\VEC{e}_0) =
(\breve{\alpha}_{j_2} \circ F^1_1)(\VEC{e}_0)$ for some $j_1 \neq j_2$,
then $\eta_{j_1,0} = \eta_{j_2,0}$.
\item If $\displaystyle (\breve{\alpha}_j \circ F^1_1)(\VEC{e}_0) = x_0$,
then $\eta_{j_1,0} = e_{x_0}$ where $e_{x_0}(t) = x_0$ for
$0 \leq t \leq 1$.
\item If $\displaystyle (\breve{\alpha}_j \circ F^0_1)(\VEC{e}_0) = x_0$,
then $\eta_{j_1,1} = e_{x_0}$.
\end{enumerate}
\pdfbox{alg_top/singhm4}
It follows from (\ref{thpi1equH1Eq3}) that
\begin{equation} \label{thpi1equH1Eq4}
  \sum_{j=1}^J a_j( \breve{\eta}_{j,1} - \breve{\eta}_{j,0} ) = 0
\end{equation}
where $\breve{\eta}_{j,i}\big((1-t)\VEC{e}_0 + t\VEC{e}_1\big)
= \eta_{j,i}(t)$ for $0 \leq t \leq 1$
because
\[
\sum_{j=1}^J a_j( \breve{\eta}_{j,1}(\VEC{e}_1) - \breve{\eta}_{j,0}(\VEC{e}_1) )
= 
\sum_{j=1}^J a_j \left( (\breve{\alpha}_j \circ F^0_1)(\VEC{e}_0)  -
(\breve{\alpha}_j \circ F^1_1)(\VEC{e}_0) \right)) = 0
\]
and
$\breve{\eta}_{j_1,1} \neq \breve{\eta}_{j_2,1}$ if and only if
$\breve{\eta}_{j_1,1}(\VEC{e}_1) \neq \breve{\eta}_{j_2,1}(\VEC{e}_1)$,
$\breve{\eta}_{j_1,1} \neq \breve{\eta}_{j_2,0}$ if and only if
$\breve{\eta}_{j_1,1}(\VEC{e}_1) \neq \breve{\eta}_{j_2,0}(\VEC{e}_1)$, and
$\breve{\eta}_{j_1,0} \neq \breve{\eta}_{j_2,0}$ if and only if
$\breve{\eta}_{j_1,0}(\VEC{e}_1) \neq \breve{\eta}_{j_2,0}(\VEC{e}_1)$
by construction.

Consider the loops at $x_0$ given by
$\displaystyle \beta_j = \eta_{j,0} \alpha_j \eta_{j,1}^{-1}$ for
all $j$.  We have from (ii) that
\begin{equation} \label{thpi1equH1Eq5}
\begin{split}
\Chi([\beta_j]) &= \Chi([\eta_{j,0}]\,[\alpha_j]\,[\eta_{j,1}^{-1}])
= \Chi([\eta_{j,0}]) + \Chi([\alpha_j]) - \Chi([\eta_{j,1}]) \\
&= [\breve{\eta}_{j,0}] + [\breve{\alpha}_j] - [\breve{\eta}_{j,1}]
\end{split}
\end{equation}
for all $j$.  In the previous equality, we have use the fact that
$\displaystyle \eta_{j,i}\eta_{j,i}^{-1} \dotsim e_{x_0}$ for $0 \leq i \leq 1$
and all $j$ implies that
\[
\Chi([\eta_{j,i}]) + \Chi([\eta_{j,i}^{-1}])
= \Chi([\eta_{j,i}]\,[\eta_{j,i}^{-1}]) = \Chi([\eta_{j,i}\eta_{j,i}^{-1}])
= \Chi([e_{x_0}]) = [\breve{e}_{x_0}] = 0
\]
in $H_1(X;\ZZ)$ for $0 \leq i \leq 1$ and all $j$.  Recall that
$e_{x_0}:[0,1] \to X$ is the loop at $x_0$ defined by
$e_{x_0}(t) = x_0$ for $0 \leq t \leq 1$.  Thus
$\displaystyle \Chi([\eta_{j,i}^{-1}]) = -\Chi([\eta_{j_i}])$
for $0 \leq i \leq 1$ and all $j$.

Consider the loop at $x_0$ given by
$\displaystyle \gamma = \prod_{j=1}^J \beta_j^{a_j}$.  It follows from
(\ref{thpi1equH1Eq4}) and (\ref{thpi1equH1Eq5}) that
\begin{align*}
\Chi([\gamma]) &= \Chi\Big( \prod_{j=1}^J [\beta_j^{a_j}]\Big)
= \sum_{j=1}^J \Chi([\beta_j^{a_j}]) = \sum_{j=1}^J \Chi([\beta_j]^{a_j})
= \sum_{j=1}^J a_j \Chi([\beta_j]) \\
&= \sum_{j=1}^J a_j \left( [\breve{\eta}_{j,0}] + [\breve{\alpha}_j] -
[\breve{\eta}_{j,1}] \right)
= \sum_{j=1}^J a_j [\breve{\alpha}_j] + \sum_{j=1}^J a_j
\left( [\breve{\eta}_{j,0}] - [\breve{\eta}_{j,1}] \right)
= \sum_{j=1}^J a_j [\breve{\alpha}_j] = [\breve{c}] \ .
\end{align*}

\stage{iv}  All elements in the commutator subgroup $G$ of $\pi_1(X.x_0)$
are of the form $\displaystyle \alpha \beta \alpha^{-1}\beta^{-1}$
where $\alpha$ and $\beta$ are two loops in $X$ at $x_0$.  For
these elements, we have
\[
\Chi\big([\alpha \beta \alpha^{-1}\beta^{-1}]\big)
= \Chi\big([\alpha]\, [\beta]\,[\alpha^{-1}]\,[\beta^{-1}]\big)
= \Chi([\alpha]) + \Chi([\beta]) - \Chi([\alpha]) - \Chi([\beta]) = 0 \ .
\]
Thus $G \subset \KE(\Chi)$.

To prove that $G = \KE(\Chi)$, we need to introduce a little tool.
Given a loop $\gamma$ in $X$ at $x_0$, suppose that
$\displaystyle \gamma = \prod_{j=1}^J \alpha_j^{\epsilon_j}$ where
the $\alpha_j$ are paths in $X$ (they may not be loops) and the
$\epsilon_j$ are either $1$ or $-1$.  We set
$\displaystyle \exp_{\gamma}(\alpha_i) = \sum_{\alpha_j=\alpha_i} \epsilon_j$
for each path $\alpha_i$ in $\gamma$.

\noindent{\bfseries Claim}: If $\exp_\gamma(\alpha_i) = 0$ for all
$\alpha_i$ in $\gamma$, then $[\gamma] \in G$.

To prove this claim, we choose the paths $\eta_{j,i}$ for
$0\leq i \leq 1$ and $1 \leq j \leq J$ as we have done in (iii).
\pdfbox{alg_top/singhm5}
Hence $\displaystyle \gamma \dotsim
\prod_{j=1}^J \big(\eta_{j,0} \alpha_j\eta_{j,1}^{-1}\big)^{\epsilon_j}$ because
$\displaystyle \eta_{j,1}^{-1}\eta_{j+1,0} \dotsim e_{x_0}$ for
$1 \leq j < J$,
$\eta_{1,0} = e_{x_0}$ and $\eta_{J,1} = e_{x_0}$.
If we set $\displaystyle \beta_j = \eta_{j,0} \alpha_j\eta_{j,1}^{-1}$ for all
$j$, then we have that
$\displaystyle \gamma \dotsim \prod_{j=1}^J \beta_j^{\epsilon_j}$ where the
$\beta_j$ are loops at $x_0$.

We consider the quotient group
$\pi_1(X,x_0)/G$.  Let $[\gamma]_G$ and $[\beta_j]_G$
denote the equivalence classes of $[\gamma]$ and $[\beta_j]$ 
in $\pi_1(X,x_0)/G$.  We note that $\pi_1(X,x_0)/G$ is a commutative
group because
$\displaystyle \big(\sigma_1 \sigma_2\big) \big(\sigma_2 \sigma_1\big)^{-1}
= \big(\sigma_1\sigma_2\big) \big(\sigma_1^{-1} \sigma_2^{-1}\big) \in G$
for any loops $\sigma_1,\sigma_2$ in $X$ at $x_0$.
Thus
$[\sigma_1]_G\,[\sigma_2]_G
= [\sigma_1\sigma_2]_G = [\sigma_2\sigma_1]_G = 
[\sigma_2]_G\,[\sigma_1]_G$.

Hence, we may reorder the factors in the product for $[\gamma]_G$ to get
\[
[\gamma]_G = \prod_{i=1}^I [\beta_{j_i}]_G^{exp_{\gamma}(\alpha_i)}
= \prod_{i=1}^I [\beta_{j_i}]_G^0 = [e_{x_0}]_G 
\]
in $\pi_1(X,x_0)/G$ where each $[\beta_{j_i}]_G$ in the product above
represent a distinct element of $\pi_1(X,x_0)/G$.  Namely, each
$\beta_{j_i}$ in the sum is a representative of all the $\beta_j$
which are homotopic to $\beta_{j_i}$, or equivalently, $\alpha_{j_i}$
is a representative of all the $\alpha_j$ which are homotopic to
$\alpha_{j_i}$.  We therefore have that $[\gamma] \in G$.

Now that this little claim is proved, we go back to the proof that
$\KE(\Chi) = G$.

If $\Chi([\gamma]) = 0$, then $\breve{\gamma} \in B_1(X;\RR)$.  Thus
\begin{equation} \label{thpi1equH1Eq6}
\begin{split}
\breve{\gamma} &= \partial_2\left(
\sum_{j=1}^J b_j \breve{\sigma}_j \right)
= \sum_{j=1}^J b_j \bigg(
\underbrace{\breve{\alpha}_j\circ F^0_2}_{=\breve{\alpha}_j^{(0)}}
- \underbrace{\breve{\alpha}_j \circ F^1_2}_{=\breve{\alpha}_j^{(1)}} 
+ \underbrace{\breve{\alpha}_j\circ F^2_2}_{=\breve{\alpha}_j^{(2)}} \bigg) \\
&= \sum_{j=1}^J b_j \left( \breve{\alpha}_j^{(0)}
- \breve{\alpha}_j^{(1)} + \breve{\alpha}_j^{(2)} \right)
\end{split}
\end{equation}
for some singular $2$-simplices $\breve{\alpha}_j$ and $b_j \in \ZZ$.
It follows that the sum of the coefficients of $\breve{\gamma}$ in the
last sum in (\ref{thpi1equH1Eq6}) is $1$ and the sum of the coefficients
for the other singular $1$-simplices in the last sum in
(\ref{thpi1equH1Eq6}) is $0$.

We choose the paths $\eta_{j,k,i}$ for
$0\leq i \leq 1$, $0 \leq k \leq 2$ and $1\leq j \leq J$ as we have
done in (iii); namely,
$\eta_{j,k,1}$ is a path in $X$ from $x_0$ to
$\displaystyle (\breve{\alpha}_j^{(k)} \circ F^0_1)(\VEC{e}_0)$ and
$\eta_{j,k,0}$ is a path in $X$ from $x_0$ to
$\displaystyle (\breve{\alpha}_j^{(k)} \circ F^1_1)(\VEC{e}_0)$
for $0 \leq k \leq 2$ and $1\leq j \leq J$.  Moreover, paths
$\eta_{j_1,k_1,i_1}$ and $\eta_{j_2,k_2,i_2}$ that end at the same
point are the same path, and paths $\eta_{j,k,i}$ that end at $x_0$
are in fact equal to the constant path $e_{x_0}$.
Note that
\begin{align*}
\eta_{j,0,1}(1) &= \big(\breve{\alpha}_j^{(0)}\circ F^0_1\big)(\VEC{e}_0)
= \breve{\alpha}_j^{(0)}(\VEC{e}_1)
= \big(\breve{\alpha}_j\circ F^0_2\big)(\VEC{e}_1)
= \breve{\alpha}_j(\VEC{e}_2) \ , \\
\eta_{j,0,0}(1) &= \big(\breve{\alpha}_j^{(0)}\circ F^1_1\big)(\VEC{e}_0)
= \breve{\alpha}_j^{(0)}(\VEC{e}_0)
= \big(\breve{\alpha}_j\circ F^0_2\big)(\VEC{e}_0)
= \breve{\alpha}_j(\VEC{e}_1) \ , \\
\eta_{j,1,1}(1) &= \big(\breve{\alpha}_j^{(1)}\circ F^0_1\big)(\VEC{e}_0)
= \breve{\alpha}_j^{(1)}(\VEC{e}_1)
= \big(\breve{\alpha}_j\circ F^1_2\big)(\VEC{e}_1)
= \breve{\alpha}_j(\VEC{e}_2) \ , \\
\eta_{j,1,0}(1) &= \big(\breve{\alpha}_j^{(1)}\circ F^1_1\big)(\VEC{e}_0)
= \breve{\alpha}_j^{(1)}(\VEC{e}_0)
= \big(\breve{\alpha}_j\circ F^1_2\big)(\VEC{e}_0)
= \breve{\alpha}_j(\VEC{e}_0) \ , \\
\eta_{j,2,1}(1) &= \big(\breve{\alpha}_j^{(2)}\circ F^0_1\big)(\VEC{e}_0)
= \breve{\alpha}_j^{(2)}(\VEC{e}_1)
= \big(\breve{\alpha}_j\circ F^2_2\big)(\VEC{e}_1)
= \breve{\alpha}_j(\VEC{e}_1) \ \text{and} \\
\eta_{j,2,0}(1) &= \big(\breve{\alpha}_j^{(2)}\circ F^1_1\big)(\VEC{e}_0)
= \breve{\alpha}_j^{(2)}(\VEC{e}_0)
= \big(\breve{\alpha}_j\circ F^2_2\big)(\VEC{e}_0)
= \breve{\alpha}_j(\VEC{e}_0) \ .
\end{align*}
All this information is summarized in the following figure.
\pdfbox{alg_top/singhm2}

Let
$\displaystyle \beta_{j,k} = \eta_{j,k,0}\,\alpha_j^{(k)} \eta_{j,k,1}^{-1}$
for $0 \leq k \leq 2$ and $1 \leq j \leq J$, and
$\displaystyle \beta_j = \beta_{j,0}\beta_{j,1}^{-1}\beta_{j,2}$ for
$1 \leq j \leq J$.  We have
\begin{align}
\beta_j &= \beta_{j,0}\beta_{j,1}^{-1}\beta_{j,2}
= \big(\eta_{j,0,0} \alpha_j^{(0)} \eta_{j,0,1}^{-1}\big)
\big(\eta_{j,1,0} \alpha_j^{(1)} \eta_{j,1,1}^{-1}\big)^{-1}
\big(\eta_{j,2,0} \alpha_j^{(2)} \eta_{j,2,1}^{-1}\big) \nonumber \\
&= \eta_{j,0,0} \alpha_j^{(0)} \big(\eta_{j,0,1}^{-1} \eta_{j,1,1}\big)
\big(\alpha_j^{(1)}\big)^{-1} \big(\eta_{j,1,0}^{-1} \eta_{j,2,0} \big)
\alpha_j^{(2)} \eta_{j,2,1}^{-1} \nonumber \\
&\dotsim \eta_{j,0,0} \alpha_j^{(0)} \big(\alpha_j^{(1)}\big)^{-1}
\alpha_j^{(2)} \eta_{j,2,1}^{-1}
\dotsim e_{x_0}  \label{thpi1equH1Eq7}
\end{align}
because $\displaystyle \eta_{j,0,1}^{-1} \eta_{j,1,1} \dotsim e_{x_0}$ and
$\displaystyle \eta_{j,1,0}^{-1} \eta_{j,2,0} \dotsim e_{x_0}$.
We also use the fact that $\hat{\alpha}_j(\Delta_2)$ is contractible
to $\breve{\alpha}_j(\VEC{e}_1)$, because $\Delta_2$ is contractible to
$\VEC{e}_1$, to obtain the second homotopy in (\ref{thpi1equH1Eq7}).

Hence $\displaystyle \xi = \prod_{j=1}^J \beta_j^{b_j} \dotsim e_{x_0}$.
It follows that $\displaystyle \gamma\,\xi^{-1} \dotsim \gamma$ and therefore
$\displaystyle [\gamma]\,[\xi^{-1}] = [\gamma]$ in $\pi_1(X,x_0)$.
However, if $\alpha$ is one of the factors of $\gamma$, then we have that
$\exp_{\xi}(\alpha) = 1$ is $\alpha = \gamma$ and
$\exp_{\xi}(\alpha) = 0$ otherwise.  This follows from the fact
to the sum of the coefficients of $\breve{\alpha}$ in the last sum in
(\ref{thpi1equH1Eq6}) is equal to $\exp_{\xi}(\alpha)$ because
$\Chi:\pi_1(X,x_0) \to H_1(X;\ZZ)$ is a group homomorphism, and
the paragraph following (\ref{thpi1equH1Eq6}).
Moreover $\exp_{\xi}(\alpha) = 0$ if $\alpha$ is one of the
$\eta_{j,k,i}$ because $\displaystyle \eta_{j,k,i}^{-1}$ is present in $\xi$ if
$\eta_{j,k,i}$ is present in $\xi$.
Therefore $\exp_{\gamma\xi^{-1}}(\sigma) = 0$
for all factors $\alpha$ in $\displaystyle \gamma \xi^{-1}$.  It
follows from the claim that
$\displaystyle [\gamma] = [\gamma \xi^{-1}]$ is in the commutator
subgroup $G$ of $\pi_1(X,x_0)$.
\end{proof}

\begin{cor}
Suppose that $X$ is a path-connected topological space and $x_0 \in X$,
then the map $\Chi$ defined in the previous theorem is a group
isomorphism between $\pi_1(X)$ and $H_1(X;\ZZ)$ if and only if
$\pi_1(X)$ is an abelian group.
\end{cor}

This corollary follows from the fact that a group is commutative if
its commutator subgroup is trivial; namely, it is composed of the unit
or identity element of the group only.

It follows from the previous corollary and Proposition~\ref{propFGSone}
that $\displaystyle H_1(S^1;\ZZ) \cong \pi_1(S^1) \cong \ZZ$.
Moreover, we get from Example~\ref{eggFGtorus} that
$\displaystyle H_1(\torus{2};\ZZ) \cong \pi_1(\torus{2}) \cong \ZZ^2$
where, as usual, $\torus{2}$ denotes the torus in $\displaystyle \RR^3$.

\subsection{Relative Homology} \label{subsectRelHom}

Though our goal is not to provide a full introduction to algebraic
topology, we feel that we cannot at least mention some of the
important concepts and results of this subject.  The readers should
consider this subsection as an invitation to pursue their study of
algebraic topology.

Let $X$ be a topological space and $R$ be an integral domain.  Moreover,
suppose that $A$ is a subset of $X$.  We may therefore consider
the $R$-module $S_k(A;R)$.  It is the submodule of $S_k(X;R)$
consisting of all the singular $k$-chains that are linear combinations of
singular $k$-simplices with images in $A$.

Since the boundary operator $\partial_k:S_k(X;R) \to S_{k-1}(X;R)$ maps
$S_k(A;R)$ into $S_{k-1}(A;R)$, we can define the operator
$\overline{\partial}_k:S_k(X;R)/S_k(A;R) \to S_{k-1}(X;R)/S_{k-1}(A;R)$
as\\
$\overline{\partial}_k(\relC[X,A]{c}) = \relC[X,A]{\partial_k c}$ for all
$\relC[X,A]{c} \in S_k(X;R)/S_k(A;R)$, where $\relC[X,A]{c}$ denotes the
equivalence class in $S_k(X;R)/S_k(A;R)$ associated to $c \in S_k(X;R)$.
We get the following commutative diagram.
\[
\xymatrix{
S_k(X;R) \ar[d]^{\partial_k} \ar[r]^-{\pi_k} & S_k(X;R)/S_k(A;R)
\ar[d]^{\overline{\partial}_k} \\
S_{k-1}(X;R) \ar[r]^-{\pi_{k-1}} & S_{k-1}(X;R)/S_{k-1}(A;R)
}
\]
where $\pi_q:S_q(X;R) \to S_q(X;R)/S_q(A;R)$ is the projection defined
by $\pi_q(c) = \relC[X,A]{c}$ for $c \in S_q(X;R)$.

It follows from $\partial_{k-1}\partial_k = 0$ that
$\overline{\partial}_{k-1}\overline{\partial}_k = 0$.

\begin{defn}
Let $A$ be a subset of a topological space $X$ and $R$ be an integral
domain.  The {\bfseries relative (singular)
$\displaystyle \mathbf{k^{th}}$ homology module}\index{Homology
Module!Relative (Singular) $k^{th}$ Homology Module} of
$X \mod A$ is the $R$-module defined as
$H_k(X,A;R) = \KE\big(\overline{\partial}_k\big) /
\IMG\big(\overline{\partial}_{k+1}\big)$.  The equivalence class in
$H_k(X,A;R)$ associated to $c \in S_k(X;R)$ is denoted
$[c]_{X,A}$.
\end{defn}

It is possible to describe more explicitly the elements of
$H_k(X,A;R)$.  For that, we need a couple of definitions.

\begin{defn}
Let $A$ be a subset of a topological space $X$ and $R$ be an integral
domain.  Given $c_1,c_2 \in S_k(X;R)$, we say that $c_1$ and $c_2$
are {\bfseries homologous in $\bm{X \mod A}$}\index{Homologous} if
there exists $\tilde{c} \in S_{k+1}(X;R)$ such that
$c_1- c_2 - \partial_{k+1}(\tilde{c}) \in S_k(A;R)$.  We then write
that $c_1 \hsim c_2$ in $X \mod A$.
\end{defn}

\begin{defn}
Elements of
$Z_k(X,A;R) = \{ c \in S_k(X;R) : \partial_k(c) \in S_{k-1}(A;R) \}$
are called {\bfseries relative $\mathbf{k}$-cycle}\index{Relative $k$-Cycle}
of $X \mod A$ and elements of
$B_k(X,A;R) = \{ c \in S_k(X;R) : c \hsim 0 \text{ in } X \mod A\}$
are called
{\bfseries relative $\mathbf{k}$-boundary}\index{Relative $k$-Boundary}
of $X \mod A$.
\end{defn}

We have that $Z_k(X,A;R) \supset Z_k(X;R)$ and
$B_k(X,A;R) \supset B_k(X;R) \cup S_k(A;R)$.
The reader may be wondering why the boundary operator is not used in
the definition of\\
$B_k(X,A;R)$.  In fact, it is used when referring to
homologous equivalence in $X \mod A$ as it is illustrated in the proof of the
following proposition.

\begin{prop}
Let $A$ be a subset of a topological space $X$ and $R$ be an integral
domain.  Then $H_k(X,A;R) \cong Z_k(X,A;R) / B_k(X,A;R)$.
\end{prop}

\begin{proof}
We have
\begin{align*}
\KE\big(\overline{\partial}_k\big)
&= \big\{ \relC[X,A]{c} : \overline{\partial}_k(\relC[X,A]{c}) 
= \relC[X,A]{ \partial_k c} = \relC[X,A]{0} \big\} \\
&= \big\{ \relC[X,A]{c} : c \in S_k(X,R) \ \text{and}
\ \partial_k(c) \in S_k(A;R) \big\} = Z_k(X,A;R) / S_k(A;R)
\end{align*}
and
\begin{align*}
\IMG\big(\overline{\partial}_{k+1}\big)
&= \big\{ \relC[X,A]{c} : \relC[X,A]{c}
= \overline{\partial}_{k+1}(\relC[X,A]{b})
\ \text{for some} \ \relC[X,A]{b} \in S_{k+1}(X;R)/S_{k+1}(A;R) \big\} \\
&= \big\{ \relC[X,A]{c} : \relC[X,A]{c - \partial_{k+1}(b)} = \relC[X,A]{0}
\ \text{for some}\ \relC[X,A]{b} \in S_{k+1}(X;R)/S_{k+1}(A;R) \big\} \\
&= \big\{ \relC[X,A]{c} : c \in S_k(X;R) \ \text{and}
\ c - \partial_{k+1}(b) \in S_k(A;R) \ \text{for some}
\ b \in S_{k+1}(X;R) \big\} \\
&= \big\{ \relC[X,A]{c} : c \in S_k(X;R)\ \text{and}
\ c \hsim 0 \text{ in } X \mod A \big\} = B_k(X,A;R)/S_k(A;R) \ .
\end{align*}
Hence
\begin{align*}
\KE\big(\overline{\partial}_k\big) / \IMG\big(\overline{\partial}_k\big) &=
\big( Z_k(X,A;R) / S_k(A;R)\big) / \big(B_k(X,A;R) / S_k(A;R) \big) \\
&\cong Z_k(X,A;R) / B_k(X,A;R) \ . \qedhere
\end{align*}
\end{proof}

\begin{defn}
Let $A$ be a subset of a topological space $X$ and $B$ be a subset of
a topological space $Y$.  If $f:X \to Y$ is a function such that
$f(A) \subset B$, then we write $f:(X,A) \to (Y,B)$.
\end{defn}

Suppose that $f:(X,A) \to (Y,B)$ is a continuous function and $R$ is
an integral domain.  Then $S_k(f):S_k(X;R) \to S_k(X;R)$ and
$S_k(f):S_k(A;R) \to S_k(B;R)$.  This implies that
$\displaystyle \tilde{S}_k(f) : S_k(X;R)/S_k(A;R) \to S_k(X;R)/S_k(B;R)$
defined by $\tilde{S}_k(f)(\relC[X,A]{c}) = \relC[Y,B]{c}$ for
$c \in S_k(X;R)$ is well defined.  The following proposition can then
be easily proved by the reader.

\begin{prop} \label{propHkPropS}
Suppose that $A$ is a subset of a topological space $X$, $B$ is a subset of a
topological space $Y$ and $R$ is an integral domain.  If $f:(X,A)\to (Y,B)$
is continuous function, then $H_k(f): H_k(X,A;R) \to H_k(Y,B;R)$ defined by
$H_k(f)([c]_{X,A}) = [f\circ c]_{Y,B}$ for all
$[c]_{X,A} \in H_k(X,A;R)$ is a homomorphism that satisfies the
following statements.
\begin{enumerate}
\item $H_{k-1}(f) \circ \overline{\partial}_k
= \overline{\partial}_k \circ H_k(f)$.
\item If $C$ is a subset of a topological space $Z$ and
$g:(Y,B)\to (Z,C)$ is a continuous function, then
$H_k(g) \circ H_k(f) = H_k(g\circ f)$.
\end{enumerate}
\end{prop}

Suppose that $A$ and $B$ are two subsets of a topological space $X$
such that $A \subset B$, and $R$ is an integral domain.  The function
$\Id_X:X \to X$ is a continuous function such that
$\Id_X:(X,\emptyset) \to (X,A)$.  Hence
$H_k(\Id_X) : H_k(X,A;R) \to H_k(X,B;R)$ is well
defined by $H_k(\Id_X)([c]_{X,A}) = [\Id_X \circ c]_{X,B} = [c]_{X,B}$ for all
$[c]_{X,A} \in H_k(X.A;R)$.
If $A = \emptyset$, then we get that
$H_k(\Id_X) :  H_k(X;R) = H_k(X,\emptyset;R) \to H_k(X,B;R)$.
Another function that will play a significant role
in this section is the inclusion map $\iota:A \to X$ for $A \subset X$.
We have the following result.

\begin{prop} \label{propAXXAexact}
Suppose that $A$ is a subset of a topological space $X$ and $R$ is an
integral domain.  Then the sequence.
\[
\xymatrix@C+2em{
H_k(A;R) \ar[r]^{H_k(\iota)} & H_k(X;R) \ar[r]^{H_k(\Id_X)} & H_k(X,A;R)
}
\]
is exact.
\end{prop}

\begin{proof}
Since $\Id_X \circ\, \iota:A \to X$ is a continuous function such
that $(\Id_X \circ\, \iota)(\emptyset) = \emptyset$, we have
that $H_k(\Id_X \circ\, \iota):H_k(A;R) = H_k(A,\emptyset;R) \to H_k(X,A;R)$
is well defined.  Moreover,
$H_k(\Id_X \circ\, \iota)([c]_A) = [0]_{X,A} \in H_k(X,A;R)$
for all $[c]_A \in H_k(A;R)$ because
\[
c \in S_k(A;R) \Rightarrow c \hsim 0 \ \text{in} \ X \mod A
\Rightarrow c \in B_k(X,A;R) \Rightarrow
\relC[X,A]{c} \in \IMG\big(\overline{\partial}_{k+1}\big) \ .
\]
Thus $H_k(\Id_X) \circ H_k(\iota) = H_k(\Id_X \circ\, \iota) = 0$.
Hence $\IMG( H_k(\iota)) \subset \KE( H_k(\Id_X))$.

