\classheader{2012-10-10}

So far, we've seen how to start from a $\Delta$-complex $X$ and then form a complex of simplicial chains, and from there we use homological algebra to define the homology groups $H_i(x)$. But we'd like to be able to start from a topological space without a specified $\Delta$-complex structure and be guaranteed to get isomorphic homology groups. In other words, we want to show that
\[X\isom Y\implies H_i(X)\cong H_i(Y)\text{ for all }i\geq 0.\]
As is common in mathematics, this becomes easier if we enlarge our category, in this case to the homotopy category.

\begin{definition}
Let $X$, $Y$ be topological spaces, and $f,g:X\to Y$ maps. A homotopy from $f$ to $g$, denoted $f\sim_F g$, is a map $F:X\times [0,1]\to Y$ such that $F|_{X\times\{0\}}=f$ and $F|_{X\times\{1\}}=g$.
\end{definition}

For $t\in[0,1]$, we write $F_t:X\to Y$ for the function $F_t(x)=F(x,t)$, so that $F_0=f$ and $F_1=g$.

\begin{definition}
A map $f:X\to Y$ is a homotopy equivalence if there is some $g:Y\to X$ such that $f\circ g\sim \id_Y$ and $g\circ f\sim \id_X$. We say that $X\simeq Y$.
\end{definition}

\begin{remark}
$f\sim g$ is an equivalence relation, and $X\simeq Y$ is an equivalence relation.
\end{remark}

\begin{definition}
$X$ is contractible if there is a point $x\in X$ such that $i:\{x\}\hookrightarrow X$ is a homotopy equivalence.
\end{definition}

\begin{example}
Homotopy equivalence can't tell you anything about contractible spaces, not even their dimension, because $\D^n$ is contractible: the inclusion $\{0\}\hookrightarrow \D^n$ has as a homotopy inverse the map sending everything to 0. The homotopy is $F_t(v)=tv$.
\end{example}

\begin{example}
Any homeomorphism $f:X\to Y$ is also a homotopy equivalence.
\end{example}


\begin{proposition}
Suppose $H_i$ is any functor from
\[\left\{\genfrac{}{}{0pt}{}{\text{topological spaces (or $\Delta$-complexes)}}{\text{with continuous maps}}\right\}\to\left\{\genfrac{}{}{0pt}{}{\text{abelian groups}}{\text{with homomorphisms}}\right\}\]
satisfying
\begin{enumerate}
\item Functoriality: $(\id_X)_*=\id_{H_i(X)}$, $(f\circ g)_*=f_*\circ g_*$.
\item Homotopy functor: if $f\sim g$, then $f_*=g_*$.
\end{enumerate}
Then $H_*$ is a topological invariant; in fact, if $f:X\xrightarrow{\;\simeq\;}Y$, then $f_*:H_i(X)\xrightarrow{\;\cong\;}H_i(Y)$.
\end{proposition}
\begin{proof}
If $f:X\xrightarrow{\;\simeq\;}Y$ has $g:Y\to X$ as a homotopy inverse, then the fact that $f\circ g\sim \id_Y$ and $g\circ f\sim \id_X$ implies that
\[f_*\circ g_*=\id \implies f_*\text{ surjective},\qquad g_*\circ f_8=\id\implies f_*\text{ injective},\]
and hence $f_