To prove the reverse inclusion, consider $[c]_X \in \KE(H_k(\Id_X))$.
We get from $H_k(\Id_X)([c]_X) = [c]_{X,A} = [0]_{X,A} \in H(X,A;R)$
that $\relC[X,A]{c} \in \IMG\big(\overline{\partial}_{k+1}\big)$.  Thus
$\relC[X,A]{c} = \overline{\partial_{k+1}} (\relC[X,A]{b})
= \relC[X,A]{\partial_{k+1} b}$ for some $b \in S_{k+1}(X;R)$.
Hence $c - \partial_{k+1}(b) \in S_k(A;R)$ for some
$b \in S_{k+1}(X;R)$.  Since $\partial_k(c) = 0$
because $[c]_X \in H_k(X;R)$, we have that
$c - \partial_{k+1}(b)$ is a $k$-cycle in $S_k(A;R)$.
Thus $[c - \partial_{k+1}(b)]_A \in H_k(A;R)$ and
$H_k(\iota)([c - \partial_{k+1}(b)]_A)
= [c - \partial_{k+1}(b)]_X = [c]_X \in H_k(X;R)$ proving that
$[c]_X \in \IMG( H_k(\iota))$.
\end{proof}

Many of the results about singular homology modules have matching
results for regular homology modules.  For instance,
Proposition~\ref{propHkeHkJ} becomes the following proposition.

\begin{prop} \label{propXAopXjAj}
Suppose that $A$ is a subset of a topological space $X$.  If the
topological space $X$ is the union of distinct path-connected 
components $X_j$ for $j \in J$ and $A_j = A \cap X_j$ for all $j \in J$, then
$\displaystyle H_k(X,A;R) = \bigoplus_{j\in J} H_k(X_j,A_j;R)$.
\end{prop}

However, there are some differences.  For instance,
Proposition~\ref{propH0XR} is replaced by the following proposition.

\begin{prop} \label{propH0XA0}
Suppose that $A$ is a non-empty subset of a path-connected topological
space $X$.  Then $H_0(X,A;R) = 0$.
\end{prop}

\begin{proof}
Choose $x_0 \in A$ and, for each $x \in X$, choose a path $\tilde{\sigma}_x$
from $x_0$ to $x$.  For $x \in X$, let $\sigma_x$ be the singular
$1$-simplex defined by $\sigma_x((1-t)\VEC{e}_0+ t\VEC{e}_1)
= \tilde{\sigma}_x(t)$ for $0 \leq t \leq 1$.   As usual, we define
the singular $0$-simplex $e_x$ by $e_x(\VEC{e}_0) = x$.

Suppose that $c$ is a relative $0$-cycle in $X \mod A$.  We have that
$\displaystyle c =  \sum_{j\in J} a_je_{x_j}$ for some $a_j \in R$ and
$x_j \in X$, where $J$ is a finite subset of $\displaystyle \NNp$.
Since
\[
\partial_1\Big( \sum_{j\in J} a_j\sigma_{x_j} \Big)
= \sum_{j\in J} a_j \big( \sigma_{x_j}\circ F^0_1 - 
\sigma_{x_j}\circ F^1_1 \big)
= \sum_{j\in J} a_j \big( e_{x_j} - e_{x_0} \big)
= c - \sum_{j\in J} a_j e_{x_0} \ ,
\]
we have that
$\displaystyle c \hsim \sum_{j\in J} a_j e_{x_0} \in S_0(A;R)$ in $X \mod A$.
Thus $c$ is a relative $k$-boundary in $X \mod A$.
Therefore $[c]_{X,A} = [0]_{X,A} \in H_k(X,A;R)$.
\end{proof}

We extend the definition of homotopic maps between two topological
spaces to the context of relative homology modules.

\begin{defn} \label{defnRelHF}
Let $A$ be a subset of a topological space $X$, $B$ be a subset of a
topological space $Y$ and $R$ be an integral domain.  Suppose that
$f:(X,A)\to (Y,B)$ and $g:(X,A)\to (Y,B)$ are two continuous
functions.  We say that $f$ is
{\bfseries homotopic}\index{Homotopic Functions} to $g$
if there exists a continuous mapping $H:X\times[0,1] \to Y$ such that
$H(x,0) = f(x)$ and $H(x,1) = g(x)$ for all $x \in X$, and
$H(A,t) \subset B$ for $0\leq t \leq 1$.  We write $f\rsim g$.
\end{defn}

The next proposition could be proved using the same technique used to
prove Corollary~\ref{corXCHom}.  This obviously requires the use of
chain complexes for relative homology modules.

\begin{prop} \label{propfgHHkfgE}
Let $A$ be a subset of a topological space $X$, $B$ be a subset of a
topological space $Y$ and $R$ be an integral domain.  Suppose that
$f:(X,A)\to (Y,B)$ is homotopic to $g:(X,A)\to (Y,B)$.  Then
$H_k(f) = H_k(g)$.  
\end{prop}

There are many interesting results about relative homology modules.
We are going to present some of these results.  The reader should
consult \cite{GH,MUat} for more information.

The map $C_k: H_k(X,A;R) \to H_{k-1}(A;R)$ defined by
$C_k([c]_{X,A}) = [\partial_k(c)]_A$ is well defined because
\[
[c]_{X,A} \in H_k(X,A;R)
\Rightarrow \relC[X,A]{c} \in \KE\big(\overline{\partial}_k\big)
\Rightarrow \partial_k(c) \in S_{k-1}(A;R) \ .
\]
Thus $\partial_k(c) \in Z_{k-1}(A;R)$ because
$\partial_k(c) \in S_{k-1}(A;R)$ and $\partial_{k-1}(\partial_k(c)) = 0$.
Hence, $[\partial_k(c)]_A \in H_{k-1}(A;R)$.

\begin{defn}
The map $C_k$ defined above is called a
{\bfseries connecting homomorphism}\index{Connection Homomorphis}
\end{defn}

This definition is justified by the following result.

\begin{prop} \label{propConnectingH}
Let $A$ be a subset of a topological space $X$ and $R$ be an integral
domain.  Then the sequence
\begin{equation} \label{relHomEq2}
\xymatrix@C+2.5ex{
\ar[r]^-{C_{k+1}} & H_k(A;R) \ar[r]^{H_k(\iota)}
& H_k(X;R) \ar[r]^-{H_k(\Id_X)}  & H_k(X,A;R) \\
\ar[r]^-{C_k} & H_{k-1}(A;R) \ar[r]^{H_{k-1}(\iota)}
& H_{k-1}(X;R) \ar[r]^-{H_{k-1}(\Id_X)} &
}
\end{equation}
is a long exact sequence.
\end{prop}

\begin{proof}
We have already from Proposition~\ref{propAXXAexact} that
$\displaystyle \IMG(H_k(\iota)) = \KE(H_k(\Id_X))$.  The only thinks
that we have to prove is that
$\displaystyle \IMG(C_k) = \KE(H_{k-1}(\iota))$
and
$\displaystyle \IMG(H_k(\Id_X)) = \KE(C_k)$.

\stage{i} $\displaystyle \IMG(H_k(\Id_X)) \subset \KE(C_k)$.
Consider $[c]_X \in H_k(X;R)$.  We have that $\partial_k(c) = 0$.
Thus $H_k(\Id_X)([c]_X) = [c]_{X,A}$ with $\partial_k(c) = 0$.
Hence $C_k([c]_{X,A}) = [\partial_k(c)]_A = [0]_A \in H_{k-1}(A;R)$.

\stage{ii} $\displaystyle \IMG(H_k(\Id_X)) \supset \KE(C_k)$.
Suppose that $[c]_{X,A} \in H_k(X,A;R)$ is such that
$C_k([c]_{X,A}) = [\partial_k(c)]_A = [0]_A \in H_{k-1}(A;R)$.
Therefore $\partial_k(c) \in B_{k-1}(A;R)$; namely, $\partial_k(c)$ is a
$(k-1)$-boundary in $S_{k-1}(A;R)$.  Hence, there exists
$b \in S_k(A;R)$ such that $\partial_k(b) = \partial_k(c)$.
Since $c - b \in S_k(X;R)$ with
$\partial_k(c-b) = \partial_k(c) - \partial_k(b) = 0$, we have that
$c- b$ is a $k$-cycle in $S_k(X;R)$ and so $[c - b]_X \in  H_k(X;R)$.
We finally have that $H_k(\Id_X)([c-b]_X) = [c-b]_{X,A} = [c]_{X,A}$
because $b \in S_k(A;R)$.

\stage{iii} $\displaystyle \IMG(C_k) \subset \KE(H_{k-1}(\iota))$.
Consider $[c]_{X,A} \in H_k(X,A;R)$.  Then
$C_k([c]_{X,A}) = [\partial_k(c)]_A \in H_{k-1}(A;R)$.  Hence
\[
H_{k-1}(\iota)([\partial_k(c)]_A) = [S_{k-1}(i)(\partial_k(c))]_X
= [\partial_k(S_k(i)(c))]_X = [0]_X \in H_{k-1}(X;R)
\]
because
$\partial_k(S_k(i)(c)) \in B_{k-1}(X;R)$ since $S_k(i)(c) \in S_k(X;R)$.

\stage{iv} $\displaystyle \IMG(C_k) \supset \KE(H_{k-1}(\iota))$.
Suppose that $[c]_A \in H_{k-1}(A;R)$ is such that
$H_{k-1}(\iota)([c]_A) = [S_{k-1}(\iota)(c)]_X = [0]_X \in H_{k-1}(X;R)$.
Since $S_{k-1}(\iota)(c) = c \in B_{k-1}(X;R)$, there exists
$b \in S_k(X;R)$ such that $\partial_k(b) = c$.  Since
$c \in S_{k-1}(A;R)$, we have that $\partial_k(b) \in S_{k-1}(A;R)$.
Thus $b \in Z_k(X,A;R)$.  Finally, we get that
$C_k([b]_{X,A}) = [\partial_k(b)]_A = [c]_A$.
\end{proof}

Note that there is a much shorter proof of the previous proposition
using the Zig-Zag Lemma.  The reader can find it in (ii) the proof of
Lemma~\ref{lemEquDofET} given later.  However, we choose to give this
stand alone proof since it illustrates nicely the basic techniques used in
algebraic topology.

\begin{defn}
The sequence in (\ref{relHomEq2}) is called a {\bfseries homology
sequence}\index{Homology Sequence} for the pair $(X,A)$.
\end{defn}

If $B$ is a subset of a topological space $Y$ and
$f:(X,A) \to (Y,B)$ is a continuous function, then the following
diagram commutes.
\[
\xymatrix@C+2em{
\ar[r]^-{C_{k+1}} & H_k(A;R) \ar[r]^{H_k(\iota)} \ar[d]^{H_k(f)}
& H_k(X;R) \ar[r]^-{H_k(\Id_X)} \ar[d]^{H_k(f)} & H_k(X,A;R)
\ar[r]^-{C_k} \ar[d]^{H_k(f)} & \\
\ar[r]_-{C_{k+1}} & H_k(B;R) \ar[r]_{H_k(\iota)}
& H_k(Y;R) \ar[r]_-{H_k(\Id_X)} & H_k(Y,B;R) \ar[r]_-{C_k} &
}
\]

\begin{prop}  \label{propAretractSES}
Suppose that $A$ is a retract of a topological space $X$ and $R$ is an
integral domain.  Then the sequence.
\[
\xymatrix@C+2.5ex{
0 \ar[r] & H_k(A;R) \ar[r]^-{H_k(\iota)} & H_k(X;R) \ar[r]^-{H_k(\Id_X)}
& H_k(X,A;R) \ar[r] & 0
}
\]
is a short exact sequence.
\end{prop}

\begin{proof}
We have already from Proposition~\ref{propAXXAexact} that
$\displaystyle \IMG(H_k(\iota)) = \KE(H_k(\Id_X))$.

Since $A$ is a retract of $X$, there exists a map $r:X \to A$ such that
$r \circ \iota = \Id_{A}$ where $\iota: A \to X$ is the inclusion map.
It follows that $H_k(r) \circ H_k(\iota) = H_k(\Id_{A}) = \Id_{H_k(A;R)}$.
Thus $\KE(H_k(\iota)) = \{[0]_A\}$.  It is therefore justify to
introduce $0 \rightarrow H_k(A;R)$.

If we consider the long exact sequence of Proposition~\ref{propConnectingH}
and apply the previous reasoning to $k-1$, then we get that
$\IMG(C_k) = \KE(H_{k-1}(\iota)) = \{[0]\}$.  Thus, we may also
introduce $H_k(X,A;R) \rightarrow 0$.
\end{proof}

The fact that $\IMG(H_k(\iota)) = \KE( H_k(\Id_X))$ in the previous
proposition is not unique to $A$ a retract of $X$.  We have already
proved it in Proposition~\ref{propAXXAexact} in the general case of
$A$ a subset of $X$.

We have shown in the proof of Proposition~\ref{propAretractSES} that
\[
\xymatrix@C+2.5ex{
0 \ar[r] & H_k(A;R) \ar[r]^-{H_k(\iota)} & H_k(X;R) \ar[r]^-{H_k(\Id_X)}
& H_k(X,A;R) \ar[r] & 0
}
\]
is in fact a split short exact sequence if $A$ is a retract of $X$ because
$H_k(r) \circ H_k(\iota) = \Id_{H_k(A;R)}$, where $r:X \to A$ is the
map such that $r \circ \iota = \Id_A$ by definition of a retract.
Therefore $H_k(X;R) \cong H_k(A;R) \oplus H_k(X,A;R)$.

\begin{prop}  \label{propConPaConn}
Suppose that $A$ is a contractible subset a path-connected topological
space $X$ and $R$ is an integral domain.  Then the sequence.
\[
\xymatrix@C+2.5ex{
0 \ar[r] & H_k(X;R) \ar[r]^-{H_k(\Id_X)} & H_k(X,A;R) \ar[r] & 0
}
\]
is an exact sequence for $k>0$.  In particular, $H_k(X;R) \cong H_k(X,A;R)$
for $k>0$.
\end{prop}

\begin{proof}
\stage{$k>1$}
Since $\displaystyle A$ is contractible, we get from
Corollary~\ref{corContrHk} that $\displaystyle H_k(A;R) = 0$.
It follows from Proposition~\ref{propConnectingH} that
\[
\xymatrix@C+2.5ex{
0 \ar[r]^-{H_k(\iota)} & H_k(X;R) \ar[r]^-{H_k(\Id_X)}
& H_k(X,A;R) \ar[r]^-{C_k} & 0
}
\]
is exact.  Hence $\KE(H_k(\Id_X)) = \IMG(H_k(\iota)) = \{[0]_X\}$ and
$\IMG(H_k(\Id_X)) = \KE(C_k) = H_k(X,A;R)$.  Thus $H_k(X;R) \cong H_k(X,A;R)$.

\stage{$k=1$} We still have that $\displaystyle H_1(A;R) = 0$
because $A$ is contractible.  We also have from
Proposition~\ref{propH0XA0} that $\displaystyle H_0(X,A;R) = 0$
because $X$ is path-connected.  Hence, it follows from
Proposition~\ref{propConnectingH} that
\[
\xymatrix@C+2ex{
0 \ar[r]^-{H_1(\iota)}
& H_1(X;R) \ar[r]^-{H_1(\Id_X)}  & H_1(X,A;R) \ar[r]^-{C_1}
& H_0(A;R) \ar[r]^-{H_0(\iota)} & H_0(X;R) \ar[r]^-{H_0(\Id_X)} & 0
}
\]
Since $\KE(H_1(\Id_{X})) = \IMG(H_1(\iota)) = \{[0]_X\}$, we
have that $H_1(\Id_{X})$ is one-to-one.
We have that $\displaystyle H_0(A;R) \cong R$ according to
Corollary~\ref{corContrHk} and
$\displaystyle H_0(X;R) \cong R$ according to
Propositions~\ref{propH0XR}.
Therefore, since $\displaystyle \IMG(H_0(\iota)) =
\KE(H_0(\Id_X)) = H_0(X;R) \cong R$, we get that
$\displaystyle \KE(H_0(\iota)) = \{[0]_A\}$.
It follows that $\IMG(C_1) = \KE(H_0(\iota)) = \{[0]_A\}$.
Hence, we get that
\[
\xymatrix@C+2ex{
0 \ar[r]^-{H_1(\iota)}
& H_1(X;R) \ar[r]^-{H_k(\Id_X)}  & H_1(X,A;R) \ar[r]^-{C_1} & 0
}
\]
is exact.  As in (i), this implies that
$H_1(X;R) \cong H_1(X,A;R)$.
\end{proof}

The previous result is also true for $k=0$ if we consider reduced
homology.

\begin{prop}  \label{propContrXeXA}
Suppose that $A$ is a contractible subset of a topological space $X$ and
$R$ is an integral domain.  Then the sequence. 
\[
\xymatrix@C+2ex{
0 \ar[r] & H_k^\sharp(X;R) \ar[r]^-{H_k(\Id_X)} & H_k^\sharp(X,A;R) \ar[r] & 0
}
\]
is an exact sequence for $k \geq 0$.
In particular, $\displaystyle H_k^\sharp(X;R) \cong H_k^\sharp(X,A;R)$
for $k \geq 0$.
\end{prop}

\begin{proof}
Since $A$ is contractible, we have that from
Corollary~\ref{corContrHk} that $\displaystyle H_k^\sharp(A;R)
= H_k(A;R) = 0$ for $k >0$.  In the reduced homology, we
also have that $\displaystyle H_0^\sharp(A;R) = 0$.

It follows from Proposition~\ref{propConnectingH} (which is also true
for reduced homology) that
\[
\xymatrix@C+2ex{
0 \ar[r]^-{H_k(\iota)} & H_k^\sharp(X;R) \ar[r]^-{H_k(\Id_X)}
& H_k^\sharp(X,A;R) \ar[r]^(0.7){C_k} & 0
}
\]
is exact for all $k$.  Hence $\KE(H_k(\Id_X)) = \IMG(H_k(\iota)) = \{[0]_X\}$
and $\displaystyle \IMG(H_k(\Id_X)) = \KE(C_k) = H_k^\sharp(X,A;R)$ for all $k$.
Thus $\displaystyle H_k^\sharp(X;R) \cong H_k^\sharp(X,A;R)$.
\end{proof}

\subsection{Excision} \label{subsectExcis}

As for the previous subsection, the readers should consider this
subsection as an invitation to pursue their study of algebraic
topology.

The following definition has very profound applications.

\begin{defn}
Let $A$ be a subset of a topological space $X$, $U$ be a subset of $A$
and $R$ be an integral domain.  We say that the inclusion map
$\iota:(X\setminus U,A\setminus U) \to (X,A)$ is an
{\bfseries excision}\index{Excision} if
$H_k(\iota):H_k(X\setminus U,A\setminus U;R) \to H_k(X,A;R)$
is an isomorphism for all $k$.  We then say that $U$ can be
{\bfseries excised}\index{Excised} from $(X,A)$.
\end{defn}

The following theorem is one of the fundamental theorems concerning
excisions.

\begin{theorem}[Excision] \label{thmExcis}
Let $A$ be a subset of a topological space $X$, $U$ be a subset of $A$
and $R$ be an integral domain.  If $\displaystyle \overline{U} \subset A^\circ$,
then $U$ can be excised from $(X,A)$.
\end{theorem}

Recall that $\overline{U}$ denotes the closure of the set $U$ and
$\displaystyle A^\circ$ denotes the interior of the set $A$.

It is interesting to note that the proof of Theorem~\ref{thmExcis}
requires a singular version of the theory of simplicial approximation
introduced in Section~\ref{ssectSA}.  We briefly review this theory
and then prove Theorem~\ref{thmExcis}.
The first concept to generalize is the concept of barycentric
subdivision.

\begin{defn}
Let $X$ be an affine space and $\delta:\Delta_k \to X$ be an
affine function.  Given $b \in X$, the
{\bfseries join}\index{Join} $J_{k,b}(\delta)$ is the affine
function from $\Delta_{k+1}$ to $X$ defined by
\[
J_{k,b}(\delta)\left(\sum_{i=0}^{k+1} a_i \VEC{e}_i\right)
= a_0 b + \sum_{i=1}^{k+1} a_i \delta(\VEC{e}_{i-1})
\]
for $\displaystyle \sum_{i=0}^{k+1} a_i = 1$ with $a_i \geq 0$ for all
$i$ (Figure~\ref{singApprF1}).
Given a linearly combination $\displaystyle c = \sum_{j\in J} a_j \delta_j$
where $J \subset \NN$ is a finite set, $\delta_j:\Delta_k \to X$ is a
affine function and $a_j \in \RR$ for all $j \in J$, we define
$J_{k,b}(c)$ as $\displaystyle J_{k,b}(c) = \sum_{j\in J} a_j J_{k,b}(\delta_j)$.
\end{defn}

\pdfF{alg_top/singapprF1}{Graphical representation of $J_{1,b}(\delta)$}
{Graphical representation of $J_{1,b}(\delta)$ for an affine function
singular $\delta:\Delta_1 \to X$.}{singApprF1}

If $\delta:\Delta_0 \to X$, then
\begin{align*}
\big(\partial_1 (J_{0,b}(\delta))\big) (\VEC{e}_0)
&= \big(J_{0,b}(\delta) \circ F^0_1
- J_{0,b}(\delta) \circ F^1_1\big)(\VEC{e}_0)
= J_{0,b}(\delta) (\VEC{e}_1) - J_{0,b}(\delta) (\VEC{e}_0) \\
&= \delta(\VEC{e}_0) - b \ .
\end{align*}
Thus
\begin{equation} \label{singApproP1J0}
  \partial_1 (J_{0,b}(\delta)) = \delta - e_{b} \ ,
\end{equation}
where $e_{b}:\Delta_0 \to X$ is defined by $e_{b}(\VEC{e}_0) = b$.  Hence, if
$\displaystyle c = \sum_{j\in J} a_j \delta_j$ where
$\delta_j:\Delta_0 \to X$ is an affine function and $a_j \in \RR$ for all
$j\in J$,  then $\displaystyle \partial_1(J_{0,b}(c))
= \sum_{j\in J} a_j \partial_1(J_{0,b}(\delta_j))
= \sum_{j\in J} a_j (\delta_j - e_{b})
= c - \Big(\sum_{j\in J} a_j\Big) e_{b}$.

If $\delta:\Delta_k \to X$ for $k>0$ is an affine function, then
\begin{align*}
&\big(\partial_{k+1} (J_{k,b}(\delta))\big)
\left( \sum_{i=0}^k a_i\VEC{e}_i\right)
= \sum_{j=0}^{k+1} (-1)^j \big(J_{k,b}(\delta) \circ
F^j_{k+1}\big) \left( \sum_{i=0}^k a_i\VEC{e}_i\right) \\
&\qquad = \sum_{j=0}^{k+1} (-1)^j J_{k,b}(\delta)\left(
\sum_{i=0}^{j-1} a_i\VEC{e}_i + \sum_{i=j}^k a_i\VEC{e}_{i+1}\right) \\
&\qquad = \sum_{i=0}^k a_i\delta(\VEC{e}_i)
+ \sum_{j=1}^{k+1} (-1)^j \left( a_0 b +
\sum_{i=1}^{j-1} a_i\delta(\VEC{e}_{i-1})
+ \sum_{i=j}^k a_i\delta(\VEC{e}_i) \right) \\
&\qquad = \delta\left(\sum_{i=0}^k a_i\VEC{e}_i\right)
+ \sum_{j=1}^{k+1} (-1)^j \left( a_0 b +
\sum_{i=1}^{j-1} a_i\delta(\VEC{e}_{i-1})
+ \sum_{i=j}^k a_i\delta(\VEC{e}_i) \right)
\end{align*}
and
\begin{align*}
&J_{k-1,b}(\partial_k(\delta))
\left( \sum_{i=0}^k a_i\VEC{e}_i\right)
= J_{k-1,b}\left(\sum_{j=0}^k (-1)^j(\delta \circ F^j_k)\right)
\left( \sum_{i=0}^k a_i\VEC{e}_i\right) \\
&\qquad
% = \left(\sum_{j=0}^k (-1)^j J_{k-1,b}(\delta \circ F^j_k)\right)
% \left( \sum_{i=0}^k a_i\VEC{e}_i\right)
= \sum_{j=0}^k (-1)^j J_{k-1,b}(\delta \circ F^j_k)
\left( \sum_{i=0}^k a_i\VEC{e}_i\right)
= \sum_{j=0}^k (-1)^j \left( a_0 b
+ \sum_{i=1}^ka_i (\delta \circ F^j_k)(\VEC{e}_{i-1})\right) \\
&\qquad = \sum_{j=0}^k (-1)^j \left( a_0 b
+ \sum_{i=1}^ja_i \delta(\VEC{e}_{i-1})
+ \sum_{i=j+1}^ka_i \delta(\VEC{e}_i) \right) \\
&\qquad = -\sum_{j=1}^{k+1} (-1)^j \left( a_0 b
+ \sum_{i=1}^{j-1}a_i \delta(\VEC{e}_{i-1})
+ \sum_{i=j}^ka_i \delta(\VEC{e}_i) \right) \ .
\end{align*}
Thus 
\begin{equation} \label{singApproPkp1Jk}
\partial_{k+1} (J_{k,b}(\delta))
= \delta - J_{k-1,b}(\partial_k(\delta))
\end{equation}
for $k>0$.  Hence, if
$\displaystyle c = \sum_{j\in J} a_j \delta_j$ where $J \subset \NN$
is a finite set, $\delta_j:\Delta_k \to X$ is an affine function and
$a_j \in \RR$ for all $j\in J$, then $\displaystyle \partial_{k+1}(J_{k,b}(c))
= \sum_{j\in J} a_j \partial_{k+1}(J_{k,b}(\delta_j))
= \sum_{j\in J} a_j \big(\delta_j - J_{k-1,b}(\partial_k(\delta_j))\big)
= c - J_{k-1,b}(\partial_k(c))$.

If the reader has read Subsection~\ref{subsOShomo} about oriented
simplicial homology, then there is a nice way to simplify all the
computations that we have done above.  We can use the following
associative diagram.
\[
\xymatrix@R-2ex{
\text{affine functions} \ar@{}[d] & \text{oriented simplices} \ar@{}[d] \\
\gamma:\Delta_k \to X \ar@{<->}[r] \ar[d] &
\os{\gamma(\VEC{e}_0)}{\gamma(\VEC{e}_1)}{}{}{\gamma(\VEC{e}_k)}
\ar[d] \\
J_{k,b}(\gamma) \ar@{<->}[r] \ar[d] &
\os{b}{\gamma(\VEC{e}_0)}{\gamma(\VEC{e}_1)}{}{\gamma(\VEC{e}_k)}
\ar[d] \\
\partial_{k+1}(J_{k,b}(\gamma)) \ar@{<->}[r] &
\partial_{k+1}
\os{b}{\gamma(\VEC{e}_0)}{\gamma(\VEC{e}_1)}{}{\gamma(\VEC{e}_k)}
}
\]
The relations
$\partial_1 (J_{0,b}(\delta)) = \delta - e_{b}$ and
$\displaystyle \partial_{k+1} (J_{k,b}(\delta))
= \delta - J_{k-1,b}(\partial_k(\delta))$ for $k>0$ then follow
easily from
\begin{align*}
& \partial_{k+1} (J_{k,b}(\delta)) = \partial_{k+1}
\os{b}{\gamma(\VEC{e}_0)}{\gamma(\VEC{e}_1)}{}{\gamma(\VEC{e}_k)} \\
&\qquad = \os{\gamma(\VEC{e}_0)}{}{\gamma(\VEC{e}_1)}{}{\gamma(\VEC{e}_k)}
+ \sum_{j=1}^{k+1} (-1)^j
\os{b}{\gamma(\VEC{e}_0)}{\gamma(\VEC{e}_1)}
{\widehat{\gamma(\VEC{e}_{j-1})}}{\gamma(\VEC{e}_k)} \\
&\qquad
= \os{\gamma(\VEC{e}_0)}{}{\gamma(\VEC{e}_1)}{}{\gamma(\VEC{e}_k)}
- \sum_{j=0}^k (-1)^j
\os{b}{\gamma(\VEC{e}_0)}{\gamma(\VEC{e}_1)}
  {\widehat{\gamma(\VEC{e}_j)}}{\gamma(\VEC{e}_k)} \\
&\qquad
= \os{\gamma(\VEC{e}_0)}{}{\gamma(\VEC{e}_1)}{}{\gamma(\VEC{e}_k)}
- J_{k-1,b} \big(\partial_k\os{\gamma(\VEC{e}_0)}{}{\gamma(\VEC{e}_1)}{}
{\gamma(\VEC{e}_k)} \big)
= \delta - J_{k-1,b}(\partial_k(\delta)) \ .
\end{align*}

The concept of barycentric subdivision introduced in
Definition~\ref{defnBarySubd} is replaced by the following operator.

\begin{defn}
The {\bfseries barycentric subdivision operator}\index{Barycentric
Subdivision Operator} on the space of affine functions from $\Delta_k$ to
itself, denoted $\Sd_k$, is the operator defined by
\[
\Sd_k(\delta) =
\begin{cases}
\delta & \quad \text{if} \ k = 0 \\
J_{k-1,\delta(\VEC{b}_{(k)})}
(\Sd_{k-1}(\partial_k(\delta))) & \quad \text{if} \ k > 0  
\end{cases}
\]
for all affine functions $\delta:\Delta_k \to \Delta_k$
where $\displaystyle \VEC{b}_{(k)} = (1+k)^{-1} \sum_{j=0}^k \VEC{e}_k$ is
the barycentre of $\Delta_k$.
If $\displaystyle c = \sum_{j\in J} a_j \delta_j$ where
$J \subset \NN$ is a finite set, 
$\delta_j:\Delta_k \to \Delta_k$ is an affine function and $a_j \in \RR$
for all $j \in J$, then $\Sd_k(c)$ is defined as
$\displaystyle \Sd_k(c) = \sum_{j\in J} a_j \Sd_k(\delta_j)$.
\end{defn}

\begin{egg}
We have sketched in Figure~\ref{SingBSO} the action of the Barycentric
subdivision operators $\Sd_1(\delta)_1$ and $\Sd_2(\delta_2)$ where
$\delta_i$ is the identity map on $\Delta_i$ for $i=1,2$.
In the first case
\[
\Sd_1(\delta_1) = J_{0,\delta_1(\VEC{b}_{(1)})}\big(\Sd_0(\delta_1\circ F^0_1)
- \Sd_0(\delta_1 \circ F^1_1)\big)
= J_{0,\VEC{b}_{(1)}}(F^0_1) - J_{0,\VEC{b}_{(1)}}(F^1_1)
\]
In the second case
\begin{align*}
\Sd_2(\delta_2)
&= J_{1,\delta_2(\VEC{b}_{(2)})}\big(\Sd_1(\delta_2 \circ F^0_2)\big)
-J_{1,\delta_2(\VEC{b}_{(2)})}\big(\Sd_1(\delta_2 \circ F^1_2)\big)
+ J_{1,\delta_2(\VEC{b}_{(2)})}\big(\Sd_1(\delta_2\circ F^2_2)\big) \\
&= J_{1,\VEC{b}_{(2)}}(\Sd_1(F^0_2))
-J_{1,\VEC{b}_{(2)}}(\Sd_1(F^1_2)) + J_{1,\VEC{b}_{(2)}}(\Sd_1(F^2_2)) \\
&= J_{1,\VEC{b}_{(2)}}\big(J_{0,F^0_2(\VEC{b}_{(1)})}(\Sd_0(
\partial_1 F^0_2))\big)
-J_{1,\VEC{b}_{(2)}}\big(J_{0,F^1_2(\VEC{b}_{(1)})}(\Sd_0(\partial_1 F^1_2))\big)
\\
&\qquad
+J_{1,\VEC{b}_{(2)}}\big(J_{0,F^2_2(\VEC{b}_{(1)})}(\Sd_0(\partial_1 F^2_2))\big)
\\
&= J_{1,\VEC{b}_{(2)}}(J_{0,F^0_2(\VEC{b}_{(1)})}(F^0_2\circ F^0_1))
- J_{1,\VEC{b}_{(2)}}(J_{0,F^0_2(\VEC{b}_{(1)})}(F^0_2\circ F^1_1)) \\
&\qquad -J_{1,\VEC{b}_{(2)}}(J_{0,F^1_2(\VEC{b}_{(1)})}(F^1_2\circ F^0_1))
+J_{1,\VEC{b}_{(2)}}(J_{0,F^1_2(\VEC{b}_{(1)})}(F^1_2\circ F^1_1)) \\
&\qquad + J_{1,\VEC{b}_{(2)}}(J_{0,F^2_2(\VEC{b}_{(1)})}(F^2_2 \circ F^0_1))
- J_{1,\VEC{b}_{(2)}}(J_{0,F^2_2(\VEC{b}_{(1)})}(F^2_2 \circ F^1_1))\ .
\end{align*}
\end{egg}

\pdfF{alg_top/singBSO}{Barycentric subdivision operators}
{We have only sketched the action of three of the terms
for $\Sd_2(\delta_2)$.}{SingBSO}

We may extend the barycentric subdivision operator to $S_k(X;R)$ where
$X$ is a topological space and $R$ is an integral domain.

\begin{defn}
Let $X$ be a topological space and $R$ be an integral domain.
The {\bfseries barycentric subdivision operator}\index{Barycentric
Subdivision Operator} on $S_k(X;R)$, denoted $\Sd_k$, is the linear
operator from $S_k(X;R)$ into itself defined by
$\Sd_k(\sigma) = S_k(\sigma)(\Sd_k(\delta_k))$
for all singular $k$-simplices, where $\delta_k:\Delta_k \to \Delta_k$
is the identity map.
If $\displaystyle c = \sum_{j\in J} a_j \sigma_j$ where $J \subset \NN$
is a finite set, $\sigma_j$ is a singular $k$-simplex and $a_j \in R$
for all $j \in J$, then $\Sd_k(c)$ is defined as
$\displaystyle \Sd_k(c) = \sum_{j\in J} a_j \Sd_k(\sigma_j)$.
\end{defn}

Suppose that $X$ and $Y$ are two topological spaces and $R$ is an
integral domain.  If $f:X\to Y$ is a continuous function, then we have
the following commutative diagram.
\begin{equation} \label{SfCommDiagr}
\xymatrix@C+2em{
S_k(X;R) \ar[r]^{S_k(f)} \ar[d]^{\Sd_k} & S_k(Y;R) \ar[d]^{\Sd_k} \\
S_k(X;R) \ar[r]^{S_k(f)} & S_k(Y;R)
}
\end{equation}
It is effectively easy to verify this property.  If $\sigma$ is a
singular $k$-simplex in $X$, then
\begin{align*}
S_k(f)(\Sd_k(\sigma)) &= S_k(f)\big(S_k(\sigma) (\Sd_k(\delta_k))\big)
= (S_k(f)\circ S_k(\sigma))(\Sd_k(\delta_k)) \\
&= S_k(f\circ \sigma)(\Sd_k(\delta_k))
= \Sd_k(f\circ \sigma) = \Sd_k( S_k(f) (\sigma)) \ .
\end{align*}

The following result is a little harder to proof.

\begin{prop} \label{propSkm1PkpkSk}
Let $X$ be a topological space and $R$ be an integral domain.
Then $\Sd_{k-1}(\partial_k(c)) = \partial_k(\Sd_k(c))$ for all
$c \in S_k(X;R)$ and $k>0$.
\end{prop}

\begin{proof}
\stage{i} We first prove by induction that
$\Sd_{k-1}(\partial_k(\delta)) = \partial_k(\Sd_k(\delta))$
is true for affine functions $\delta:\Delta_k \to \Delta_k$.

For $k=1$, we have
\begin{align*}
\partial_1(\Sd_1(\delta))
&= \partial_1\big(J_{0,\delta(\VEC{b}_{(1)})}(\Sd_0(\partial_1(\delta)))\big)
= \partial_1\big(J_{0,\delta(\VEC{b}_{(1)})}(\Sd_0(\delta\circ F^0_1 - \delta\circ
F^1_1))\big) \\
&= \partial_1\big(J_{0,\delta(\VEC{b}_{(1)})}
(\delta\circ F^0_1 - \delta\circ F^1_1)\big)
= \big(\delta\circ F^0_1 - e_{\delta(\VEC{b}_{(1)})}\big)
- \big(\delta\circ F^1_1 - e_{\delta(\VEC{b}_{(1)})}\big) \\
&= \delta\circ F^0_1 - \delta\circ F^1_1
= \partial_1(\delta) = \Sd_0(\partial_1(\delta))
\end{align*}
where the third equality comes from the definition of $\Sd_0$ and the
fourth equality comes from (\ref{singApproP1J0}).

Suppose that the result is true for $k \geq 1$ and 
consider affine functions $\delta:\Delta_{k+1} \to \Delta_{k+1}$.  Then
\begin{align*}
\partial_{k+1}(\Sd_{k+1}(\delta))
&= \partial_{k+1}\big(J_{k,\delta(\VEC{b}_{(k+1)})}
(\Sd_k(\partial_{k+1}(\delta)))\big) \\
& = \Sd_k(\partial_{k+1}(\delta))
- J_{k-1,\delta(\VEC{b}_{(k+1)})}
\big(\partial_k(\Sd_k(\partial_{k+1}(\delta)))\big) \\
&= \Sd_k(\partial_{k+1}(\delta)) - J_{k-1,\delta(\VEC{b}_{(k+1)})}
\big(\Sd_{k-1}(\partial_k(\partial_{k+1}(\delta)))\big) \\
&= \Sd_k(\partial_{k+1}(\delta))
- J_{k-1,\delta(\VEC{b}_{(k+1)})}\big(\Sd_{k-1}(0)\big)
= \Sd_k(\partial_{k+1}(\delta))
\end{align*}
where we have used (\ref{singApproPkp1Jk}) to get the second equality
and our hypothesis of induction to get the third equality.

\stage{ii} We can now prove that the statement of the proposition is
true for a singular $k$-simplex $\sigma$.
\begin{align*}
&\partial_k(\Sd_k(\sigma))
= \partial_k\big(S_k(\sigma)(\Sd_k(\delta_k))\big)
= S_{k-1}(\sigma)\big(\partial_k(\Sd_k(\delta_k))\big)
= S_{k-1}(\sigma)\big(\Sd_{k-1}(\partial_k(\delta_k))\big) \\
&\quad = S_{k-1}(\sigma)\left(\Sd_{k-1}
\left(\sum_{j=0}^k(-1)^j\delta_k\circ F^j_k\right)\right)
= S_{k-1}(\sigma)\left(\sum_{j=0}^k(-1)^j\Sd_{k-1}(F^j_k)\right) \\
&\quad = S_{k-1}(\sigma)\left(\sum_{j=0}^k(-1)^j
S_{k-1}(F^j_k)(\Sd_{k-1}(\delta_{k-1}))\right)
= \sum_{j=0}^k(-1)^j S_{k-1}(\sigma \circ F^j_k)(\Sd_{k-1}(\delta_{k-1})) \\
&\quad = \sum_{j=0}^k(-1)^j \Sd_{k-1}(\sigma \circ F^j_k)
= \Sd_{k-1}\left(\sum_{j=0}^k(-1)^j \sigma \circ F^j_k\right)
= \Sd_{k-1}(\partial_k \sigma)
\end{align*}
where we have used (i) to get the third equality and the fact that
$\delta_k:\Delta_k \to \Delta_k$ is the identity map to get the fifth
equality.
\end{proof}

Let $X$ be a topological space and $R$ be an integral domain.
To obtain a theory of singular approximation, we need to introduce
a map $T_k:S_k(X;R) \to S_{k+1}(X;R)$.

\begin{defn}
We define a linear function, denoted $T_k$, that maps an affine function
$\delta:\Delta_k \to \Delta_k$ to an affine function
$T_k(\delta):\Delta_{k+1} \to \Delta_{k+1}$ as it follows.
\[
T_k(\delta) = \begin{cases}
0 & \quad \text{if} \ k = 0 \\
J_{k,\delta(\VEC{b}_{(k)})}\big(\delta - \Sd_k(\delta)
- T_{k-1}(\partial_k(\delta))\big)
& \quad \text{if} \ k > 0
\end{cases}
\]
for all affine functions $\delta:\Delta_k \to \Delta_k$,
where $\displaystyle \VEC{b}_{(k)} = (1+k)^{-1} \sum_{j=0}^k \VEC{e}_k$ is
the barycentre of $\Delta_k$.
If $\displaystyle c = \sum_{j\in J} a_j \delta_j$ where $J \subset \NN$
is a finite set, $\delta_j:\Delta_k \to \Delta_k$ is an affine
function and $a_j \in \RR$ for $j\in J$, then $T_k(c)$ is defined as
$\displaystyle T_k(c) = \sum_{j\in J} a_j T_k(\delta_j)$.
\end{defn}

We extend the function $T_k$ to $S_k(X;R)$ where
$X$ is a topological space and $R$ is an integral domain.

\begin{defn}
Let $X$ be a topological space and $R$ be an integral domain.
We define a linear function $T_k: S_k(X;R) \to S_{k+1}(X;R)$ by
$T_k(\sigma) = S_{k+1}(\sigma)(T_k(\delta_k))$
for all singular $k$-simplices, where $\delta_k:\Delta_k \to \Delta_k$
is the identity map.
If $\displaystyle c = \sum_{j\in J} a_j \sigma_j$ where $J \subset
\NN$ is a finite set, $\sigma_j$ is a singular $k$-simplex and
$a_j \in R$ for all $j\in J$, then $T_k(c)$ is 
defined as $\displaystyle T_k(c) = \sum_{j\in J} a_j T_k(\sigma_j)$.
\end{defn}

We note that $S_{k+1}(\sigma)(T_k(\delta_k))$ is well defined because
$T_k(\delta_k)(\Delta_{k+1}) \subset \Delta_k$ and
$\sigma:\Delta_k \to X$.  In fact, we may extend $\sigma$ to
$\Delta_{k+1}$ by setting $\sigma(\VEC{e}_{k+1}) = 0$.

Suppose that $X$ and $Y$ are two topological spaces and $R$ is an
integral domain.  If $f:X\to Y$ is a continuous function, then the
reader can proceed as we did for $\Sd_k$ to verify that we have the
following commutative diagram.
\begin{equation} \label{TfCommDiagr}
\xymatrix@C+2em{
S_k(X;R) \ar[r]^{S_k(f)} \ar[d]^{T_k} & S_k(Y;R) \ar[d]^{T_k} \\
S_{k+1}(X;R) \ar[r]^{S_{k+1}(f)} & S_{k+1}(Y;R)
}
\end{equation}

\begin{prop} \label{proppTTpISd}
Let $X$ be a topological space and $R$ be an integral domain.  Then
$\partial_{k+1}(T_k(c)) + T_{k-1}(\partial_k(c)) = c - \Sd_k(c)$ for all
$c \in S_k(X;R)$ and $k>0$.
\end{prop}

\begin{proof}
\stage{i} We first prove by induction the
$\partial_{k+1}(T_k(\delta)) +T_{k-1}(\partial_k(\delta)) = \delta -
\Sd_k(\delta)$ is true for affine functions $\delta:\Delta_k \to \Delta_k$.

For $k=1$, we have
\begin{align*}
&\partial_2(T_1(\delta))
= \partial_2\left( J_{1,\delta(\VEC{b}_{(k)})}\big(\delta - \Sd_1(\delta)
- T_0(\partial_1(\delta))\big)\right)
= \partial_2\left( J_{1,\delta(\VEC{b}_{(k)})}\big(\delta
- \Sd_1(\delta)\big)\right) \\
&\qquad = \delta - \Sd_1(\delta)
- J_{0,\delta(\VEC{b}_{(k)})}\big(\partial_1(\delta - \Sd_1(\delta))\big)
= \delta - \Sd_1(\delta) - J_{0,\delta(\VEC{b}_{(k)})}\big(\partial_1(\delta) -
\partial_1(\Sd_1(\delta))\big) \\
&\qquad
= \delta - \Sd_1(\delta) - J_{0,\delta(\VEC{b}_{(k)})}\big(\partial_1(\delta)
- \Sd_0(\partial_1(\delta))\big)
= \delta - \Sd_1(\delta) - J_{0,\delta(\VEC{b}_{(k)})}\big(\partial_1(\delta)
- \partial_1(\delta)\big) \\
&\qquad
= \delta - \Sd_1(\delta) = \delta - \Sd_1(\delta) - T_0(\partial_1(\delta))
\end{align*}
for all affine functions $\delta:\Delta_1 \to \Delta_1$, where we have
used (\ref{singApproPkp1Jk}) to get the third equality and
Proposition~\ref{propSkm1PkpkSk} to get the fifth equality.

Suppose that the result is true for $k > 0$ and 
consider affine functions $\delta:\Delta_{k+1} \to \Delta_{k+1}$.  Then
\begin{align*}
&\partial_{k+2}(T_{k+1}(\delta)))
= \partial_{k+2} \left( J_{k+1,\delta(\VEC{b}_{(k+1)})}
\big( \delta - \Sd_{k+1}(\delta)
- T_k(\partial_{k+1}(\delta))\big)\right) \\
&\qquad = \big( \delta - \Sd_{k+1}(\delta) - T_k(\partial_{k+1}(\delta))\big)
- J_{k,\delta(\VEC{b}_{(k+1)})} \big(\partial_{k+1}\big( \delta - \Sd_{k+1}(\delta)
- T_k(\partial_{k+1}(\delta))\big)\big) \\
&\qquad = \big( \delta - \Sd_{k+1}(\delta) - T_k(\partial_{k+1}(\delta))\big)
- J_{k,\delta(\VEC{b}_{(k+1)})}\big(\partial_{k+1}(\delta)
- \partial_{k+1}(\Sd_{k+1}(\delta)) \\
&\qquad\qquad - \partial_{k+1}(T_k(\partial_{k+1}(\delta)))\big) \\
&\qquad = \big( \delta - \Sd_{k+1}(\delta) - T_k(\partial_{k+1}(\delta))\big)
- J_{k,\delta(\VEC{b}_{(k+1)})}\big(\partial_{k+1}(\delta)
- \Sd_k(\partial_{k+1}(\delta)) \\
&\qquad \qquad
- \big(\partial_{k+1}(\delta) - \Sd_k(\partial_{k+1}(\delta))
  - T_{k-1}(\partial_k(\partial_{k+1}(\delta))) \big) \big) \\
&\qquad = \delta - \Sd_{k+1}(\delta) - T_k(\partial_{k+1}(\delta))
\end{align*}
where we have used (\ref{singApproPkp1Jk}) to get the second equality,
and Proposition~\ref{propSkm1PkpkSk} and the hypothesis of induction to get the
fourth equation.

\stage{ii} We can now prove that the statement of the proposition is
true for a singular $k$-simplex $\sigma$.
\begin{align*}
&\partial_{k+1}(T_k(\sigma))
= \partial_{k+1}(S_{k+1}(\sigma)(T_k(\delta_k)))
= S_k(\sigma)(\partial_{k+1}(T_k(\delta_k))) \\
& = S_k(\sigma)\big(\delta_k - \Sd_k(\delta_k)
-T_{k-1}(\partial_k(\delta_k))\big)
= S_k(\sigma)\bigg(\delta_k - \Sd_k(\delta_k)
-T_{k-1}\bigg(\sum_{j=0}^k (-1)^j \delta_k\circ F^j_k\bigg)\bigg) \\
&= S_k(\sigma)\bigg(\delta_k - \Sd_k(\delta_k)
- \sum_{j=0}^k (-1)^j T_{k-1}(F^j_k)\bigg) \\
&= S_k(\sigma)\bigg(\delta_k - \Sd_k(\delta_k)
- \sum_{j=0}^k (-1)^j S_k(F^j_k) (T_{k-1}(\delta_{k-1}))\bigg) \\
&= S_k(\sigma)(\delta_k) - S_k(\sigma)(\Sd_k(\delta_k))
- S_k\bigg(\sum_{j=0}^k (-1)^j \sigma \circ F^j_k\bigg)(T_{k-1}(\delta_{k-1})) \\
&= \sigma - \Sd_k(\sigma) - S_k(\partial_k \sigma) (T_{k-1}(\delta_{k-1}))
= \sigma - \Sd_k(\sigma) -T_{k-1}(\partial_k(\sigma))
\end{align*}
where we have used (i) to get the third equality.  We have also used
several times the fact that $\delta_q$ is the identity map on
$\Delta_q$ for $q=k$ and $k-1$.
\end{proof}

Since the topological spaces $X$ that we are considering for the singular 
$k$-chains may not have a metric, we have to formulate the distance
between singular $k$-chains in terms of open sets.

\begin{defn}
Let $\U = \{U_i\}_{i\in I}$ be an open cover of a topological space $X$.
We say that a singular $k$-simplex $\sigma$ in $X$ is {\bfseries small
of order $\mathbf{\U}$}\index{Small of Order} if there is an $i \in I$
such that $\sigma(\delta_k) \subset U_i$.
\end{defn}

We associate to every linear combination
$\displaystyle c = \sum_{j\in J} a_j \delta_j$ where $J \subset \NN$
is a finite set, $a_j \in \RR$ and $\delta_j:\Delta_k \to \Delta_k$ is
an affine function for all $j \in J$, the value
$\displaystyle \Dm(c) = \max_{j\in J}\, \diam(\delta_j(\Delta_k))$.
We have a result similar to Proposition~\ref{propMeshMBS}.  The proof
is very similar.

\begin{prop}
Let $\delta:\Delta_k \to \Delta_k$ be an affine function.  Then
$\displaystyle \Dm\big(\Sd_k(\delta)\big) \leq \frac{k}{k+1} \Dm(\delta)$.
\end{prop}

\begin{proof}[Proof (Sketch)]
There is a result equivalent to Lemma~\ref{lemDiams} for affine
functions $\delta:\Delta_k \to \Delta_k$.  Namely, 
$\diam(\delta(\Delta_k)) =
\|\delta(\VEC{e}_{i_1})  - \delta(\VEC{e}_{i_2})\|$
for some $i_1,i_2 \in \{0,1,2,\ldots,k\}$.  The proof is basically
identical because $\delta(\VEC{x}) = t\delta(\VEC{x}) + (1-t)
\delta(\VEC{x})$ for all $\VEC{x} \in \Delta_k$.

By definitions of $\displaystyle \Sd_k(\delta)$ and the previous
paragraph, it follows that
$\Dm(c) = \|\delta(\VEC{b}_{(p)}) - \delta(\VEC{b}_{(q)}\|$
for some $p,q \in \{0,1,\ldots,k\}$ where $\VEC{b}_{(p)}$ and
$\VEC{b}_{(q)}$ are vertices on the image of the same function
\[
\Sd_k(\delta_k) =
J_{k-1,\VEC{b}_{(k)}}\big(J_{k-2, F^{s_k}_k(\VEC{b}_{(k-1)})}\big( \ldots
\big(J_{0,F^{s_k}_k \circ \cdots \circ F^{s_2}_2(\VEC{b}_{(1)})}
\big(F_k^{s_k} \circ F_{k-1}^{s_{k-1}} \circ
\ldots \circ F_1^{s_1}\big)\big)\big)\big) : \Delta_k \to \Delta_k
\]
for some $0\leq s_j \leq j$ with $1 \leq j \leq k$, where $\delta_k$ is
the identity map.  We may assume that $p < q$,
$\displaystyle \VEC{b}_{(p)} = \frac{1}{1+p} \sum_{j=0} \VEC{e}_{i_j}$
and
$\displaystyle \VEC{b}_{(q)} = \frac{1}{1+q} \sum_{j=0} \VEC{e}_{i_j}$.
Note that the image of $\Sd_k(\delta_k)$ is nothing else than
$\displaystyle K^{[1]}$ defined in Subsection~\ref{ssectBarSubd} where
$K$ is the $k$-simplex $[\VEC{e}_0,\VEC{e}_1,\ldots,\VEC{e}_k]$.
We can then proceed as in the proof of Proposition~\ref{propMeshMBS}
to show that
\begin{align*}
&\Dm(c) = \|\delta(\VEC{b}_{(p)}) - \delta(\VEC{b}_{(q)})\|
= \left\| \frac{1}{p+1} \sum_{j=0}^p \delta(\VEC{e}_{i_j}) 
- \frac{1}{q+1} \sum_{j=0}^q \delta(\VEC{e}_{i_j}) \right\| \\
&\qquad \leq \frac{1}{(q+1)(p+1)} \sum_{j=0}^q \sum_{j=0}^p
  \left\|  \delta(\VEC{e}_{i_j}) - \delta(\VEC{e}_{i_j})\right\|
\leq \frac{q(p+1)}{(q+1)(p+1)} \Dm(\delta) \\
&\qquad = \frac{q}{q+1} \Dm(\delta) \leq \frac{k}{k+1} \Dm(\delta)
\end{align*}
because
$\left\|  \delta(\VEC{e}_{i_j}) - \delta(\VEC{e}_{i_j})\right\|
\leq \Dm(\delta)$ for all $i$ and $j$, and $q \leq k$.
\end{proof}

Given an affine function $\delta:\Delta_k \to \Delta_k$, it follows
from the previous proposition that
$\displaystyle \Dm\big((\Sd_k)^q(\delta)\big)
\leq \left(\frac{k}{k+1}\right)^q \Dm(\delta) \to 0$
as $q \to \infty$, where
\[
(\Sd_k)^q(\delta)
= \underbrace{\Sd_k(\Sd_k(\ldots(\Sd_k(\delta))))}_{q\text{ times}} \ .
\]
for $q \in \NN$.

The theorem in singular homology that corresponds to
Theorem~\ref{thESAf} is the following theorem.

\begin{theorem}  \label{thSdkqSmall}
Let $\U = \{U_i\}_{i\in I}$ be an open cover of a topological space
$X$ and $R$ be an integral domain.  Given a singular $k$-simplex
$\sigma$ in $X$, there exists $q \in \NN$ such that
$\displaystyle (\Sd_k)^q(\sigma)$ is a linear combination of singular
$k$-simplices small of order $\U$.
\end{theorem}

\begin{proof}
Since $\displaystyle \{\sigma^{-1}(U_i)\}_{i\in I}$ is an open cover
of the compact set $\Delta_k$, there exists a Lebesgue number
$\epsilon$ such that, for every $\VEC{x} \in \Delta_k$, we have that
$\displaystyle B_\epsilon(\VEC{x}) \cap \Delta_k
\subset \sigma^{-1}(U_{i_\VEC{x}})$ for some $i_{\VEC{x}} \in I$;
namely,
$\displaystyle \sigma(B_\epsilon(\VEC{x}) \cap \Delta_k)
\subset U_{i_\VEC{x}}$ for some $i_{\VEC{x}} \in I$.

Choose $q$ large enough to have
$\displaystyle \Dm\big((\Sd_k)^q(\delta_k)\big) < \epsilon$ where
$\delta_k$ is the identity map on $\Delta_k$.
It follows from the commutative diagram in (\ref{SfCommDiagr}) that
$\displaystyle (\Sd_k)^q(\sigma) = S_k(\sigma)\big((\Sd_k)^q(\delta_k)\big)$.
So $\displaystyle (\Sd_k)^q(\sigma)$ is the linear combinations of
singular $k$-simplices of the form $\sigma \circ \delta$, where
$\delta:\Delta_k \to \Delta_k$ is an affine function such that
$\diam(\delta(\Delta_k)) < \epsilon$.  Since
$\delta(\Delta_k) \subset B_\epsilon(\VEC{x}) \cap \Delta_k$ for some
$\VEC{x} \in \Delta_k$, we have that
$(\sigma \circ \delta)(\Delta_k) \subset U_{i_{\VEC{x}}}$ for some
$i_{\VEC{x}} \in I$.
\end{proof}

\begin{prop} \label{propHkXAsmall}
Let $A$ be a subset of a topological space $X$ and $R$ be an integral
domain.  Moreover, let $\U = \{U_i\}_{i\in I}$ be an open cover of $X$.
Every homology class in $H_k(X,A;R)$ can be represented by a relative
$k$-cycle of $X \mod A$ which is a linear combination of singular
$k$-simplices small of order $\U$.
\end{prop}

\begin{proof}
Given $[c]_{X,A} \in H_k(X,A;R)$, we have that
$c$ is a relative $k$-cycle of $X \mod A$.  Recall that this means that
$\partial_k(c) \in S_{k-1}(A;R)$.

It follows from Proposition~\ref{proppTTpISd} that
\begin{equation} \label{propHkRCsmallEq1}
\partial_{k+1}(T_k(c)) + T_{k-1}(\partial_k(c)) = c - \Sd_k(c) \ .
\end{equation}
Since $\partial_k(c) \in S_{k-1}(A;R)$, we have that
$T_{k-1}(\partial_k(c)) \in S_k(A;R)$ follows from the
commutative diagram in (\ref{TfCommDiagr}) with $f$ replaced by the
inclusion map $\iota:A \to X$.  Moreover,
$\Sd_k(c)$ is a relative $k$-cycle of $X \mod A$
because $\partial_k(\Sd_k(c)) = \Sd_{k-1}(\partial_k(c))$
according to Proposition~\ref{propSkm1PkpkSk} and
$\partial_k(c) \in S_{k-1}(A;R)$ implies
that $\Sd_{k-1}(\partial_k(c)) \in S_{k-1}(A;R)$ due to 
the commutative diagram in (\ref{SfCommDiagr}) with $f$ replaced by the
inclusion map $\iota:A \to X$.
It therefore follows from (\ref{propHkRCsmallEq1}) and the definition
of homologous in $X \mod A$ that
$c \hsim \Sd_k(c)$ in $X \mod A$.   We get that
$[c]_{X,A} = [\Sd_k(c)]_{X,A}$ in $H_k(X,A;R)$.

It follows by induction that
$\displaystyle [c]_{X,A} = [(\Sd_k)^q(c)]_{X,A}$ for all $q \in \NN$.
If we apply Theorem~\ref{thSdkqSmall} to each singular $k$-simplex of
the linear combination of $c$, then we can find $q$ large enough to
have that $\displaystyle (\Sd_k)^q(c)$ a linear combination of 
singular $k$-simplices small of order $\U$.
\end{proof}

We can finally prove the excision theorem.

\begin{proof}[Proof (of Theorem~\ref{thmExcis})]
Let $\iota:A \to X$ be the inclusion map.  We prove that
$H_k(\iota):H_k(X\setminus U,A\setminus U;R) \to H_k(X,A;R)$ is an
isomorphism.

\stage{i} We first prove that $H_k(\iota)$ is onto.
Suppose that $[c]_{X,A} \in H_k(X,A;R)$.  It follows from
Proposition~\ref{propHkXAsmall} that we may assume that
$\displaystyle c = \sum_{j\in J} a_j \sigma_j$ is a relative $k$-cycle
where each $\sigma_j$ is small of order
$\displaystyle \U = \{ X \setminus \overline{U}, A^\circ\}$.

If $\sigma_j(\Delta_k) \not\subset X \setminus U$, then
we must have that $\displaystyle \sigma_j(\Delta_k) \subset A^\circ$ because
$X \setminus \overline{U} \subset X \setminus U$.
It follows that $\sigma_j \in S_k(A;R)$.  In other words, we have that
$\sigma_j \in B_k(X,A;R)$ and so $[\sigma_j]_{X,A} = [0]_{X,A}$.  We can
then remove all this terms from the summation and still get a
summation homologous in $X \mod A$ to $c$.    We may therefore assume
that $\sigma_j(\Delta_k) \subset X \setminus U$ for all $j$.

Since $c$ is a relative $k$-cycle and $c \in S_k(X \setminus U;R)$, we
get that \\
$\partial_k(c) \in S_k(A;R) \cap S_k(X \setminus U;R)$.
Thus $\partial_k(c) \in S_k(A\setminus U;R)$.  Hence $c$ is a
relative $k$-cycle in $X\setminus U \mod A \setminus U$.
We have that $[c]_{X\setminus U,A\setminus U}
\in H_k(X\setminus U;A\setminus U;R)$ and
$H_k(\iota)([c]_{X\setminus U,A\setminus U}) = [c]_{X,A} \in H_k(X,A;R)$.

\stage{ii} We now prove that $H_k(\iota)$ is one-to-one.
Suppose that $[c]_{X\setminus U,A\setminus U}
\in H_k(X \setminus U,A\setminus U;R)$
satisfies $H_k(\iota)([c]_{X\setminus U,A\setminus U})
= [c]_{X,A} = [0]_{X,A} \in H_k(X,A;R)$.
Since $\relC[X,A]{c} \in \IMG\big(\overline{\partial}_k\big)
= B_k(X,A;R)/S_k(A;R)$, we have that $c$ is a relative $k$-boundary in
$X \mod A$.  Therefore, there exists $b \in S_{k+1}(X;R)$ such that
$\tilde{c} = c + \partial_{k+1}(b) \in S_k(A;R)$.

We get from Proposition~\ref{propSkm1PkpkSk} that
\begin{equation} \label{thmExcisEq1}
(\Sd_k)^q(\tilde{c}) = (\Sd_k)^q(c) +
\partial_{k+1}\left((\Sd_{k+1})^q(b)\right)
\end{equation}
for $q \geq 0$.
According to Proposition~\ref{propHkXAsmall}, we may choose $q$ large
enough such that $\displaystyle (\Sd_k)^q(b)$ is small of order
$\displaystyle \U = \{X\setminus \overline{U}, A^\circ\}$.  Thus, we may write
$\displaystyle (\Sd_k)^q(b) = b_1 + b_2$ for some $b_1 \in
S_{k+1}(X\setminus \overline{U};R) \subset S_{k+1}(X\setminus U;R)$
and $\displaystyle b_2 \in S_{k+1}(A^\circ;R) \subset S_{k+1}(A;R)$.
We get from (\ref{thmExcisEq1}) that
\[
(\Sd_k)^q(\tilde{c}) - \partial_{k+1}(b_2) = (\Sd_k)^q(c) +
\partial_{k+1}(b_1) \ ,
\]
where the right hand side is in $S_k(X\setminus U;R)$ and the left
hand side is in $S_k(A;R)$.  Therefore, both sides of the previous
equality must be in $S_k(A\setminus U;R)$.  We conclude that
$\displaystyle (\Sd_k)^q(c) \hsim
(\Sd_k)^q(\tilde{c}) - \partial_{k+1}(b_2) \in S_k(A\setminus U;R)$.
Thus $\displaystyle (\Sd_k)^q(c)$ is a relative $k$-boundary in
$X\setminus U \mod A \setminus U$ because
$\displaystyle (\Sd_k)^q(c) \in S_k(X\setminus U;R)$
since $c \in S_k(X\setminus U;R)$.  It follows that
$\displaystyle [c]_{X\setminus U,A\setminus U} =
[(\Sd_k)^q(c)]_{X\setminus U,A\setminus U}
= [0]_{X\setminus U,A \setminus U} \in H_k(X\setminus U,A\setminus U;R)$.
Note that $\displaystyle [c]_{X\setminus U,A\setminus U} =
[(\Sd_k)^q(c)]_{X\setminus U,A\setminus U}$ as we saw in the context
of the proof of Proposition~\ref{propHkXAsmall}.
\end{proof}

Before closing this subsection, we should mention some results to
illustrate the power of the technique of excision.  First, we have to
strengthen the concept of retraction introduced in
definition~\ref{defnRetract} to the context of relative homology.

\begin{defn}
Let $A$ and $Y$ be two subsets of a topological space $X$ and let
$B = Y \cap A$ (Figure~\ref{DefRetr1}).  We say that $(Y,B)$ is a
{\bfseries deformation retract}\index{Deformation Retract} of $(X,A)$
if there exists a function $r:X \to Y$ such that
$r(A) \subset B$, $r \circ \iota = \Id_Y$ where $\iota:Y \to X$ is the
inclusion map (i.e. $Y$ is a retract of $X$), and
$\Id_X \rsim \iota \circ r$.
The homotopy $H:X \times [0,1] \to H$ from $\Id_X$ to $\iota \circ r$
is called {\bfseries deformation retraction}\index{Deformation Retraction}.
\end{defn}

\pdfF{alg_top/defretr1}{Deformation retraction}{$(Y,B)$ is a
deformation retract of $(X,A)$.}{DefRetr1}

In the previous definition, to apply the definition of homotopic
functions given in Definition~\ref{defnRelHF}, we consider
$\Id_X:(X,A) \to (X,A)$ and $\iota \circ r:(X,A) \to (X,A)$.  To get
$(\iota \circ r)(A) \subset B$, we have to require that $r(A) \subset B$.

\begin{prop} \label{propVUAX}
Let $X$ be a topological space and $R$ be an integral domain.  Assume that
$V \subset U \subset A \subset X$ (Figure~\ref{Excision1}).  If $V$
can be excised from $(X,A)$ and $(X\setminus U,A \setminus U)$ is
a deformation retract of $(X \setminus V, A \setminus V)$, then $U$
can be excised from $(X,A)$.
\end{prop}

\begin{proof}
We have that $X \setminus U \subset X \setminus V$,
$A \setminus U \subset A \setminus V$ and
$(A \setminus V) \cap (X \setminus U) = A \setminus U$
(Figure~\ref{Excision1}).  As usual, let $\iota:Y \to X$ be the
inclusion map.  Let $r:X\setminus V \to X \setminus U$ be the
function satisfying $r(x) = x$ for all $x \in X\setminus U$,
$r(A\setminus V) \subset A \setminus U$, and
$\Id_{X \setminus V} \rsim \iota \circ r$.

It follows from $r \circ \iota = \Id_{X\setminus U}$,
$\Id_{X \setminus V} \rsim \iota \circ r$, and
Propositions~\ref{propHkPropS} and \ref{propfgHHkfgE}, that
$H_k(r) \circ H_k(\iota) = H_k(\Id_{X\setminus U}) =
\Id_{H_k(X\setminus U,A\setminus U;R)}$ and
$H_k(\iota) \circ H_k(r) = H_k(\Id_{X\setminus V}) =
\Id_{H_k(X\setminus V,A\setminus V;R)}$.  Thus
$H_k(\iota):H_k(X\setminus U,A\setminus U;R) \to 
H_k(X\setminus V,A\setminus V;R)$ is an isomorphism.

Since $V$ can be excised from $(X,A)$, we have that
$H_k(\iota):H_k(X\setminus V,A\setminus V;R) \to H_k(X,A;R)$ is an
isomorphism.  It follows that
$H_k(\iota):H_k(X\setminus U,A\setminus U;R) \cong H_k(X,A;R)$.
\end{proof}

\pdfF{alg_top/excision1}{Excision theorem}{Representation of the sets
involved in the statement of Proposition~\ref{propVUAX}.
$X\setminus V$ is the union of the pale and darker blue regions.
$A \setminus V$ is the darker blue region.  $X\setminus U$ is the
union of the regions with horizon dashed lines and the region with
horizontal and vertical dashed lines.  $A\setminus U$ is the region with
horizontal and vertical dashed lines.}{Excision1}

\begin{egg}
Let                      \label{eggSpnSmn}
$\displaystyle S^n_+ = \{ \VEC{x} \in S^n : x_{n+1} \geq 0 \}$
and
$\displaystyle S^n_- = \{ \VEC{x} \in S^n : x_{n+1} \leq 0 \}$.
We use the previous proposition to prove below that 
$\displaystyle (S^n_-)^\circ = \{ \VEC{x} \in S^n : x_{n+1} < 0 \}$ 
can be excised from $\displaystyle (S^n,S^n_-)$.  Hence
$\displaystyle H_k(S^n_+,S^{n-1};R) \cong H_k(S^n,S^n_-;R)$ because
$\displaystyle S^n \setminus (S^n_-)^\circ = S^n_+$ and
$\displaystyle S^n_- \setminus (S^n_-)^\circ = S^{n-1}$.

We cannot directly use the excision theorem, Theorem~\ref{thmExcis},
because we are in the situation where
$\displaystyle U = (S^n_-)^\circ$ and $\displaystyle A = S^n_-$ in the 
statement of the theorem.  Therefore
$\displaystyle \overline{U} \not\subset A^\circ$.
We have to use the previous proposition with
$\displaystyle V = \{ \VEC{x} \in S^n : x_{n+1} \leq -\epsilon \}$ and
$0 \leq \epsilon < 1$ (Figure~\ref{Excision2}).  We have from
Theorem~\ref{thmExcis} that $V$ can be excised from
$\displaystyle (S^n,S^n_-)$ because
$\displaystyle \overline{V} \subset (S^n_-)^\circ$.  Moreover,
$\displaystyle (S^n_+,S^{n-1})
= \big(S^n \setminus (S^n_-)^\circ, S^n_- \setminus (S^n_-)^\circ\big)$
is a deformation retract of
$\displaystyle (S^n \setminus V,S^n_- \setminus V)$.
A possible function $\displaystyle r: S^n \setminus V \to S^n_+$ for
the deformation retract is represented by the arrows along the great circles in
Figure~\ref{Excision2}.  Only the $\displaystyle \VEC{x} \in (S_-^n)^\circ$ are
moving along the great circles toward $\displaystyle S^{n-1}$.
\end{egg}

\pdfF{alg_top/excision2}{Example of excision theorem}{Representation
of the set $V$ involved in Example~\ref{eggSpnSmn}.  The arrows
illustrate the map $r$ use to show that
$\displaystyle (S^n \setminus (S^n_-)^\circ,S^n_- \setminus (S^n_-)^\circ)$
is a deformation retract of $\displaystyle (S^n \setminus V,S^n_- \setminus V)$.
Only the $\displaystyle \VEC{x} \in (S_-^n)^\circ$ are moving along
the great circles toward $\displaystyle S^{n-1}$.}{Excision2}

\begin{egg}
We may use the tools that we have introduced     \label{eggHkSq}
so far to compute $\displaystyle H_k(S^n)$ for $k,n>0$.

\stage{i}
Let $\displaystyle B^n = \overline{B_1(\VEC{0})} \subset \RR^n$.
So $\displaystyle \partial B^n = S^{n-1}$.
We get from Proposition~\ref{propConnectingH} that
\begin{equation} \label{eggHkSn}
\xymatrix@C+2.5ex{
\ar[r]^-{C_{k+1}} & H_k(S^{n-1};R) \ar[r]^{H_k(\iota)}
& H_k(B^n;R) \ar[r]^-{H_k(\Id_{B^n})} & H_k(B^n,S^{n-1};R) \\
\ar[r]^-{C_k} & H_{k-1}(S^{n-1};R) \ar[r]^-{H_{k-1}(\iota)}
& H_{k-1}(B^n;R) \ar[r]^-{H_{k-1}(\Id_{B^n})} &
}
\end{equation}

\stage{$\mathbf{k>1}$}
Since $\displaystyle B^n$ is contractible, we get from
Corollary~\ref{corContrHk} that $\displaystyle H_k(B^n;R) = 0$
and $\displaystyle H_{k-1}(B^n;R) = 0$.  We deduce from
(\ref{eggHkSn}) that
\[
\xymatrix@C+2.5ex{
0  \ar[r]^-{H_k(\Id_{B^n})} & H_k(B^n,S^{n-1};R) \ar[r]^-{C_k} &
H_{k-1}(S^{n-1};R) \ar[r]^-{H_{k-1}(\iota)} & 0
}
\]
Since $\KE(C_k) = \IMG(H_k(\Id_{B^n})) = \{[0]_{B^n,S^{n-1}}\}$
and $\displaystyle \IMG(C_k) = \KE(H_{k-1}(\iota)) = H_{k-1}(S^{n-1};R)$,
we find that $C_k$ is
a isomorphism between $\displaystyle H_k(B^n,S^{n-1};R)$ and
$\displaystyle H_{k-1}(S^{n-1};R)$.

\stage{$\mathbf{k=1}$} We consider two cases: $n>1$ and $n=1$.

For $n>1$, we still have that $\displaystyle H_1(B^n;R) = 0$
because $\displaystyle B^n$ is contractible.  Moreover, we have from
Proposition~\ref{propH0XA0} that
$\displaystyle H_0(B^n,S^{n-1};R) = 0$ because
$\displaystyle B^n$ is path-connected.  Hence, we get from
(\ref{eggHkSn}) that
\[
\xymatrix@C+3ex{
0 \ar[r]^-{H_1(\Id_{B^n})} & H_1(B^n,S^{n-1};R)
\ar[r]^-{C_1} & H_0(S^{n-1};R) \ar[r]^{H_0(\iota)}
& H_0(B^n;R) \ar[r]^-{H_0(\Id_{B^n})} & 0
}
\]
Since $\KE(C_1) = \IMG(H_1(\Id_{B^n})) = \{[0]_{B^n,S^{n-1}}\}$, we
still have that $C_1$ is one-to-one.
We now have that $\displaystyle H_0(B^n;R) \cong R$ and
$\displaystyle H_0(S^{n-1};R) \cong R$ according to
Proposition~\ref{propH0XR} because $\displaystyle B^n$ and
$\displaystyle S^{n-1}$ are path-connected.  Hence, since
$\displaystyle \IMG(H_0(\iota)) = \KE(H_0(\Id_{B^n})) = H_0(B^n;R) \cong R$,
we get
that $\displaystyle \KE(H_0(\iota)) = \{[0]\} \in H_0(S^{n-1};R) \cong R$.
It follows that $\IMG(C_1) = \KE(H_0(\iota)) = \{[0]_{S^{n-1}}\}$.  Since $C_1$
is one-to-one, we must have that
$\displaystyle H_1(B^n,S^{n-1};R) = 0$.

For $n=1$, we still have that $\displaystyle H_1(B^1;R) = 0$
because $\displaystyle B^1 = [-1,1]$ is contractible and
$\displaystyle H_0(B^1,S^0;R) = 0$ because $B^1$ is
path-connected.  Hence, we get from (\ref{eggHkSn}) that
\[
\xymatrix@C+3ex{
0 \ar[r]^-{H_1(\Id_{B^1})} & H_1(B^1,S^0;R)
\ar[r]^-{C_1} & H_0(S^0;R) \ar[r]^{H_0(\iota)}
& H_0(B^1;R) \ar[r]^-{H_0(\Id_{B^1})} & 0
}
\]
Since $\KE(C_1) = \IMG(H_1(\Id_{B^1})) = \{[0]_{B^1,S^0}\}$, we
still have that $C_1$ is one-to-one.
We now have that $\displaystyle H_0(B^1;R) \cong R$ and
$\displaystyle H_0(S^0;R) \cong R^2$ according to
Propositions~\ref{propH0XR} and \ref{propHkeHkJ}.  Hence, since
$\displaystyle \IMG(H_0(\iota)) = \KE(H_0(\Id_{B^1})) = H_0(B^1;R) \cong R$,
we get that $\displaystyle \KE(H_0(\iota)) \cong R$.
It follows that $\IMG(C_1) = \KE(H_0(\iota)) \cong R$.  Since $C_1$
is one-to-one, we have that $\displaystyle H_1(B^1,S^0;R) \cong R$.

In summary, we have found so far that
$\displaystyle H_k(B^n,S^{n-1};R) \cong H_{k-1}(S^{n-1};R)$ for
$k>1$ and $n>0$, $\displaystyle H_1(B^n,S^{n-1};R) = 0$ for
$n>1$ and $\displaystyle H_1(B^1,S^0;R) \cong R$.

\stage{ii}
The projection $\displaystyle \pi:\RR^{n+1} \to \RR^n$ defined by
$\pi(\VEC{x}) = (x_1,x_2,\ldots,x_n)$ for
$\displaystyle \VEC{x} \in \RR^{n+1}$ and $n>0$ provides a homeomorphism
$\displaystyle \pi:(S^n_+,S^{n-1}) \to (B^n,S^{n-1})$.  Hence
$\displaystyle H_k(S^n_+,S^{n-1};R) \cong H_k(B^n,S^{n-1};R)$ for $n>0$.

Moreover, we get from Example~\ref{eggSpnSmn} that
$\displaystyle H_k(S^n_+,S^{n-1};R) \cong H_k(S^n,S^n_-;R)$ for $n>0$.
Thus $\displaystyle H_k(B^n,S^{n-1};R) \cong H_k(S^n,S^n_-;R)$ for $n>0$.

It follows from Proposition~\ref{propConPaConn} that
$\displaystyle H_k(S^n,S^n_-;R) \cong H_k(S^n;R)$ for $k>0$ and $n>0$ because
$\displaystyle S^n$ is path connected and $\displaystyle S^n_-$
is contractible.  Therefore
$\displaystyle H_k(B^n,S^{n-1};R) \cong H_k(S^n;R)$ for $k>0$

In conclusion, we have found that
\begin{enumerate}
\item $\displaystyle H_0(S^0;R) \cong R^2$ comes from
Propositions~\ref{propH0XR} and \ref{propHkeHkJ} because
$\displaystyle S^0 = \{-1,1\}$ has two components,
\item $\displaystyle H_0(S^n;R) \cong R$ for $n>0$ comes
from Propositions~\ref{propH0XR} because $\displaystyle S^n$ is
path-connected,
\item $\displaystyle H_1(S^0;R) = 0$ comes from
Example~\ref{eggHk1} and Proposition~\ref{propHkeHkJ}
because $\displaystyle S^0 = \{-1,1\}$,
\item $\displaystyle H_1(S^1;R) \cong H_1(B^1,S^0;R) \cong R$,
\item $\displaystyle H_1(S^n;R) \cong H_1(B^n,S^{n-1};R) = 0$ for $n>1$, and
\item $\displaystyle H_k(S^n;R) \cong H_k(B^n,S^{n-1};R) \cong
H_{k-1}(S^{n-1};R)$ for $k>1$ and $n>0$.
\end{enumerate}
\end{egg}

\subsection{Mayer-Vietoris Sequence}

The Mayer-Vietoris sequence alluded to in
Subsection~\ref{subsectMVarg} is an import topic.  To prove
Mayer-Vietoris theorem below, we need along the way a very famous
result.  The proof of this result is amusing.

\begin{theorem}[Five Lemma]
Let $R$ be an integral domain.            \index{Five Lemma}
Suppose that we have the following commutative diagram
\[
\xymatrix{
A_1 \ar[r]^{f_1} \ar[d]^{h_1} & A_2 \ar[r]^{f_2} \ar[d]^{h_2}
& A_3 \ar[r]^{f_3} \ar[d]^{h_3} & A_4 \ar[r]^{f_4} \ar[d]^{h_4} &
A_5 \ar[d]^{h_5} \\
B_1 \ar[r]_{g_1} & B_2 \ar[r]_{g_2} & B_3 \ar[r]_{g_3}
& B_4 \ar[r]_{g_4} & B_5
}
\]
where the $A_j$ and $B_j$ are $R$-module, and the $f_j$, $g_j$ and
$h_j$ are $R$-module homomorphism.
Moreover, suppose that each row represents an exact sequence and that
the $h_j$ for $j\neq 3$ are isomorphism.  Then $h_3$ is also an
isomorphism.
\end{theorem}

\begin{proof}
\stage{i} $h_3$ is one-to-one.

Suppose that $h_3(a_3) = 0$.  Using the diagram above, we get that
$g_3(h(a_3)) = 0 = h_4(f_3(a_3))$.  Thus
$f_3(a_3) = 0$ because $h_4$ is an isomorphism.
Therefore, there exists $a_2 \in A_2$ such that $f_2(a_2) = a_3$ because
$\IMG(f_2) = \KE(f_3)$.  Using the diagram above, we get that
$h_3(a_3) = 0 = h_3(f_2(a_2)) = g_2(h_2(a_2))$.  Since
$\IMG(g_1) = \KE(g_2)$, there exists $b_1 \in B_1$ such that
$g_1(b_1) = h_2(a_2)$.  Since $h_1$ is an isomorphism, there exists
$a_1 \in A_1$ such that $h_1(a_1)=b_1$.  Hence, using the diagram
above, we get that $h_2(a_2) = g_1(h_1(a_1)) = h_2(f_1(a_1))$.
Since $h_2$ is an isomorphism, we get that $a_2 = f_1(a_1)$.
Finally, $a_3 = f_2(a_2) = f_2(f_1(a_1)) = 0$ because
$\IMG(f_1) = \KE(f_2)$.  This proves that $h_3$ is one-to-one.

\stage{ii} $h_3$ is onto.

Given $b_3 \in B_3$, there exists $a_4 \in A_4$ such that
$h_4(a_4) = g_3(b_3)$ because $h_4$ is an isomorphism.
Using the diagram above and $\IMG(g_3) = \KE(g_4)$, we then get that
$h_5(f_4(a_4)) = g_4(h_4(a_4)) = g_4(g_3(b_3)) = 0$.
Since $h_5$ is an isomorphism, we get that $f_4(a_4) = 0$.
Since $\IMG(f_3) = \KE(f_4)$, there exists $a_3 \in A_3$ such that
$f_3(a_3) = a_4$.  Hence, using the diagram above, we get that
$g_3(h_3(a_3)) = h_4(f_3(a_3)) = h_4(a_4) = g_3(b_3)$.  Thus
$h_3(a_3) - b_3 \in \KE(g_3)$.  Since $\IMG(g_2) = \KE(g_2)$, there
exists $b_2 \in B_2$ such that $g_2(b_2) = h_3(a_3) - b_3$.
Since $h_2$ is an isomorphism, there
exists $a_2 \in A_2$ such that $h_2(a_2) = b_2$.  Using the diagram
above for the last time, we get that
$h_3(f_2(a_2)) = g_2(h_2(a_2)) = g_2(b_2) = h_3(a_3) - b_3$.  Therefore
$h_3( a_3 - f_2(a_2)) = b_3$.  Since $b_3$ is arbitrary, we have shown
that $\IMG(h_3) = B_3$.
\end{proof}

\begin{defn}
Suppose that $X_1$ and $X_2$ are two topological spaces.  The triad
$(X_1 \cup X_2,X_1,X_2)$ is said to be {\bfseries Exact}\index{Exact Triad} if
$X_1 \setminus X_2$ can be excised from $(X_1 \cup X_2,X_1)$
and $X_2 \setminus X_1$ can be excised from $(X_1 \cup X_2,X_2)$.
\end{defn}

In other words, The triad $(X_1 \cup X_2,X_1,X_2)$ is exact if the
inclusion maps $\iota_2 : (X_2, X_1 \cap X_2) \to (X_1 \cup X_2, X_1)$
and $\iota_1 : (X_1, X_1 \cap X_2) \to (X_1 \cup X_2, X_2)$ are excisions.
For instance, using the notation introduced in
Example~\ref{eggSpnSmn}, we have that
$\displaystyle (S^n,S_n^+,S_n^-)$ is an exact triad.

\begin{lemma} \label{lemEquDofET}
Suppose that $X_1$ and $X_2$ are two topological spaces.
Let
\[
\tilde{H}_k = \KE\big(\partial_k\big|_{S_k(X_1;R) + S_k(X_2;R)}\big)
\big/
\KE\big(\partial_{k+1}\big|_{S_{k+1}(X_1;R) + S_{k+1}(X_2;R)}\big) \ .
\]
Then $(X_1 \cup X_2,X_1,X_2)$ is an exact triad if and only if
$h_k:\tilde{H}_k \to H_k(X_1 \cup X_2;R)$ defined by
$h_k([c_1+c_2]) = [c_1 +c_2]_X$ for all $c_1 \in S_k(X_1;R)$ and
$c_2 \in S_k(X_2;R)$ is an isomorphism.
\end{lemma}

\begin{proof}
Let $X = X_1 \cup X_2$ and $B = X_1 \cap X_2$.

We first note that $h_k$ is well defined.  Suppose that
$[c_1+c_2]=[\tilde{c}_1+\tilde{c}_2]$ with $c_1, \tilde{c}_1 \in S_k(X_1;R)$ and
$c_2, \tilde{c}_2 \in S_k(X_2;R)$.  Then
$(c_1 + c_2) - (\tilde{c}_1 + \tilde{c}_2) = \partial_{k+1}(b_1 + b_2)$ 
for some $b_1 \in S_{k+1}(X_1;R)$ and $b_2 \in S_{k+1}(X_2;R)$.  Since
$b_1 + b_2 \in S_{k+1}(X;R)$, we get that
$[c_1+c_2]_X=[\tilde{c}_1+\tilde{c}_2]_X$.

\stage{i} We prove that the sequence
\begin{equation} \label{triadEqEq1}
\xymatrix{
0 \ar[r] & S_k(X_2;R) \ar[r]^-{\iota}
& S_k(X_1;R) + S_k(X_2;R) \ar[r]^-{\pi_1}
& S_k(X_1;R)/S_k(B;R) \ar[r] & 0
}
\end{equation}
is exact where $\iota(c) = c$ for all $c \in S_k(X_2;R)$ and
$\pi_1(c_1 + c_2) = \relC[X_1,B]{c_1}$ for all
$c_1 \in S_k(X_1;R)$ and $c_2 \in S_k(X_2;R)$.

We first have to prove that $\pi_1$ is well defined.  Suppose
$c_1, \tilde{c}_1\in S_k(X_1;R)$ and $c_2, \tilde{c}_2 \in S_k(X_2;R)$
are such that $c_1 + c_2 = \tilde{c}_1 + \tilde{c}_2$.  Then
$c_1 - \tilde{c}_1 = \tilde{c}_2 - c_2$ where the right side is in
$S_k(X_1;R)$ and the left side in $S_k(X_2;R)$.  Hence
$c_1 - \tilde{c}_1 \in S_k(B;R)$.
Thus $\relC[X_1.B]{c_1} = \relC[X_1.B]{\tilde{c}_1}$.

The function $\iota$ is clearly one-to-one and the function $\pi_1$ is
clearly onto.  We also have that $\IMG(\iota) = S_q(X_2;R) = \KE(\pi_1)$.
Thus the sequence (\ref{triadEqEq1}) is exact.

\stage{ii} Using the Zig-Zag lemma, Lemma~\ref{lemZigZag}, we get from
(\ref{triadEqEq1}) the long exact sequence
\begin{equation} \label{triadEqEq2}
\xymatrix@C+2ex{
\ar[r]^-{\tilde{\partial}_{\,k+1}} & H_k(X_2;R) \ar[r]^-{H_k(\iota)}
& \tilde{H}_k \ar[r]^-{H_k(\pi_1)}
& H_k(X_1,B;R) \ar[r]^-{\tilde{\partial}_k}
& H_{k-1}(X_2;R) \ar[r]^-{H_{k-1}(\iota)} &
}
\end{equation}

Moreover, using again the Zig-Zag lemma with the exact sequence
\[
\xymatrix{
0 \ar[r] & S_k(X_2;R) \ar[r]^-{\iota}
& S_k(X;R) \ar[r]^-{\pi_2} & S_k(X;R)/S_k(X_2;R) \ar[r] & 0
}
\]
we get the long exact sequence
\begin{equation} \label{triadEqEq3}
\xymatrix@C+1em{
\ar[r]^-{\tilde{\partial}_{\,k+1}} & H_k(X_2;R) \ar[r]^-{H_k(\iota)}
& H_k(X;R) \ar[r]^-{H_k(\pi_2)}
& H_k(X,X_2;R) \ar[r]^-{\tilde{\partial}_k} & \\
& H_{k-1}(X_2;R) \ar[r]^-{H_{k-1}(\iota)} & & &
}
\end{equation}
Note that we could also have referred to
Proposition~\ref{propConnectingH} to get (\ref{triadEqEq3}).
In fact, this is an equivalent proof of Proposition~\ref{propConnectingH}.

\stage{iii}
Suppose that $(X,X_1,X_2)$ is an exact triad.  Thus
$H_k(\iota_1):H_k(X_1,B;R) \to H_k(X,X_2;R)$ is an isomorphism.
We join the long exact sequences in (\ref{triadEqEq2}) and
(\ref{triadEqEq3}) to get the following commutative diagram.
\[
\xymatrix@C+2ex{
\ar[r]^-{H_{k+1}(\pi_1)}
& H_{k+1}(X_1,B;R) \ar[r]^-{\tilde{\partial}_{\,k+1}}\ar[d]^{H_{k+1}(\iota_1)}
& H_k(X_2;R) \ar[r]^-{H_k(\iota)}\ar[d]^{\Id}
& \tilde{H}_k \ar[r]^-{H_k(\pi_1)}\ar[d]^{h_k}
& H_k(X_1,B;R) \ar[r]^-{\tilde{\partial}_k}\ar[d]^{H_k(\iota_1)} & \\
\ar[r]^-{H_{k+1}(\pi_2)}
&H_{k+1}(X,X_2;R) \ar[r]^-{\tilde{\partial}_{\,k+1}}
& H_k(X_2;R) \ar[r]^-{H_k(\iota)}
& H_k(X;R) \ar[r]^-{H_k(\pi_2)}
& H_k(X,X_2;R) \ar[r]^-{\tilde{\partial}_k} & \\
& H_{k-1}(X_2;R) \ar[r]^-{H_{k-1}(\iota)}\ar[d]^{\Id} & & & & \\
& H_{k-1}(X_2;R) \ar[r]^-{H_{k-1}(\iota)} & & & &
}
\]
where $\iota_1 : (X_1,B) \to (X,X_2)$ is the inclusion map.
The only condition that may require a little bit attention in order
to prove that we really have a commutative diagram is
$H_k(\iota_1) \circ H_k(\pi_1) = H_k(\pi_2) \circ h_k$.
Given $[c_1 + c_2] \in \tilde{H}_k$, we have that
$H_k(\pi_2)(h_k([c_1+c_2])) = H_k(\pi_2)([c_1+c_2]_X) 
= [c_1 + c_2]_{X,X_2} = [c_1]_{X,X_2}$ because $c_2 \in S_k(X_2;R)$.
Moreover, $H_k(\iota_1)(H_k(\pi_1)([c_1 + c_2]) = H_k(\iota_1)([c_1]_{X_1,B})
= [c_1]_{X,X_2}$.

Since $H_k(\iota_1)$, $H_{k+1}(\iota_1)$ and all the identity maps are
isomorphism, we get from the Five Lemma that $h_k$ is an isomorphism.

\stage{iv} Suppose $h_k:\tilde{H}_k \to H_k(X_1 \cup X_2;R)$ defined by
$h_k([c_1+c_2]) = [c_1 +c_2]_X$ for all $c_1 \in S_k(X_1;R)$ and
$c_2 \in S_k(X_2;R)$ is an isomorphism.
We join the long exact sequences (\ref{triadEqEq2}) and
(\ref{triadEqEq3}) to get the following commutative diagram.
\[
\xymatrix@C+2ex{
\ar[r]^-{\tilde{\partial}_{\,k+1}}
& H_k(X_2;R) \ar[r]^-{H_k(\iota)}\ar[d]^{\Id}
& \tilde{H}_k \ar[r]^-{H_k(\pi_1)}\ar[d]^{h_k}
& H_k(X_1,B;R) \ar[r]^-{\tilde{\partial}_k}\ar[d]^{H_k(\iota_1)}
& H_{k-1}(X_2;R) \ar[r]^-{H_{k-1}(\iota)}\ar[d]^{\Id} & \\
\ar[r]^-{\tilde{\partial}_{\,k+1}}
& H_k(X_2;R) \ar[r]^-{H_k(\iota)}
& H_k(X;R) \ar[r]^-{H_k(\pi_2)}
& H_k(X,X_2;R) \ar[r]^-{\tilde{\partial}_k}
& H_{k-1}(X_2;R) \ar[r]^-{H_{k-1}(\iota)} & \\
& \tilde{H}_{k-1} \ar[r]^-{H_{k-1}(\pi_1)}\ar[d]^{h_{k-1}} & & & & \\
& \tilde{H}_{k-1} \ar[r]^-{H_{k-1}(\pi_1)} & & & & 
}
\]

Since $h_k$, $h_{k-1}$ and all the identity mappings are
isomorphism, we get from the Five Lemma that $H_k(\iota_1)$ is an isomorphism.
\end{proof}

\begin{prop}[Mayer-Vietoris]  \label{propMayVistHom}
Suppose that $(X_1 \cup X_2,X_1,X_2)$ is an exact triad.  Then
the sequence
\[
\xymatrix{
\ar[r]^-{H_{k+1}} & H_k(X_1 \cap X_2;R) \ar[r]^-{F_k}
& H_k(X_1;R) \oplus H_k(X_2;R) \ar[r]^-{G_k} & \\
& H_k(X_1\cup X_2;R) \ar[r]^-{H_k} & H_{k-1}(X_1\cap X_2;R) \ar[r]^-{F_{k-1}} & 
}
\]
is a long exact sequence where
$F_k([c]_{X_1\cap X_2}) = ([c]_{X_2},-[c]_{X_1})$,
$G_k([c_2]_{X_2},[c_1]_{X_1}) = [c_1]_{X_1\cup X_2} + [c_2]_{X_1 \cup X_2}$ and
$H_k:H_k(X_1 \cup X_2;R) \to H_k(X_1 \cap X_2;R)$ is defined in the
proof below.  This sequence is called a
{\bfseries Mayer-Vietoris sequence}\index{Mayer-Vietoris Sequence}
\end{prop}

\begin{proof}
Let $B= X_1\cap X_2$ and $X = X_1 \cup X_2$.

We consider the following three chain complexes.
$\displaystyle \A = \{(S_k(B;R),\partial_k)\}_{k\in \ZZ}$,\\
$\displaystyle \BB = \{(S_k(X_1;R) \oplus S_k(X_2;R),\partial_k)\}_{k\in \ZZ}$
and
$\displaystyle \C = \{(S_k(X_1;R) +S_k(X_2;R),\partial_k\}_{k\in \ZZ}$.
Let $\F = \{f_k\}_{k\in\ZZ}$ be the chain map between $\A$ and $\BB$
defined by $f_k(c) = (c,-c)$ for all $c \in S_k(B;R)$, and
$\GG = \{g_k\}_{k\in\ZZ}$ is a chain map between $\BB$ and $\C$
defined by $g_k(c_1,c_2) = c_1+c_2$ for all $c_1 \in S_k(X_1;R)$
and $c_2 \in S_k(X_2;R)$.
Note that $\partial_k$ is defined on $S_k(X_1;R) \oplus S_k(X_2;R)$
by $\partial_k(c_1,c_2) = (\partial_k(c_1),\partial_k(c_2))$ for all
for all $c_1 \in S_k(X_1;R)$ and $c_2 \in S_k(X_2;R)$.

The sequence
\[
\xymatrix{
0 \ar[r] & S_k(B;R) \ar[r]^-{f_k} & S_k(X_1;R) \oplus S_k(X_2;R) \ar[r]^-{g_k}
& S_k(X_1;R) + S_k(X_2;R) \ar[r] & 0
}
\]
is an exact sequence for all $k$ because $f_k$ is one-to-one, $g_k$ is
onto and $\IMG(f_k) = \KE(g_k) = \{ (c,-c) : c \in S_k(B;R) \}$.
It follows from the Zig-Zag lemma, Lemma~\ref{lemZigZag}, that
\[
\xymatrix{
\ar[r]^-{H_{k+1}} & H_k(\A) \ar[r]^-{F_k}
& H_k(\BB) \ar[r]^-{G_k}
& H_k(\C) \ar[r]^-{H_k} & H_{k-1}(\A) \ar[r]^-{F_{k-1}} & 
}
\]
is a long exact sequence where $H_k(\A) = H_k(B;R)$,
$H_k(\C) \cong H_k(X;R)$ according to the previous lemma, and
\begin{align*}
H_k(\BB)
&= \KE\big(\partial_k\big|_{S_k(X_1;R) \oplus S_k(X_2;R)}\big) \big/
\IMG\big(\partial_{k+1}\big|_{S_{k+1}(X_1;R) \oplus S_{k+1}(X_2;R)}\big) \\
& \cong \KE\big(\partial_k\big|_{S_k(X_1;R)}\big) \big/
\IMG\big(\partial_{k+1}\big|_{S_{k+1}(X_1;R)}\big)
\oplus
\KE\big(\partial_k\big|_{S_k(X_2;R)}\big) \big/
\IMG\big(\partial_{k+1}\big|_{S_{k+1}(X_2;R)}\big) \\
&\cong H_k(X_1;R) \oplus H_k(X_2;R)
\end{align*}
because $\KE\big(\partial_k\big|_{S_k(X_1;R) \oplus S_k(X_2;R)}\big)
= \KE\big(\partial_k\big|_{S_k(X_1;R)}\big) \oplus
\KE\big(\partial_k\big|_{S_k(X_2;R)}\big)$ and\\
$\IMG\big(\partial_{k+1}\big|_{S_{k+1}(X_1;R) \oplus S_{k+1}(X_2;R)}\big)
= \IMG\big(\partial_{k+1}\big|_{S_{k+1}(X_1;R)}\big) \oplus
\IMG\big(\partial_{k+1}\big|_{S_{k+1}(X_2;R)}\big)$.
Note that $F_k = H_k(f_k)$ and $G_k = H_k(g_k)$ in the notation of the
Zig-Zag lemma.

It follows from the Zig-Zag lemma that
$H_k:H_k(X_1 \cup X_2;R) \to H_{k-1}(X_1 \cap X_2;R)$ is defined by
$H_k([c]_{X_1 \cup X_2}) = [a]_{X_1 \cap X_2}$ where
$a \in Z_{k-1}(X_1 \cap X_2;R)$ is such that
$f_{k-1}(a) = (a,-a) = (\partial_k (b_1), \partial_k(b_2))$
for $b_1 \in Z_k(X_1;R)$ and $b_2 \in Z_k(X_2;R)$ with
$g_k(b_1,b_2) = b_1 + b_2 = c$.
\end{proof}

\begin{egg}
The torus $\displaystyle \torus{2} \subset \RR^3$ is the union of two annulus,
$X_1$ and $X_2$, as in Figure~\ref{Torus}.  We have that
$X_1 \cap X_2$ is the union of two disjoint circles.

We prove that $(\torus{2},X_1,X_2)$ is an exact triad.  We first prove that
$\displaystyle U = X_2^\circ = X_2 \setminus X_1$ can be excised from
$(\torus{2},X_2)$.  We cannot use Theorem~\ref{thmExcis} to justify this
operation because $\displaystyle \overline{U} \not\subset X_2^\circ$.
To justify the excision of $U$ from $(\torus{2},X_2)$, we have to use
Proposition~\ref{propVUAX} with $V$ as in Figure~\ref{Torus}.  We have
that $\displaystyle \overline{V} \subset X_2^\circ$.  Therefore, 
$V$ can be excised from $(\torus{2},X_2)$ according to Theorem~\ref{thmExcis}.
Moreover, we have that
$(\torus{2}\setminus U,X_2\setminus U) = (X_1,\partial X_1)$
is a deformation retract of $(\torus{2}\setminus V, X_2 \setminus V)$.  It then
follows from Proposition~\ref{propVUAX} that $\displaystyle X_2^\circ$
can be excised from $(\torus{2},X_2)$.  Similarly,
$\displaystyle X_1^\circ = X_1 \setminus X_2$
can be excised from $(\torus{2},X_1)$.
This proves that $(\torus{2},X_1,X_2)$ is a triad.

\pdfF{alg_top/torus}{Torus}{The torus is the union of two annulus,
$X_1$ and $X_2$, whose intersection is the union of two transversal
sections of the torus homeomorphic to a circle.}{Torus}

We get the following Mayer-Vietoris sequence from the Mayer-Vietoris
theorem.
\[
\xymatrix{
\ar[r]^-{H_{k+1}} & H_k(\partial X_1;R) \ar[r]^-{F_k}
& H_k(X_2;R) \oplus H_k(X_1;R) \ar[r]^-{G_k}
& H_k(\torus{2};R) \ar[r]^-{H_k} & H_{k-1}(\partial X_1;R) \ar[r]^-{F_{k-1}}  &
}
\]
In particular, for $k=1$ and using reduced homology, we get the exact
sequence
\[
\xymatrix{
0 \ar[r]^-{G_2} & H_2(\torus{2};R)  \ar[r]^-{H_2}
& H_1(\partial X_1;R) \ar[r]^-{F_1}
& H_1(X_2;R) \oplus H_1(X_1;R) \ar[r]^-{G_1} & \\
& H_1(\torus{2};R) \ar[r]^-{H_1} &
H_0^\sharp(\partial X_1;R) \ar[r]^-{F_0}  & 0 &
}
\]
because $\displaystyle H_0^\sharp(X_2;R) \cong H_0^\sharp(S^1;R) = 0$,
$\displaystyle H_0^\sharp(X_1;R) \cong H_0^\sharp(S^1;R) = 0$,
$\displaystyle H_2(X_2;R) \cong H_2(S^1;R) = 0$ and
$\displaystyle H_2(X_1;R) \cong H_2(S^1;R) = 0$ according to
the information collected at the end of Example~\ref{eggHkSq}.
We also have that $\displaystyle H_1(X_2;R) \cong H_1(S^1;R) \cong R$ and
$\displaystyle H_1(X_1;R) \cong H_1(S^1;R) \cong R$.  Moreover, we get from
Proposition~\ref{propHkeHkJ} that
$\displaystyle H_1(\partial X_1;R) \cong H_1(S^1;R) \oplus H_1(S^1;R)
\cong R \oplus R$.  Finally, we have that 
$\displaystyle H_0^\sharp(\partial X_1;R) \cong R$ by direct computation.

In summary, the sequence above is equivalent to
\begin{equation} \label{sequTorus}
\xymatrix{
0 \ar[r]^-{G_2} & H_2(\torus{2};R)  \ar[r]^-{H_2}
& R \oplus R \ar[r]^-{F_1} & R \oplus R \ar[r]^-{G_1}
& H_1(\torus{2};R) \ar[r]^-{H_1} & R \ar[r]^-{F_0}  & 0
}
\end{equation}
If we determine the kernel and range of $F_1$, then we will be able to
determine the singular $\displaystyle 1^{st}$ and $\displaystyle 2^{nd}$
homology module of the torus $\torus{2}$.  Recall that
$F_1 = (H_1(\iota) \times H_1(\iota)) \circ \eta_1$ where
the first inclusion map is from $\partial X_1$ into $X_2$, the
second inclusion map is from $\partial X_1$ into $X_1$, and
$\eta_1:H_1(\partial X_1;R) \to H_1(\partial X_1;R) \times
H_1(\partial X_1;R)$ is defined by
$\eta_1([c]_{\partial X_1}) = \big([c]_{\partial X_1},[c]_{\partial X_1}\big)$
for all $[c]_{\partial X_1} \in H_1(\partial X_1;R)$.
Since $\displaystyle \partial X_1 = S^1$, we have from
Example~\ref{eggHkSq} that $H_1(\partial X_1;R) \cong R$.
Thu $\IMG(F_1) \cong R$.  This implies that $\KE(F_1) \cong R$.

We may now combine this information with the sequence in
(\ref{sequTorus}).  Since $H_2$ is one-to-one and
$\IMG(H_2) = \KE(F_1) \cong \RR$, we get that $H_2(\torus{2};R) \cong R$. 
Since $\KE(G_1) = \IMG(F_1) \cong R$, we have that
$\IMG(G_1) \cong R$.  Since $\KE(H_1) = \IMG(G_1) \cong R$ and
$\IMG(H_1) \cong R$, we get that $H_1(\torus{2};R) \cong R \oplus R$. 
\end{egg}

\begin{egg}
To give another example of how Mayer-Vietoris     \label{eggFlower}
sequences can be used,
suppose that $S$ and $T$ are two topological spaces, and that
$s_0 \in S$ and $t_0 \in T$.  Let $X_1 = \{(s,t_0) : s \in S\}$ and
$X_2 = \{(s_0,t) : t \in T\}$.  They are both subsets of $S \times T$.
If $S_0$ is an open neighbourhood of $s_0$ such that
$S \setminus S_0$ is a deformation retract of
$S \setminus \{s_0\}$ and $T_0$ is an open neighbourhood of $t_0$
such that $T \setminus T_0$ is a deformation retract of
$T \setminus \{t_0\}$, then it follows from Proposition~\ref{propVUAX}
that $(X_1 \cup X_2, X_1,X_2)$ is an exact triad.
To be more precise, it follows from Proposition~\ref{propVUAX} with
$A = X_2$, $X = X_1 \cup X_2$,
$U = X_2 \setminus \{(s_0,t_0)\} = X \setminus X_1$ and
$V= X_2 \setminus (\{s_0\}\times T_0)$ that the inclusion map
$\iota_1: (X\setminus U, A \setminus U) = (X_1, X_1 \cap X_2) \to
(X,A) = (X_1 \cup X_2,X_2)$ is an excision.  Note that
$(X \setminus U, A \setminus U) = (X_1, \{(s_0,t_0)\})$ is a
deformation retract of $(X \setminus V,A \setminus V) =
(X_1 \cup (\{s_0\} \times T_0), \{s_0\} \times T_0)$ because
$T \setminus T_0$ is a deformation retract of
$T \setminus \{t_0\}$.  Similarly, we have that
the inclusion map
$\iota_2: (X_2, X_1 \cap X_2) \to (X_1 \cup X_2,X_1)$ is an excision. 

Since $(X_1 \cup X_2, X_1,X_2)$ is an exact triad, we get the
Mayer-Vietoris sequence
\[
\xymatrix{
\ar[r]^-{H_{k+1}} & H_k(\{(s_0,t_0)\};R) \ar[r]^-{F_k}
& H_k(X_2;R) \oplus H_k(X_1;R) \ar[r]^-{G_k} & \\
& H_k(X_1 \cup X_2;R) \ar[r]^-{H_k}
& H_{k-1}(\{(s_0,t_0)\};R) \ar[r]^-{F_{k-1}}  &
}
\]
because $X_1 \cap X_2 = \{(s_0,t_0)\}$.
All the previous theory is also valid for reduced homology.
Since $\displaystyle H_k^\sharp(\{(s_0,t_0)\};R) = 0$ for
all $k \geq 0$, we get the exact sequence
\[
\xymatrix{
0 \ar[r]^-{F_k} & H_k^\sharp(X_2;R) \oplus H_k^\sharp(X_1;R) \ar[r]^-{G_k}
& H_k^\sharp(X_1 \cup X_2;R) \ar[r]^-{H_k} & 0
}
\]
Thus $\displaystyle H_k^\sharp(X_1 \cup X_2;R) \cong
H_k^\sharp(X_2;R) \oplus H_k^\sharp(X_1;R)$.

Suppose that $S$ and $T$ are two subsets of $\displaystyle \RR^n$ that
are homeomorphic to $\displaystyle S^1$ and intersect at a single
point $\VEC{x}_0$ that we may assume to be associated to $(s_0,t_0)$.
The set $G_2 = X_1 \cup X_2$ is represented in Figure~\ref{Flower}.
We get from the previous discussion that
$\displaystyle H_k^\sharp(G_2;R) \cong
H_k^\sharp(S^1;R) \oplus H_k^\sharp(S^1;R)$.

We can generalize this construction by induction.  Suppose that
$G_{q+1}$ is the union of $G_q$ with another ``leave'' $L$ which is
homeomorphic to $\displaystyle S^1$ and such that
$L \cap G_q = \{\VEC{x}_0\}$ as in Figure~\ref{Flower} for $q=4$.
We can show that $(G_{q+1},G_q,L)$ is an exact triad and use the 
Mayer-Vietoris sequence to show that
$\displaystyle H_k^\sharp(G_{k+1};R) = H_k^\sharp(S^1;R)
\oplus H_k^\sharp(G_q;R)$.

Using the information collected at the end of Example~\ref{eggHkSq},
we conclude that
\[
H_k(G_q) \cong 
\begin{cases}
0 & \quad \text{if} \ k > 1 \\
\displaystyle  R^q & \quad \text{if} \ k = 1 
\end{cases}
\]
\end{egg}

\pdfF{alg_top/flower}{Flowers}{The topological spaces $G_2$ and $G_4$
referred to in Example~\ref{eggFlower}}{Flower}

\subsection{Further Studies}

There are a lot more results about singular homology that could be
presented in a first course.  We list a few of them below.

\subsubsection{Vector Fields}

Singular homology theory can be used to give another proof of
Proposition~\ref{propNNVFodd} stating that there is not a
nowhere null vector field on $\displaystyle S^{k-1}$ if $k$ is odd.

\subsubsection{Jordan-Brouwer Separation Theorem}

The ``obvious'' but non-trivial Jordan-Brouwer separation theorem can be
proved using singular homology.  Let
$\displaystyle s_{n-1} \subset S^n$ be a set homeomorphic to
$\displaystyle S^{n-1}$.  The Jordan-Brouwer separation theorem states that
$\displaystyle S^n \setminus s_{n-1}$ is the union of two components
having $s_{n-1}$ as boundary.  In particular, suppose that
$\displaystyle \sigma:[0,1] \to \RR^2$ is a closed path such that
$\sigma(t_1) \neq \sigma(t_2)$ for $t_1 \neq t_2$.  Since the sphere
$\displaystyle S^2$ is homeomorphic to the extended space
$\displaystyle \RR^2$ through stereographic projection
(Figure~\ref{Stereo}), we get that $\sigma([0,1])$ splits
$\displaystyle \RR^2$ into two components whose boundary is
$\sigma([0,1])$ because $\sigma([0,1])$ lifts to a closed path $s_1$ on
$\displaystyle S^2$.  In fact, the result is also true 
for a curve $\displaystyle \sigma:]0,1[ \to \RR^2$ that does not
intersect itself and such that $\sigma(t) \to \infty$ as $t\to 0$ and
$t\to 1$ because this curve can also be lifted to a closed path on
$\displaystyle S^2$.

\pdfF{alg_top/stereo}{Jordan-Brouwer separation theorem}{
The stereographic projection of $\displaystyle S^2$ on
$\displaystyle \RR^2$.}{Stereo}

\section{Singular Cohomology} \label{sectSingCohom}

\subsection{Cohomotopy Modules}

\begin{defn}
Let $X$ be a topological space and $R$ be an integral domain.  The dual
space $\displaystyle S_k(X;R)^\ast$ of $S_k(X;R)$ is denoted
$\displaystyle S^k(X;R)$.  Namely, $\displaystyle S^k(X;R)$ is the
$R$-module of all $R$-linear functionals on $S_k(X;R)$.  The elements
of $\displaystyle S^k(X,R)$ are called
{\bfseries cochaines}\index{Cochaines}. 
\end{defn}

\begin{defn}
Let $X$ be a topological space and $R$ be an integral domain.  The
adjoint of the boundary map $\partial_k: S_k(X;R) \to S_{k-1}(X;R)$ is
the map $\displaystyle \dfC_{k-1}: S^{k-1}(X;R) \to S^k(X;R)$ defined
by
\begin{equation} \label{singCoOpEq1}
\big(\dfC_{k-1}(\phi)\big)(c) = \phi\big(\partial_k(c)\big)
\end{equation}
for all $c \in S_k(X;R)$ and $\displaystyle \phi \in S^{k-1}(X;R)$.
The map $\displaystyle \dfC_k$ is called
the {\bfseries coboundary operator}\index{Coboundary Operator}.
\end{defn}

It is clear that $\displaystyle \dfC_k$ is well and uniquely
defined by (\ref{singCoOpEq1}).  Since
$\partial_{k+1} \circ \partial_{k+2} = 0$, we have that
$\displaystyle \dfC_{k+1} \circ \dfC_k = 0$.

Let $X$ and $Y$ be two topological spaces and $R$ be an integral domain.
Given $f:X\to Y$, we define an homomorphism.
$\displaystyle S^k(f):S^k(Y;R) \to S^k(X;R)$ by
$\displaystyle \big(S^k(f)(\phi)\big)(c) = \phi\big(S(f)(c)\big)$ for all
$\displaystyle c \in S^k(X;R)$ and $\displaystyle \phi \in S^k(Y;R)$.
It is easy to verify that
$\displaystyle \dfC_k \circ S^k(f) = S^{k+1}(f) \circ \dfC_k$.

\begin{defn}
Let $X$ be a topological space and $R$ be an integral domain.  The
$R$-modules of {\bfseries cocycles}\index{Cocycle} and
{\bfseries coboundaries}\index{Coboundary} are respectively defined by
$\displaystyle Z^k(X;R) = \{ \phi \in S^k(X;R) : \dfC_k(\phi) = 0 \}$
and $\displaystyle B^k(X;\RR) = \{ \dfC_{k-1}(\phi) : \phi \in
S^{k-1}(X,R)\}$.
The {\bfseries $\displaystyle \mathbf{k^{th}}$ cohomology
module}\index{Cohomology Module!$k^{th}$ Cohomology Module}
of $X$ is the quotient space $\displaystyle H^k(X;R) = Z^k(X;R) / B^k(X;R)$.
The equivalence class of $\displaystyle H^k(X;R)$ associated to
$\displaystyle \phi \in Z^k(X;R)$ is denoted $[\phi]_X$.
\end{defn}

\begin{rmk}
Let $X$ be a topological space and $R$ be an integral domain.
Consider $\displaystyle [\phi] \in H^k(X;R)$.  If
$\displaystyle \tilde{\phi} \in Z^k(X;R)$ is
another member of the equivalence class $\displaystyle [\phi] \in H^k(X;R)$,
then there exists $\displaystyle \psi \in S^{k-1}(X;R)$ such that
$\displaystyle \phi = \tilde{\phi} + \dfC_{k-1}(\psi)$.
Despite the fact that $\phi$ and $\tilde{\phi}$ may not be equal on
$S_k(X;R)$, we have that
$\displaystyle \phi(c) = \tilde{\phi}(c)
+ \big(\dfC_{k-1}(\psi)\big)(c)
= \tilde{\phi}(c) + \psi(\partial_k(c)) = \tilde{\phi}(c)$ for all
$c \in Z_k(X;R)$.
\end{rmk}

Many of the results in singular homology have equivalent results in 
singular cohomology.  For instance Proposition~\ref{propHkeHkJ}
becomes the following proposition.  The proof of this proposition is
almost identical to the proof of Proposition~\ref{propHkeHkJ}.

\begin{prop} \label{propHkeHkJCoho}
If the topological space $X$ is the union of distinct (connected)
components $X_j$ for $j \in J$, then
$\displaystyle H^k(X;R) = \bigoplus_{j\in J}H^k(X_j;R)$.
\end{prop}

We have a result similar to Theorem~\ref{thmHupeHdk} for simplicial
cohomology.

\begin{theorem} \label{thmHupeHdkS}
Let $X$ be a topological space and $R$ be an integral domain.
Then $\displaystyle H^k(X;R)$ is isomorphic to
$\displaystyle (H_k(X;R))^\ast$.
\end{theorem}

\begin{proof}
The proof is basically identical to the proof of Theorem~\ref{thmHupeHdk}.
We repeat it here for the sake of those who skipped the sections on
simplicial homology and cohomology.
 
The map $\displaystyle h:H^k(X;R) \to (H_k(X;R))^\ast$ defined by
$h([\phi]_X) = \phi$ for $\displaystyle \phi \in Z^k(X;R)$ is well
defined because
\begin{align*}
\phi \in Z^k(X;R) & \Rightarrow \dfC_k(\phi) = 0
\Rightarrow \phi(\partial_{k+1}(c)) = (\dfC_k \phi)(c) =  0
\ \text{for all} \ c \in S_{k+1}(X;R) \\
&\Rightarrow \phi = 0 \ \text{on} \ B_k(X;R)
\Rightarrow \phi \in (H_k(X;R))^\ast \ .
\end{align*}

We have that $h$ is one-to-one.  Suppose that $[\phi_1]_X = [\phi_2]_X$.
Then $\displaystyle \phi_1 - \phi_2 = \psi \in B^k(X;R)$.  Hence
$h([\phi_1]_X) = h([\phi_2]_X) + h([\psi]_X) = h([\phi_2]_X)$
because
\begin{align*}
\psi \in B^k(X;R)
&\Rightarrow \psi = \dfC_{k-1} \eta \ \text{for some}
\ \eta \in S^{k-1}(X;R) \\
&\Rightarrow \psi(c) = (\dfC_{k-1} \eta)(c)
= \eta(\partial_k(c)) = 0 \ \text{for all} \ c \in Z_k(X;R) \\
&\Rightarrow \psi = 0 \ \text{on}\ Z_k(X;R)
\Rightarrow h([\psi]_X) = 0 \ .
\end{align*}

We have that $h$ is onto.  Given $\displaystyle \phi \in (H_k(X;R))^\ast$,
then $\displaystyle [\phi]_X \in H^k(X;R)$ because
\begin{align*}
\phi(c) = 0 \ \text{for all}\ c \in B_k(X;R)
&\Rightarrow (\dfC_k \phi)(c) = \phi(\partial_{k+1}(c)) = 0
\ \text{for all} \ c \in S_{k+1}(X;R) \ \text{since} \\
&\qquad \ \partial_{k+1}(c) \in B_k(X;R) \ \text{for all}
\ c \in S_{k+1}(X;R) \\
&\Rightarrow \dfC_k \phi = 0 \Rightarrow \phi \in Z^k(K;R) \ .
\end{align*}
Thus $h([\phi]_K) = \phi$.

Hence, the map $h$ is an isomorphism between $\displaystyle H^k(X;R)$ and
$\displaystyle (H_k(X;R))^\ast$.
\end{proof}

\begin{cor} \label{corHupeHdkS}
Let $X$ be a topological space and $R$ be an integral domain.
Then $\displaystyle H^k(X;R)$ is isomorphic to
$\displaystyle H_k(X;R)$.
\end{cor}

\begin{proof}
This follows from the previous theorem and the fact that
$\displaystyle H_k(X;R) \cong (H_k(X;R))^\ast$.
\end{proof}

Let $X$ and $Y$ be two topological spaces and $R$ be an integral domain.
Given $f:X\to Y$, we define an homomorphism.
$\displaystyle H^k(f):H^k(Y;R) \to H^k(X;R)$ by
$\displaystyle H^k(f)([\phi]_Y) = [S^k(f)(\phi)]_X$ for all
$\displaystyle [\phi]_Y \in H^k(Y;R)$.  This map is well defined because
$\displaystyle S^k(f)$ maps $\displaystyle Z^k(Y;R)$ to
$\displaystyle Z^k(X;R)$ and
$\displaystyle B^k(Z;R)$ to $\displaystyle B^k(X;R)$.

\subsection{Cochain Complexes}

It should not be a surprise that there is a version of the Zig-Zag
lemma for cohomology.  But before stating it, we need a couple of
definitions.

\begin{defn}
Let $R$ be an integral domain.  A
{\bfseries cochain complex}\index{Cochain Complex} is a sequence
$\displaystyle \C = \{ (C^k,\dfC_k) \}_{k\in \ZZ}$ where the sets
$\displaystyle C^k$ are
$R$-modules and the maps $\dfC_k:C_k \to C_{k+1}$ are homomorphism
such that $\dfC_{k+1} \circ \dfC_k = 0$ for $k \in \ZZ$.
\end{defn}

\begin{defn}
A {\bfseries cochain map}\index{Cochain Map} between two chain complexes
given by $\displaystyle \C = \{(C^k,\dfC_k)\}_{k\in \ZZ}$ and
$\displaystyle \tilde{\C} = \{ (\tilde{C}^k,\tilde{\dfC}_k)\}_{k\in \ZZ}$
is a sequence $\F = \{f_k\}_{k\in \ZZ}$ where the maps
$\displaystyle f_k:C^k \to \tilde{C}^k$ are homomorphism such that
$\tilde{\dfC}_k \circ f_k = f_{k+1} \circ \dfC_k$ for $k \in \ZZ$.
\end{defn}

\begin{defn}
Consider a cochain complex given by
$\displaystyle \C = \{(C^k,\dfC_k)\}_{k\in \ZZ}$.
A {\bfseries $\mathbf{k}$-cocycle}\index{$k$-Cocycle}
in $\C$ is an element $\displaystyle c \in C^k$ such that $\dfC_k(c) = 0$.
The set of all $k$-cocycles in $\displaystyle C^k$ is a $R$-module
denoted $\displaystyle Z^k(\C)$.

A {\bfseries $\mathbf{k}$-coboundary}\index{$k$-Coboundary}
in $\C$ is an element $\displaystyle c \in C^k$ with the property that
there exists an element $\displaystyle \tilde{c} \in C^{k-1}$ such that
$\dfC_{k-1}(\tilde{c}) = c$.
The set of all $k$-coboundaries in $\displaystyle C^k$ is a $R$-module denoted
$\displaystyle B^k(\C)$.

The {\bfseries singular $\displaystyle \mathbf{k^{th}}$ cohomology
module}\index{Cohomology Module!Singular $k^{th}$ Cohomology Module}
of $\C$ is the $R$-module
defined as $\displaystyle H^k(\C) = Z^k(\C) / B^k(\C)$.  The
equivalence class of $\displaystyle H^k(\C)$ associated to
$\displaystyle c \in Z^k(\C)$ is denoted $[c]$.
\end{defn}

It follows from the definition that a cochain map
$\F = \{f_k\}_{k\in\ZZ}$ between two chain complexes
$\displaystyle \C = \{(C^k,\dfC_k)\}_{k\in \ZZ}$ and
$\displaystyle \tilde{\C} = \{ (\tilde{C}^k,\tilde{\dfC}_k)\}_{k\in \ZZ}$
sends $k$-cocycles in $\C$ to $k$-cocycles in $\tilde{\C}$, and
$k$-coboundaries in $\C$ to $k$-coboundaries in $\tilde{\C}$.
Hence, we may define $\displaystyle H^k(\F): H^k(\C) \to H^k(\tilde{\C})$ by
$\displaystyle H^k(\F)([c]) = [f_k(c)]$ for all
$\displaystyle c \in Z^k(\C)$.

We can know state the version of the Zig-Zag lemma for cohomology.

\begin{lemma}[Zig-Zag Lemma]  \label{lemZigZagCohom}
Suppose that $\F = \{f_k\}_{k\in\ZZ}$ is a    \index{Zig-Zag Lemma}
cochain map between the cochain
complexes $\displaystyle \A = \{(A^k,\dfC_k^{\A})\}_{k\in \ZZ}$ and
$\displaystyle \BB = \{(B^k,\dfC_k^{\BB})\}_{k\in \ZZ}$, and
$\GG = \{g_k\}_{k\in\ZZ}$ is a cochain map between the cochain complexes
$\displaystyle \BB = \{(B^k,\dfC_k^{\BB})\}_{k\in \ZZ}$ and
$\displaystyle \C = \{(C^k,\dfC_k^{\C})\}_{k\in \ZZ}$.
If
\[
\xymatrix{
0 \ar[r] & A^k \ar[r]^{f_k} & B^k \ar[r]^{g_k} & C^k \ar[r] & 0
}
\]
is an exact sequence for all $k$, then there exists a long exact
sequence
\[
\xymatrix@C+2ex{
& \ar[r]^-{\tilde{\dfC}_{\,k-1}} & H^k(\A) \ar[r]^{H^k(\F)}
& H^k(\BB) \ar[r]^{H^k(\GG} & H^k(\C) \ar[r]^-{\tilde{\dfC}_{\,k}}
& H^{k+1}(\A) \ar[r]^-{H^k(\F)} &
}
\]
where $\tilde{\dfC}_k$ is induced by $\displaystyle \dfC_k^{\BB}$.
\end{lemma}

The proof of this lemma is very similar (dual) to the proof of
Lemma~\ref{lemZigZag} and is left as exercise to the reader.

There is also a dual to Proposition~\ref{propMayVistHom} which could
be proved in a similar fashion or using a duality argument (see
\cite{LJM,MU}) \footnote{Duality and category theories are necessary
tools in the study of cohomology and would have simplified the proofs
of some results.  Since we are not planning to systematically study
cohomology theory, the inclusion of these subjects in these lecture
notes would have added too much extra material to these lecture
notes; in particular, to this chapter which is already longer than
initially intended.}.

\begin{prop}[Mayer-Vietoris]  \label{propMayVistCoHom}
Suppose that $(X_1 \cup X_2,X_1,X_2)$ is an exact triad.  Then 
the sequence
\[
\xymatrix{
\ar[r]^-{H_{k-1}^\ast} & H^k(X_1 \cup X_2;R) \ar[r]^-{F_k^\ast}
& H^k(X_2;R) \oplus H^k(X_1;R) \ar[r]^-{G_k^\ast} & \\
& H^k(X_1\cap X_2;R) \ar[r]^-{H_k^\ast}
& H^{k+1}(X_1\cup X_2;R) \ar[r]^-{F_{k+1}^\ast} & 
}
\]
is an exact sequence where
$\displaystyle F_k^\ast([\phi]_{X_1\cup X_2}) = ([\phi]_{X_2},[\phi]_{X_1})$,
$G_k([\phi_2]_{X_2},[\phi_1]_{X_1}) = [\phi_1]_{X_1\cap X_2}
- [\phi_2]_{X_1 \cap X_2}$ and
$\displaystyle H_k^\ast: H^k(X_1 \cap X_2;R) \to H^{k}(X_1 \cup X_2;R)$
is the dual of
$H_k:H_k(X_1 \cup X_2;R) \to H_k(X_1 \cap X_2;R)$; namely,\\
$\displaystyle H_k^\ast([\phi]_{X_1 \cap X_2})([c]_{X_1\cup X_2})
= [\phi]_{X_1 \cap X_2}\big(H_{k+1}([c]_{X_1\cup X_2})\big)$ for all
$[c]_{X_1 \cup X_2} \in H_{k+1}(X_1 \cup X_2;R)$ and
$\displaystyle [\phi]_{X_1\cap X_2} \in H^k(X_1 \cap X_2;R)$.
\end{prop}

\subsection{Relative Cohomology} \label{ssectRCsingT}

It is also possible to define relative $\displaystyle k^{th}$ cohomology
modules.

\begin{defn}
Let $A$ be a subset of a topological space $X$ and $R$ be an integral
domain.  The adjoint of the boundary operator 
$\overline{\partial}_k : S_k(X;R)/S_k(A;R) \to
S_{k-1}(X;R)/S_{k-1}(A;R)$ is the operator
$\overline{\dfC}_{k-1} : \big(S_{k-1}(X;R)/S_{k-1}(A;R)\big)^\ast
\to \big(S_k(X;R)/S_k(A;R)\big)^\ast$ defined
\begin{equation} \label{relsingCoOpEq1}
\big(\overline{\dfC}_{k-1}(\phi)\big)(\relC[X,A]{c})
= \phi\big(\overline{\partial}_k(\relC[X,A]{c})\big)
\end{equation}
for all $\relC[X,A]{c} \in S_k(X;R)/S_k(A;R)$ and
$\displaystyle \phi \in \big(S_{k-1}(X;R)/S_{k-1}(A;R)\big)^\ast$.
\end{defn}

It is clear that $\displaystyle \overline{\dfC}_k$ is well
and uniquely defined by (\ref{relsingCoOpEq1}).

\begin{defn}
Let $A$ be a subset of a topological space $X$ and $R$ be an integral
domain.  The {\bfseries relative $\displaystyle \mathbf{k^{th}}$ cohomology
module}\index{Cohomology Module!Relative $k^{th}$ Cohomology Module}
of $X \mod A$ is the
quotient space $\displaystyle H^k(X,A;R) = \KE(\overline{\dfC}_k)/
\IMG(\overline{\dfC}_{k-1})$.
The equivalence class of $\displaystyle H^k(X,A;R)$ associated to
$\phi \in \KE(\overline{\dfC}_k)$ is denoted $[\phi]_{X,A}$.
\end{defn}

\begin{prop} \label{propCohoSES}
Let $A$ be a subset of a topological space $X$ and $R$ be an integral
domain.  The short exact sequence
\[
\xymatrix{
0 \ar[r] & S_k(A;R) \ar[r]^-{S_k(\iota)} & S_k(X;R) \ar[r]^-{\pi_k}
& S_k(X;R)/S_k(A;R) \ar[r] & 0
}
\]
yields the short exact sequence
\[
\xymatrix{
0 \ar[r] & (S_k(X;R)/S_k(A;R))^\ast \ar[r]^-{\pi_k^\ast}
& S^k(X;R) \ar[r]^-{S_k(\iota)^\ast} & S^k(A;R) \ar[r] & 0
}
\]
where $\displaystyle S_k(\iota)^\ast : S^k(X;R) \to S^k(A;R)$ is defined by
$\displaystyle \big(S_k(\iota)^\ast(\phi)\big)(c) = \phi\big(S_k(\iota)(c)\big)$
for all $c \in S_k(A;R)$ and $\displaystyle \phi \in S^k(X;R)$, and
$\displaystyle \pi_k^\ast : \big(S_k(X;R)/S_k(A;R)\big)^\ast
\to S^k(X;R)$ is defined by
$\displaystyle \big(\pi_k^\ast(\phi)\big)(c) = \phi(\relC[X,A]{c})$ for all
$c \in S_k(X;R)$ and $\displaystyle \phi \in (S_k(X;R)/S_k(A;R))^\ast$.
\end{prop}

\begin{proof}
\stage{i} $\displaystyle S_k(\iota)^\ast$ is onto.  If
$\displaystyle \phi \in S^k(A;R)$, then we can extend it to a linear
functional $\tilde{\phi}$ on $S_k(X;R)$ by setting
$\tilde{\phi}(\sigma) = \phi(\sigma)$ for all singular $k$-simplices
with image in $A$ and $\tilde{\phi}(\sigma) = 0$ for all singular
$k$-simplices whose image is not in $A$.  Hence,
$\displaystyle \big(S_k(\iota)^\ast(\tilde{\phi})\big)(c)
= \tilde{\phi}( S_k(\iota)(c)) = \phi(c)$ for all
$c \in S_k(A;R)$.

\stage{ii} $\displaystyle \pi_k^\ast$ is obviously one-to-one because
$\displaystyle \big(\pi_k^\ast(\phi)\big)(c) = 0$ for all $c \in S_k(X;R)$
means that $\phi(\relC[X,A]{c}) = 0$ for all
$\relC[X,A]{c} \in S_k(X;R)/S_k(A;R)$.  Thus $\phi = 0$.

\stage{iii} If $\displaystyle \phi \in \KE(S_k(\iota)^\ast)$, then
$\displaystyle 0 = \big(S_k(\iota)^\ast(\phi)\big)(c) =
\phi\big(S_k(\iota)(c)\big)$ for all $c \in S_k(A;R)$.  Therefore
$\phi(c) = 0$ for all $c \in S_k(A;R)$.  Hence, if we set
$\tilde{\phi}(\relC[X,A]{c}) = \phi(c)$ for all
$\relC[X,A]{c} \in S_k(X;R)/S_k(A;R)$, then we get a well defined linear
functional on $S_k(X;R)/S_k(A;R)$.  Since
$\displaystyle \big(\pi_k^\ast(\tilde{\phi})\big)(c) =
\tilde{\phi}(\relC[X,A]{c}) = \phi(c)$ for all $c \in S_k(X;R)$, we get that
$\displaystyle \big(\pi_k^\ast(\tilde{\phi})\big) = \phi$ and so
$\displaystyle \phi \in \IMG(\pi_k^\ast)$.

We have shown that
$\displaystyle \KE(S_k(\iota)^\ast) \subset \IMG(\pi_k^\ast)$.  To
prove the reverse inequality, suppose that
$\displaystyle \phi = \pi_k^\ast(\tilde{\phi})$ for
$\displaystyle \tilde{\phi} \in (S_k(X;R)/S_k(A;R))^\ast$.
Then
\begin{align*}
\big(\big(S_k(\iota)^\ast\big)(\phi)\big)(c)
&= \big(\big(S_k(\iota)^\ast\big)\big(\pi_k^\ast(\tilde{\phi})\big)\big)(c)
= \big(\pi_k^\ast(\tilde{\phi})\big)\big(S_k(\iota)(c)\big)
= \tilde{\phi}\big(\pi_k(S_k(\iota)(c))\big) \\
&= \tilde{\phi}(\relC[X,A]{c}) = 0
\end{align*}
for all $c \in S_k(A;R)$.
Thus $\displaystyle \phi \in \KE(S_k(\iota)^\ast)$.  Hence
$\displaystyle \IMG(\pi_k^\ast) \subset \KE(S_k(\iota)^\ast)$.
\end{proof}

It follows from the previous proposition that
$\displaystyle \pi_k^\ast$ is an isomorphism between \\
$\displaystyle (S_k(X;R)/S_k(A;R))^\ast$ and the set
$\displaystyle \{ \phi \in S^k(X;R) : \phi(c) = 0 \ \text{for all}
\ c \in S_k(A;R)\}$.

\begin{defn}
The set $\displaystyle S^k(X,A;R) =
\{ \phi \in S^k(X;R) : \phi(c) = 0 \ \text{for all} \ c \in S_k(A;R)\}$ is the
{\bfseries annihilator}\index{Annihilator} of $S_k(A;R)$.
The set of {\bfseries relative cocycles}\index{Relative Cocycle}
is the set $\displaystyle Z^k(X,A;R) = \{ \phi \in S^k(X,A;R) :
\phi(c) = 0 \ \text{for all} \ c \in B_k(X,A;R) \}$ and
the set of {\bfseries relative coboundaries}\index{Relative Coboundary}
is the set $\displaystyle B^k(X,A;R) = \{ \dfC_{k-1}(\phi) :
\phi \in S^{k-1}(X,A;R) \}$.
\end{defn}

As for the relative homology, we have that
$\displaystyle H^k(X,A;R) \cong  Z^k(X,A;R)/B^k(X,A;R)$.

We can use $\displaystyle \pi_k^\ast: (S_k(X;R)/S_k(A;R))^\ast\to S^k(X;R)$
in Proposition~\ref{propCohoSES} to obtain an isomorphism
between $\displaystyle (S_k(X;R)/S_k(A;R))^\ast$ and
$\displaystyle S^k(X,A;R)$.  Using this isomorphism, then
$\displaystyle \overline{\dfC}_k$ is a restriction 
of $\displaystyle \dfC_k$.

We have that $\displaystyle B^k(X,A;R) \subset Z^k(X,A;R)$ because
$\displaystyle \big(\dfC_{k-1}(\phi)\big)(c)
= \phi(\partial_k(c)) = 0$ for all $c \in B_k(X,A;R)$
if $\displaystyle \phi \in B^k(X,A;R)$.  To be more precise,
$c \in B_k(X,A;R)$ implies that $c = b + \partial_{k+1}(a)$ with
$b \in S_k(B;R)$ and $a \in S_{k+1}(X;R)$.  Thus
$\phi(\partial_k(c)) = \phi(\partial_k(b)) = 0$ because
$\phi(g) = 0$ for all $g \in S_{k-1}(A;R)$.

Unfortunately, we do not have a result as strong as in
Theorem~\ref{thmHupeHdkS} for relative homology.

\begin{theorem}
Let $A$ be a subset of a topological space $X$ and $R$ be a principal
ideal domain.  Then $\displaystyle {\cal H}: H^k(X,A;R) \to (H_k(X,A;R))^\ast$
defined by ${\cal H}([\phi]_{X,A})([c]_{X,A}) = \phi(c)$ for all
$\displaystyle [c]_{X,A} \in H_k(X,A;R)$ and
$\displaystyle [\phi]_{X,A} \in H^k(X,A;R)$ is an epimorphism; namely,
a surjective homomorphism.
\end{theorem}

Instead of giving the definition of principal ideal domain, let us
just say that $\ZZ$ and $\RR$ are principal ideal domain.  The proof
of the previous theorem can be found in (\cite{GH}).

It is probably clear to the reader that we could go on an develop a
full theory of cohomology and relative cohomology as we have done for
singular homology and relative singular homology.  We will not do so
and refer the interested reader to \cite{GH,MUat}.  We have already
gone much further than our initial intention at the beginning of the
chapter to provide a short survey of algebraic topology.

\subsection{Cup Product}

There is however another topic that needs to be mentioned.  There is a
feature of singular cohomology theory that does not have any equivalent in
singular homology theory.  This feature is present in our study of differential
forms on manifolds.  It is the wedge product that combines a
differential $k_1$-form with a differential $k_2$-form to produce a
differential $(k_1+k_2)$-form.  In cohomology theory, this new feature
is called the ``cup product.''\  We have already seen it in our study
of simplicial cohomology.

Let $\delta_1:\Delta_{k_1} \to \Delta_{k_1+k_2}$ and
$\delta_2:\Delta_{k_2} \to \Delta_{k_1+k_2}$ be the affine maps
defined by $\delta_1(\VEC{e}_j) = \VEC{e}_j$ for $0 \leq j \leq k_1$
and $\delta_2(\VEC{e}_j) = \VEC{e}_{k_1+j}$ for $0 \leq j \leq k_2$.

\begin{defn}
Let $X$ be a topological space and $R$ an integral domain, and
let $\displaystyle S^\ast(X;R) = \bigoplus_{k\geq 0} S^k(X;R)$.
The {\bfseries cup product}\index{Cup Product} of
$\displaystyle \phi_1 \in S^{k_1}(X;R)$ with
$\displaystyle \phi_2 \in S^{k_2}(X;R)$ is the linear functional
$\displaystyle \phi_1 \cup \phi_2 \in S^{k_1+k_2}(X;R)$ defined by
$(\phi_1 \cup \phi_2)(\sigma) = \phi_1(\sigma\circ \delta_1) \, 
\phi_2(\sigma\circ \delta_2)$ for all singular $(k_1+k_2)$-simplices
$\sigma$ and extended linearly to $\displaystyle S_{k_1+k_2}(S;R)$.
\end{defn}

Given $\displaystyle \phi = \sum_{j\in \NN} a_j\phi_j$ and 
$\displaystyle \psi = \sum_{j\in \NN} b_j\psi_j$, where
all but a finite number of $a_j\in \RR$ and $b_j\in \RR$ are null,
and $\displaystyle \phi_j,\psi_j \in S^j(X;R)$ for all $j$.  We have
by definition that
\[
\phi \cup \psi = \sum_{i,j\in \NN} a_jb_i \, \phi_j \cup \psi_i \ .
\]
It is not hard to prove that the cup product on $\displaystyle S^\ast(X;R)$
is bilinear, associative and that the identity element in
$\displaystyle S^\ast(X;R)$ is the linear functional
$\displaystyle \phi \in S^0(X;R)$
defined by $\phi(\sigma) = 1$ for all singular $0$-simplices
$\sigma$.  The cup product also satisfies the relation
\begin{equation} \label{partial1U2}
\dfC_{k_1+k_1}(\phi_1 \cup \phi_2) =
\dfC_{k_1}(\phi_1) \cup \phi_2 + (-1)^{k_1} \phi_1 \cup
\dfC_{k_2}(\phi_2)
\end{equation}
for all $\displaystyle \phi_i \in S^{k_1}(X;R)$ and
$\displaystyle \phi_2 \in S^{k_2}(X;R)$.  This is similar to item (2)
of Theorem~\ref{stokesDF} about differential forms.

Let $\displaystyle Z^\ast(X;R) = \bigoplus_{k\geq 0} Z^k(X;R)$ and
$\displaystyle B^\ast(X;R) = \bigoplus_{k\geq 0} B^k(X;R)$.  We have
that $\displaystyle B^\ast(X;R)$ is a (two-sided) ideal of
$\displaystyle Z^\ast(X;R)$.  Note that the cup product of two
elements in $\displaystyle Z^\ast(X;R)$ yields an element in
$\displaystyle Z^\ast(X;R)$.  Hence
$\displaystyle H^\ast(X;R) = \bigoplus_{k\geq 0} H^k(X;R)$
is a graded $R$-algebra.  It is graded because it is the direct sums of
$R$-modules and it is an algebra because of the cup product.

Let $X$ and $Y$ be two topological spaces and $R$ be an integral domain.
Given $f:X\to Y$, we define an homomorphism
$\displaystyle S^\ast(f):S^\ast(Y;R) \to S^\ast(X;R)$ as it follows.  If
$\displaystyle \phi = \sum_{j\in \NN} a_j\phi_j$ where
all but a finite number of $a_j\in \RR$ are null and
$\displaystyle \phi_j\in S^j(X;R)$ for all $j$, then we set
$\displaystyle S^\ast(f)(\phi) = \sum_{j\in \NN} a_j S^j(f)(\phi_j)$.
We may use $\displaystyle S^\ast(f)$ to define
$\displaystyle H^\ast(f):H^\ast(Y;R) \to H^\ast(X;R)$.  Namely, we set
$\displaystyle H^\ast(f)([\phi]_Y) = [ S^\ast(f)(\phi)]_X$ for all
$\displaystyle [\phi]_Y \in H^\ast(Y;R)$.

We define the cup product of
$\displaystyle [\phi_1]_X,[\phi_2]_X \in H^\ast(X;R)$ as
$\displaystyle [\phi_1]_X\cup[\phi_2]_X = [\phi_1 \cup \phi_2]_X$.  To prove
that this cup product is well defined, it suffices to consider
$\displaystyle [\phi_i]_X \in H^{k_i}(X;R)$ with $k_i\in \NN$ for
$i =1,2$.  Suppose that $\phi_i = \psi_i + \dfC_k(\tau_i)$ for some
$\displaystyle \psi_i \in Z^{k_i}(X;R)$ and
$\displaystyle \tau_i \in S^{k_i-1}(S;R)$ for
$i =1,2$.  It follows from (\ref{partial1U2}) that
\begin{align*}
\big(\psi_1 + \dfC_{k_1}(\tau_1)\big) \cup \phi_2
&= \psi_1 \cup \phi_2 + \dfC_{k_1}(\tau_1) \cup \phi_2 \\
&= \psi_1 \cup \phi_2 + \left( \dfC_{k_1+k_1}(\tau_1 \cup \phi_2)
+ (-1)^{k_1+1} \tau_1 \cup \dfC_{k_2}(\phi_2) \right) \\
&= \psi_1 \cup \phi_2 + \dfC_{k_1+k_1}(\tau_1 \cup \phi_2)
\end{align*}
because $\displaystyle \phi_2 \in Z^{k_2}(X;R)$.  Similarly,
\begin{align*}
\psi_1 \cup \big(\psi_2 + \dfC_{k_2}(\tau_2)\big)
&= \psi_1 \cup \psi_2 + \psi_1 \cup \dfC_{k_2}(\tau_2) \\
&= \psi_1 \cup \psi_2 + (-1)^{k_2}\left( \dfC_{k_1+k_1}(\psi_1 \cup \tau_2)
- \dfC_{k_1}(\psi_1) \cup \tau_2 \right) \\
&= \psi_1 \cup \psi_2 + (-1)^{k_2}\dfC_{k_1+k_1}(\psi_1 \cup \tau_2)
\end{align*}
because $\displaystyle \psi_1 \in Z^{k_1}(X;R)$.  Hence, by
transitivity, we have
\[
[\phi_1 \cup \phi_2]_X
= \big[\big(\psi_1 + \dfC_{k_1}(\tau_1)\big) \cup \phi_2\big]_X
= [ \psi_1 \cup \phi_2 ]_X
= \big[\psi_1 \cup \big(\psi_2 + \dfC_{k_2}(\tau_2)\big)\big]_X
= [\psi_1 \cup \psi_2]_X \ .
\]

The relation in singular cohomology that is equivalent to item (4) in
Proposition~\ref{propWedgeR} is
\[
  [\phi_1]_X\cup[\phi_2]_X = (-1)^{k_2} [\phi_2]_X \cup [\phi_1]_X
\]
for $\displaystyle [\phi_i]_X \in H^{k_i}(X;R)$ with $k_i\in \NN$.  To
quote \cite{GH}, ``The proof of this theorem is surpringly
complicated.''\ The proof can be found in \cite{GH}.

\subsection{Relation Between Singular Cohomology and
de Rham Cohomology}  \label{ssectdeRhamSing}

For this section, we follow the approach in \cite{LJM}.

Suppose that $\displaystyle S \subset \RR^n$ is a $k$-dimensional
smooth manifold.  We are going to prove that $\displaystyle H^q(S)$ is
(linearly) isomorphic to $\displaystyle H^q(S;\RR)$ for all $q \in \NN$.
From now on, $\displaystyle H^q(S)$ without any
mention to the integral domain $\RR$ will always refer to the
$\displaystyle q^{th}$ (de Rham) Cohomology Module of $S$ defined in
Definition~\ref{defnqdeRhamCM}.

We define a linear map $\displaystyle \JJ : H^q(S) \to H^q(S;\RR)$
for $q \in \NN$ as it follows.  Given
$\displaystyle [\omega] \in H^q(S)$, we set
\begin{equation} \label{defnCalJ}
  \JJ([\omega])([c]_S) = \int_c \omega
\end{equation}
for all $\displaystyle [c]_S \in H_q(S;\RR)$.  This map is well defined
because it is independent of the representative of the equivalence
class $[c]_S \in H_q(S;\RR)$ and the representative of the equivalent
class $\displaystyle [\omega] \in H^q(S)$ used to define the integral.
Suppose that $\tilde{c} \hsim c$.
Then $\tilde{c} = c + \partial_{q+1}b$ for some
$b \in S_{q+1}(S;\RR)$.  Using Stokes' theorem for chains, we get
\[
\int_{\tilde{c}} \omega = \int_c \omega + \int_{\partial_{q-1} b} \omega
= \int_c \omega + \int_b \df{\omega} = \int_c \omega
\]
because $\df{\omega} = 0$ since $\displaystyle [\omega] \in H^q(S)$.
Suppose that $\tilde{\omega} \sim \omega$.  Then $\tilde{\omega} =
\omega + \df{\eta}$ for some differential $(q-1)$-form $\eta$ on $S$.
Again, using Stokes' theorem for chains, we get
\[
\int_c \tilde{\omega} = \int_c \omega + \int_c \df{\eta}
= \int_c \omega + \int_{\partial_q c} \eta = \int_c \omega
\]
because $\partial_q c = 0$ since $[c]_S \in H_q(S;\RR)$.  
It is easy to verify that $\displaystyle \JJ(\omega) \in H^q(S;\RR)$.

There is a version of Proposition~\ref{propMayVistCoHom} for
cohomology on manifolds.
Suppose that $S$ is a $k$-dimensional smooth manifold.
To reformulate our definition of the cohomology module
$\displaystyle H^q(S)$ given
in Definition~\ref{defnqdeRhamCM} of Chapter~\ref{chaptCohom}, we set
$\displaystyle Z^k(\Omega^{\,q}(S),\df{}) = \{ \omega \in \Omega^{\,q}(S)
:\df{\omega} = 0 \}$ and
$\displaystyle B^k(\Omega^{\,q}(S),\df{}) = \{ \df{\omega}
: \omega \in \Omega^{\,{q-1}}(S) \}$.  Then
$\displaystyle H^q(S) = Z^q(S) / B^q(S)$.

\begin{prop}[Mayer-Vietoris] \label{propMayVistdeR}
Suppose that $S$ is a $k$-dimensional smooth manifold and that
$S = U \cup V$ where $U$ and $V$ are two open subsets of $S$.  Then
the sequence
\[
\xymatrix{
\ar[r]^-{\tilde{H}_{q-1}^\ast} & H^q(S) \ar[r]^-{\tilde{F}_q^\ast}
& H^q(U) \oplus H^q(V) \ar[r]^-{\tilde{G}_q^\ast}
& H^q(U \cap V) \ar[r]^-{\tilde{H}_q^\ast}
& H^{q+1}(S) \ar[r]^-{\tilde{F}_{q+1}^\ast} & 
}
\]
is an exact sequence where
$\displaystyle \tilde{F}_q^\ast([\omega]) = ([\omega\big|_U],[\omega\big|_V])$,
$\displaystyle \tilde{G}_q([\omega_1],[\omega_2]) = [\omega_1\big|_{U\cap V}]
- [\omega_2\big|_{U\cap V}]$ and
$\displaystyle \tilde{H}_q^\ast: H^q(U \cap V) \to H^{q+1}(S)$ is
defined in proof below.
\end{prop}

\begin{proof}
Let $B= U\cap V$.
We consider the following three cochain complexes.
$\displaystyle \A = \{(\Omega^{\,q}(S),\df{})\}_{k\in \ZZ}$,
$\displaystyle \BB = \{(\Omega^{\,q}(U) \oplus \Omega^{\,q}(V),
\df{})\}_{k\in \ZZ}$
and
$\displaystyle \C = \{(\Omega^{\,q}(B),\df{})\}_{k\in \ZZ}$.
Let $\F = \{\tilde{f}_k\}_{k\in\ZZ}$ be the cochain map between $\A$ and $\BB$
defined by $\tilde{f}_k(\eta) = \big(\eta\big|_U,\eta\big|_V\big)$ for all
$\displaystyle \eta \in \Omega^{\,q}(S)$, and
$\GG = \{\tilde{g}_k\}_{k\in\ZZ}$ be the cochain map between $\BB$ and $\C$
defined by
$\tilde{g}_k(\eta_1,\eta_2) = \eta_1\big|_{U\cap V}-\eta_2\big|_{U\cap V}$
for all $\displaystyle \eta_1 \in \Omega^{\,q}(U_1)$ and
$\displaystyle \eta_2 \in \Omega^{\,q}(U_2)$.
Note that the differential operator $\df{}$ is defined on
$\displaystyle \Omega^{\,q}(U) \oplus \Omega^{\,q}(V)$
by $\df{(\eta_1,\eta_2)} = (\df(\eta_1),\df(\eta_2))$ for all
$\displaystyle \eta_1 \in \Omega^{\,q}(U_1)$ and
$\displaystyle \eta_2 \in \Omega^{\,q}(U_2)$.

Our goal is to prove that the sequence
\[
\xymatrix{
0 \ar[r]
& \Omega^{\,q}(S) \ar[r]^-{\tilde{f}_q}
& \Omega^{\,q}(U) \oplus \Omega^q(V) \ar[r]^-{\tilde{g}_q}
& \Omega^{\,q}(B) \ar[r] & 0
}
\]
is an exact sequence.  The conclusion of the proposition will then
follow from the Zig-Zag lemma, Lemma~\ref{lemZigZagCohom}.

It follows from the Zig-Zag lemma that
$\displaystyle \tilde{H}_q^\ast:H^q(U \cap V) \to H^{q+1}(S)$ is defined by
$\displaystyle \tilde{H}_q^\ast([\gamma]) = [\alpha]$ where
$\displaystyle \alpha \in Z^{q+1}(S)$ is such that
$\tilde{f}_{k+1}(\alpha) = (\alpha\big|_U, \alpha\big|_V) =
(\df{\beta_1}, \df{\beta_2})$
for $\displaystyle \beta_1 \in Z^q(U)$ and
$\displaystyle \beta_2 \in Z^q(V)$ with
$\tilde{g}_k(\beta_1,\beta_2) = \beta_1\big|_{U\cap V} - \beta_2\big|_{U\cap V}
= \gamma$.

\stage{i} We have that $\tilde{f}_q$ is one-to-one because
$\tilde{f}_q(\eta) = \big(\eta\big|_U,\eta\big|_V\big) = (0,0)$ implies that
$\eta\big|_S = 0$ since $S = U \cup V$.

\stage{ii} To prove that $\IMG(\tilde{f}_q) = \KE(\tilde{g}_k)$, we note that
\[
  \tilde{g}_q(\tilde{f}_q(\eta)) = \tilde{g}_q\big(\eta\big|_U ,\eta\big|_V\big)
= \eta\big|_B - \eta\big|_B = 0
\]
for all $\displaystyle \eta \in \Omega^{\,q}(S)$.  Thus
$\IMG(\tilde{f}_q) \subset \KE(\tilde{g}_k)$.

Suppose that $(\eta_1,\eta_2) \in \KE(\tilde{g}_k)$.  Then
$\eta_1\big|_B - \eta_2\big|_B = 0\big|_B$.
Let
\[
\eta = \begin{cases}
\eta_1 & \quad \text{on}\ U \\
\eta_2 & \quad \text{on}\ U\setminus V  
\end{cases}
\]
We have that $\displaystyle \eta \in \Omega^{\,q}$ because
$\eta_1 = \eta_2$ on $B = U\cap V$.  Since
$\tilde{f}_k(\eta) = \big(\eta\big|_U,\eta\big|_V\big) = (\eta_1,\eta_2)$,
we have that $(\eta_1,\eta_2) \in \IMG(\tilde{f}_k)$.  This proves that
$\KE(\tilde{g}_k) \subset \IMG(\tilde{f}_q)$.

\stage{iii} We now prove that $\tilde{g}_q$ is onto.
Given $\displaystyle \eta \in \Omega^{\,q}(B)$, we need to find
$\displaystyle \eta_1 \in \Omega^{\,q}(U)$ and
$\displaystyle \eta_2 \in \Omega^{\,q}(V)$ such that
$\tilde{g}_q(\eta_1,\eta_2) = \eta_1\big|_B - \eta_2\big|_B = \eta$.

Since $\{U,V\}$ is an open cover of $S$, we may use Theorem~\ref{cov1}
to get a partition of unity $\displaystyle \{\psi_j\}_{j\in \NNp}$
subordinate to the open cover $\{U,V\}$ of $\supp \eta$.
Let $\displaystyle \phi_1 = \sum_{\supp \psi_j \subset U} \psi_j$
and $\displaystyle \phi_2 = \sum_{\supp \psi_j \subset V} \psi_j$.
The function $\phi_1$ and $\phi_2$ are well defined and of class
$\displaystyle C^\infty$ because the sums are finite on compact sets
according to Proposition~\ref{cov4}.
We have that $\supp \phi_1 \subset U$ and $\supp \phi_2 \subset V$.
Let
\[
\eta_1 = \begin{cases}
\phi_2 \eta & \quad \text{on} \ U \cap \supp \phi_2 \\
0 & \quad \text{on} \ U \setminus \supp \phi_2
\end{cases}
\qquad \text{and} \qquad
\eta_2 = \begin{cases}
-\phi_1 \eta & \quad \text{on} \ V \cap \supp \phi_1 \\
0 & \quad \text{on} \ V \setminus \supp \phi_1  
\end{cases}
\]
Since $U \cap \supp \phi_2 \subset U \cap V$ and
$V \cap \supp \phi_1 \subset V \cap U$ where $\eta$ is defined,
we get that $\eta_1$ and $\eta_2$ are well defined,
$\displaystyle \eta_1 \in \Omega^{\,q}(U)$ and
$\displaystyle \eta_2 \in \Omega^{\,q}(V)$.
We also have that
$\tilde{g}_k(\eta_1,\eta_2) = \eta_1\big|_B - \eta_2\big|_B
= \phi_2 \eta + \phi_1\eta =(\phi_2+\phi_1)\eta = \eta$ on $B$.
\end{proof}

The operator $\JJ$ has some natural properties.

\begin{prop} \label{propFqJcomm}
Suppose that $S_1$ and $S_2$ are $k$-dimensional smooth manifolds.
If $f:S_1 \to S_2$ is a smooth function, then
\[
\xymatrix@C+2em{
H^q(S_2) \ar[r]^-{f^\sharp}\ar[d]^{\JJ} & H_k^q(S_1) \ar[d]^-{\JJ} \\
H^q(S_2;\RR) \ar[r]^-{H^q(f)} & H^q(S_1;\RR)
}
\]
is a commutative diagram for all $q \in \NN$, where
$\displaystyle f^\sharp$ is defined in Subsection~\ref{ssubCMGcase}.
\end{prop}

\begin{proof}
If $\displaystyle [\omega] \in H^q(S_2)$ and $\sigma$ is a $q$-cycle
in $S_1$, then $\displaystyle f^\sharp([\omega]) = [f^\ast(\omega)]$ and
\begin{align*}
\JJ(f^\sharp([\omega]))([\sigma]_{S_1})
&= \int_\sigma f^\ast(\omega)
= \int_{\Delta_q} \sigma^\ast(f^\ast(\omega)
= \int_{\Delta_q} (f\circ \sigma)^\ast(\omega)
= \int_{f \circ \sigma} \omega \\
&= \JJ([\omega])([f\circ \sigma]_{S_2})
= \JJ([\omega])\big( [S_q(f)(\sigma)]_{S_2}\big)
= H^q(f)\big(\JJ([\omega])\big)([\sigma]_{S_1}) \ .
\end{align*}
Hence, by linearity,
$\displaystyle \JJ(f^\sharp([\omega]))([c]_{S_1})
= H^q(f)\big(\JJ([\omega])\big)([c]_{S_1})$ for all
$[c]_{S_1} \in H_q(S_1;\RR)$.
\end{proof}

\begin{lemma} \label{lemSUVET}
Suppose that $S$ is a $k$-dimensional smooth manifold and that
$S = U \cap V$ where $U$ and $V$ are open subsets of $S$.
Then $(S,U,V)$ is an exact triad.
\end{lemma}

\begin{proof}
We have that $S\setminus U = V \setminus U$ is a closed subset
(in the topology induced on $S$) of the open set $V$.  It follows
from Excision theorem, Theorem~\ref{thmExcis}, that $V \setminus U$
can be excised from $(S,V)$; namely,
$H_k(\iota):H_k(U, U \cap V;\RR) \to H_k(S,V;\RR)$ is
an isomorphism.  Similarly, $U \setminus V$ can be excised from
$(S,U)$; namely, $H_k(\iota):H_k(V, U \cap V;\RR) \to H_y(S,V;\RR)$.
\end{proof}

\begin{prop} \label{propHqJcomm}
Suppose that $S$ is $k$-dimensional smooth manifold and that
$S = U \cup V$ where $U$ and $V$ are open subsets of $S$.  Then 
\[
\xymatrix@C+2em{
H^{q-1}(U\cap V) \ar[r]^-{\tilde{H}_{q-1}^\ast}\ar[d]^{\JJ}
& H^q(S) \ar[d]^-{\JJ} \\
H^{q-1}(U\cap V;R) \ar[r]^-{H_{q-1}^\ast} & H^q(S;\RR)
}
\]
is a commutative diagram for all $q \in \NN$ where
$\displaystyle H_q^\ast$ and $\displaystyle \tilde{H}_q^\ast$ are defined in
Propositions~\ref{propMayVistCoHom} and \ref{propMayVistdeR}
respectively.
\end{prop}

\begin{proof}
Note that we can refer to Propositions~\ref{propMayVistCoHom} in the
statement of the proposition because $(S,U,V)$ is an exact triad
according to Lemma~\ref{lemSUVET}.

The diagram states that
\begin{equation} \label{propHqJEq1}
\JJ\big(\tilde{H}_{q-1}^\ast([\omega])\big)([c]_S)
= H_{q-1}^\ast\big(\JJ([\omega])\big)([c]_S)
\end{equation}
for all $\displaystyle [c]_S \in H_q(S;\RR)$ and
$\displaystyle [\omega] \in H^{q-1}(U\cap V)$.  From the duality
mentioned in Proposition~\ref{propMayVistCoHom}, we get that
(\ref{propHqJEq1}) is equivalent to
\begin{equation} \label{propHqJEq2}
\JJ\big(\tilde{H}_{q-1}^\ast([\omega])\big)([c]_S)
= \JJ([\omega])(H_q([c]_S))
\end{equation}
for all $\displaystyle [c]_S \in H_q(S;\RR)$ and
$\displaystyle [\omega] \in H^{q-1}(U\cap V)$.
We can restate (\ref{propHqJEq2}) as
\begin{equation} \label{propHqJEq3}
\int_c \eta = \int_a \omega
\end{equation}
where $\displaystyle [\eta] = \tilde{H}_{q-1}^\ast([\omega]) \in H^q(S)$
and $\displaystyle [a]_{U\cap V} = H_q([c]_S) \in H_q(U\cap V)$
for all $\displaystyle [c]_S \in H_q(S;\RR)$ and
$\displaystyle [\omega] \in H^{q-1}(U\cap V)$.
We prove (\ref{propHqJEq3}).

If we refer to the definition of $H_q$ given in the proof of
Proposition~\ref{propMayVistHom}, then we have that
$[a]_{U\cap V} = H_q([c]_S)$ is given by
$a \in Z_{q-1}(U\cap V;\RR)$ such that
$f_{k-1}(a) = (a,-a) = (\partial_k (b_1), \partial_k(b_2))$ for
$b_1 \in Z_k(U;R)$ and $b_2 \in Z_k(V;R)$ with
$g_k(b_1,b_2) = b_1 + b_2 = c$ in $S$.
In particular, we have that $a = \partial_k(b_1) = \partial_k(b_2)$
with $\partial_k(b_1)(\Delta_k) \subset U \cap V$ and
$\partial_k(b_2)(\Delta_k) \subset U \cap V$. 

If we now refer to the definition of
$\displaystyle \tilde{H}_{q-1}^\ast$ given in
the proof of Proposition~\ref{propMayVistHom}, then we have that
$\displaystyle [\eta] = \tilde{H}_{q-1}^\ast([\omega])$ is given by
$\displaystyle \eta \in Z^{q+1}(S)$ is such that
$\tilde{f}_{k+1}(\eta) = (\eta\big|_U, \eta\big|_V) =
(\df{\beta_1}, \df{\beta_2})$
for $\displaystyle \beta_1 \in Z^q(U)$ and
$\displaystyle \beta_2 \in Z^q(V)$ with
$\tilde{g}_k(\beta_1,\beta_2) = \beta_1\big|_{U\cap V} - \beta_2\big|_{U\cap V}
= \omega$.  In particular, we have that
$\eta\big|_U = \df{\beta_1}$ and $\eta\big|_V = \df{\beta_2}$.

Since $\partial_q(b_1) + \partial_q(b_2) = \partial_q(c) = 0$ because
$c \in Z_q(S;\RR)$ and $\df{\beta_1}\big|_{U\cap V}
- \df{\beta_2}\big|_{U\cap V} = \df{\omega} = 0$
because $\displaystyle \omega \in Z^{q-1}(U\cap V)$, we get from
Stokes' theorem that
\begin{align*}
\int_a \omega = \int_{\partial_k(b_1)} \omega
&= \int_{\partial_k(b_1)} (\beta_1 - \beta_2)
= \int_{\partial_k(b_1)} \beta_1 - \int_{\partial_k(b_1)} \beta_2
= \int_{\partial_k(b_1)} \beta_1 + \int_{\partial_k(b_2)} \beta_2 \\
&= \int_{b_1} \df{\beta_1} + \int_{b_2} \df{\beta_2}
= \int_{b_1} \eta + \int_{b_2} \eta = \int_c \eta \ .
\end{align*}
This proves (\ref{propHqJEq3}).
\end{proof}

\begin{theorem}[de Rham]  \label{deRhamThmSing}
Let $\displaystyle S \subset \RR^n$ be a $k$-dimensional smooth
manifold and $q \in \NN$.  The map
$\displaystyle \JJ : H^q(S) \to H^q(S;\RR)$ defined in  
(\ref{defnCalJ}) is an isomorphism.
\end{theorem}

\begin{proof}
For convenience, we will say that a smooth manifold $M$ is a
{\bfseries de Rham} manifold if
$\displaystyle \JJ : H^q(M) \to H^q(M;\RR)$ is an isomorphism.

\stage{Claim 1} If $U$ is a convex open subset of $\displaystyle \RR^k$,
then $U$ is a de Rham manifold.

It follows from Corollary~\ref{corHqg0e0} (see comments after the
corollary) and Corollary~\ref{corContrHk} that $\displaystyle H^q(U) = 0$
and $\displaystyle H^q(U;\RR) \cong H_q(U;\RR) = 0$ for $q>0$,
where Corollary~\ref{corHupeHdkS} was used for the last relation.  The
condition for being a de Rahm manifold is therefore trivially satisfied.

If $q=0$, then we get from Corollaries~\ref{corHqe0e1} and
\ref{corContrHk} that $\displaystyle H^0(U) \cong \RR$
and $\displaystyle H^0(U;\RR) \cong H_0(U;\RR) \cong \RR$.
Hence $\displaystyle H^0(U)$ is the set of constant functions on $U$
and $\displaystyle H^0(U;\RR)$ is the set of constant functions on
$\{ e_{\VEC{u}} : \VEC{u} \in U\}$ where $e_{\VEC{u}}:\Delta_0 \to U$ is
defined by $e_{\VEC{u}}(\VEC{e}_0) = \VEC{u}$ for all $\VEC{u} \in U$.
If we identify $e_{\VEC{u}}$ to $\VEC{u}$, then we have the set of constant
functions $f$ on $U$ in both cases.  Hence
\[
\JJ(f)(\sigma) = \int_\sigma f = \int_{\Delta_0} \sigma^\ast(f)
= f(\sigma(\VEC{e}_0)) = f(\VEC{u}) \neq 0
\]
for all singular $0$-simplices $\sigma = e_{\VEC{u}}$ in $U$ unless $f = 0$.
Thus $\displaystyle \JJ : H^0(U) \to H^0(U;\RR)$ is an isomorphism.

\stage{Claim 2} If $\displaystyle \{S_j\}_{j=0}^\infty$ is a
collections of disjoint de Rahm manifolds where each $S_j$ is
connected, then $\displaystyle \bigcup_{j=0}^\infty S_j$ is a de Rham
manifold.

This follows from the fact that the inclusion
$\displaystyle \iota : S_i \to \bigcup_{j=0}^\infty S_j$ induces an
isomorphism $\displaystyle \tilde{\iota} : \bigoplus_{j=0}^\infty S_j
\to \bigcup_{j=0}^\infty S_j$.
Hence, using Propositions~\ref{propHkeHkJ} and \ref{propHkeHkJCoho},
we get that\\
$\displaystyle H_q(\tilde{\iota}) : \bigoplus_{j=0}^\infty H_q(S_j;\RR)
\to H_q\Big(\bigcup_{j=0}^\infty S_j;\RR \Big)$ and
$\displaystyle \tilde{\iota}^{\, \sharp}:\bigoplus_{j=0}^\infty H^q(S_j)
\to H^q\Big(\bigcup_{j=0}^\infty S_j\Big)$ are isomorphism
(see comment after Proposition~\ref{propFsharpP}).
We have from Proposition~\ref{propFqJcomm} that
\[
\xymatrix@C+2em{
H^q\Big(\bigcup_{j=0}^\infty S_j\Big)
\ar[r]^{\tilde{\iota}^{\,\sharp}}\ar[d]^{\JJ} & \bigoplus_{j=0}^\infty H^q(S_j)
\ar[d]^-{\JJ} \\
H^q\Big(\bigcup_{j=0}^\infty S_j;\RR\Big) \ar[r]^{H_q(\tilde{\iota})} &
\bigoplus_{j=0}^\infty H^q(S_j;\RR)
}
\]
is a commutative diagram.  Since $H_q(\tilde{\iota})$ and
$\displaystyle \tilde{\iota}^{\,\sharp}$ are isomorphism and
$\displaystyle \JJ : H^q(S_j) \to H^q(S_j;\RR)$ 
is an isomorphism for all $j$, we get that
$\displaystyle \JJ : H^q\Big(\bigcup_{j=0}^\infty S_j\Big)
\to H^q\Big(\bigcup_{j=0}^\infty S_j;\RR\Big)$ is an isomorphism.

\stage{Claim 3} If $\displaystyle U = U_1 \cup U_2$ where $U_1$ and
$U_2$ are two open subsets of $S$, and the sets $U_1$, $U_2$ and
$U_1 \cap U_2$ are de Rham manifolds, then $U$ is a de Rham manifold.

According to Lemma~\ref{lemSUVET}, $(U_1 \cup U_2,U_1,U_2)$ is an exact triad.
Hence, we get from Proposition~\ref{propMayVistCoHom} and
Proposition~\ref{propMayVistdeR} that each row of the following
diagram is exact.
\[
\xymatrix{
\ar[r]^-{\tilde{F}_{q-1}^\ast}
& H^{q-1}(U_1) \oplus H^{q-1}(U_2) \ar[r]^-{\tilde{G}_{q-1}^\ast}
\ar[d]^{\JJ}
& H^{q-1}(U_1 \cap U_2) \ar[r]^-{\tilde{H}_{q-1}^\ast}\ar[d]^{\JJ}
& H^q(U) \ar[r]^-{\tilde{F}_q^\ast}\ar[d]^{\JJ} & \\
\ar[r]^-{F_{q-1}^\ast}
& H^{q-1}(U_2;R) \oplus H^{q-1}(U_1;R) \ar[r]^-{G_{q-1}^\ast}
& H^{q-1}(U_1\cap U_2;R) \ar[r]^-{H_{k-1}^\ast}
& H^k(U;R) \ar[r]^-{F_k^\ast} & \\
& H^q(U_1) \oplus H^q(U_2) \ar[r]^-{\tilde{G}_q^\ast}\ar[d]^{\JJ}
& H^q(U_1 \cap U_2) \ar[r]^-{\tilde{H}_q^\ast}\ar[d]^{\JJ}
& & \\
& H^k(U_2;R) \oplus H^k(U_1;R) \ar[r]^-{G_k^\ast}
& H^k(U_1\cap U_2;R) \ar[r]^-{H_k^\ast}
& &
}
\]
It follows from Propositions~\ref{propFqJcomm} and \ref{propHqJcomm}
that this diagram is commutative.  Since the first, second, fourth
and fifth maps $\JJ$ are isomorphism by assumption, we get from
the Five Lemma that $\displaystyle \JJ: H^k(U) \to H^k(U;R)$ is also an
isomorphism.

\stage{Claim 4} If $\displaystyle U = \bigcup_{j=1}^J U_j$
where each $U_j$ is an open subset of $S$ and a de Rham manifold, and
any finite intersection of $U_j$ with $1\leq j \leq J$ is a de Rham
manifold, then $U$ is a de Rham manifold.

The proof is by induction on $J$.  The result is trivial if $J=1$.
Suppose that the result is true for $J$ and that
$\displaystyle U = \bigcup_{j=1}^{J+1} U_j$
where each $U_j$ is an open subset of $S$ and a de Rham manifold, and
any finite intersection of $U_j$ for $1 \leq j \leq J+1$ is a de Rham
manifold. 

By induction, we have that $\displaystyle \tilde{U} = \bigcup_{j=1}^J U_j$
is a de Rham manifold.  If we apply Claim 3 to $\tilde{U}$ and $U_{J+1}$,
then we get that $U = \tilde{U} \cup U_{J+1}$ is a de Rham manifold.
Note that $\displaystyle \tilde{U} \cap U_{J+1}
= \bigcup_{j=1}^J (U_j \cap U_{J+1})$ is a de Rham manifold because
each $U_j \cap U_{j+1}$ is a de Rahm manifold by assumption and
any finite intersection of $U_j \cap U_{J+1}$ is a finite intersection
of $U_j$ with $1 \leq j \leq J+1$ which is also a de Rham manifold by
assumption.  Thus, by induction hypothesis with $U_j \cap U_{J+1}$
with $1 \leq j \leq J$ instead of $U_j$ with $1\leq j \leq J$, we get
that $U = \tilde{U} \cup U_{J+1}$ is a de Rham manifold.

\stage{Claim 5} If $S$ has a basis of open sets $\{ U_i\}_{i \in I}$
such that each $U_i$ is a connected, bounded and open set
which is also a de Rham manifold, and finite intersections of sets
from $\{ U_i\}_{i \in I}$ are de Rham manifolds, then $S$ is a de Rham
manifold.

Let $\displaystyle \{\phi_j\}_{j \in \NNp}$ be a partition of unity
subordinated to $\{ U_i\}_{i \in I}$.  Consider the function
$\displaystyle g = \sum_{j=1}^\infty j \phi_j$ on $S$.  Note that the
sum $\displaystyle \sum_{j=1}^\infty j \phi_j$ is finite on compact
subsets of $S$ according to Proposition~\ref{cov4}.
Thus $g$ is well defined.  We also have that
$\displaystyle g(\VEC{u}) = \sum_{j=1}^\infty j \phi_j(\VEC{u})
\geq \sum_{j=1}^\infty \phi_j(\VEC{u}) = 1$ for all $\VEC{u} \in S$.

Moreover
$\displaystyle g^{-1}([0,a]) \subset \bigcup_{j=1}^{J_a} \overline{U}_{i_j}$
for some $i_j \in I$ such that $\supp \phi_j \subset U_{i_j}$
and $J_a$ is a positive integer greater or equal to $a$.
To prove this statement, we prove that, if
$\displaystyle \VEC{u} \not\in \bigcup_{j=1}^{J_a} \overline{U}_{i_j}$, then
$d(\VEC{u}) > 1$.  Since $\phi_j(\VEC{u}) = 0$ for $1 \leq j \leq j_a$
because $\VEC{u} \not\in U_{i_j} \supset \supp \phi_j$ for $1 \leq j \leq j_a$,
we get that
$\displaystyle g(\VEC{u}) = \sum_{j=1}^\infty j \phi_j(\VEC{u})
= \sum_{j=J_a+1}^\infty j \phi_j(\VEC{u})
\geq (J_a +1) \sum_{j=J_a+1}^\infty \phi_j(\VEC{u})
= J_a +1 > c$.

Note that $\displaystyle g^{-1}([0,a])$ is a closed subset of the
compact set $\displaystyle \bigcup_{j=1}^{J_a} \overline{U}_{i_j}$.
Thus $\displaystyle g^{-1}([0,a])$ is a compact subset of $S$ for all
$a >0$.

Let $\displaystyle V_m = g^{-1}([m,m+1])$ and
$\displaystyle W_m = g^{-1}(]m-1/2,m+3/2[)$ for $m \in \NN$.
Since $W_m$ is an open subset of $S$, it is the union of some open sets
from $\{ U_i\}_{i \in I}$.  Since $V_m \subset W_m$, these subsets
form an open cover of the compact set $V_m$.  Thus, there exists a
finite subcover
$\displaystyle \{ U_{i_{m,r}}\}_{r=1}^{R_m}$ of $V_m$.
We have from Claim 4
that $\displaystyle G_m = \bigcup_{r=1}^{R_m} U_{i_{m,r}}$ is a de
Rham manifold.  Note that $G_m$ is the union of more than one distinct
connected de Rham manifolds if
$\displaystyle \bigcap_{r=1}^{R_m} U_{i_{m,r}} = \emptyset$.

By construction, we have that $V_m \subset G_m \subset W_m$.
Thus $G_{m_1} \cap G_{m_2} = \emptyset$ for $|m_2 - m_1| > 1$.
Hence $\displaystyle G_e = \bigcup_{m\ \text{even}} G_m$ and
$\displaystyle G_o = \bigcup_{m\ \text{odd}} G_m$ are both the union
of distinct connected de Rham manifolds.  It follows from Claim 2 that
$G_e$ and $G_o$ are de Rham manifolds.

Since $\displaystyle G_m \cap G_{m+1}
= \bigcup_{r=1}^{R_m} \bigcup_{\tilde{r}=1}^{R_{m+1}} U_{i_{m,r}}
\cap U_{i_{m+1,\tilde{r}}}$, we get that $G_m \cap G_{m+1}$ is the finite
union of open and connected de Rham manifolds because any finite
intersection of $U_i$ is de Rham manifolds by assumption.
Thus $G_m \cap G_{m+1}$ is the disjoint union of open and connected de
Rham Manifolds according to Claim 4.
Since $(G_{m_1} \cap G_{m_1+1}) \cap
(G_{m_2} \cap G_{m_2+1}) = \emptyset$ for $m_1 \neq m_2$, it follows
that $\displaystyle G_e \cap G_o = \bigcup_{m=1}^\infty (G_m \cap G_{m+1})$
is the disjoint union of open and connected de Rham Manifolds.  We get
from Claim 2 that $G_e \cap G_o$ is a de Rham manifold.
Finally,  it follows from Claim 3 that $S = G_e \cup G_o$ is a de Rham
manifold.
\pdfbox{alg_top/derham}

\stage{Claim 6} If $\displaystyle U \subset \RR^k$ is an open set,
then $U$ is a de Rham manifold.

We have that $\displaystyle U = \bigcup_{i\in I} U_i$ where each $U_i$
is an open ball in $\displaystyle \RR^k$.  It follows from Claim 1
that the $U_i$ are de Rham manifold since open balls are convex.
The $U_i$ are also connected and bounded.  Since the finite
intersection of open balls is also a convex set, we have that finite
intersections of sets from $\{ U_i\}_{i \in I}$ are de Rham manifolds.
Therefore the collection $\{ U_i\}_{i \in I}$ meets the requirements
of Claim 5.  Hence $U$ is de Rham.

\stage{Claim 7} Finally, if $\displaystyle S \subset \RR^n$ is a
$k$-dimensional smooth manifold, then $S$ is a de Rham manifold.

The manifold $S$ has a basis of open set $\{ U_i\}_{i \in I}$ where
each $U_i$ is associated to a local chart $(W_i,U_i,\phi_i)$ of $S$.
We may select the $U_i$ to be bounded and connected \footnote{For
instance, for each $\VEC{u}$, we may use $U_{\VEC{u}} = S \cap B_r(\VEC{u})$
for $r$ small enough, and use $I = S$ as set of indices.}.
Since each $U_i$ is diffeomorphic to the open set
$\displaystyle W_i \subset \RR^k$ and $W_i$ is a de Rham manifold
according to Claim 6, we get from Proposition~\ref{propFqJcomm} that
$U_i$ is a de Rham manifold.   This is also true for the finite
intersection of set from $\{ U_i\}_{i \in I}$ be definition of local
charts.  It follows from Claim 5 that $S$ is a de Rham manifold.
\end{proof}

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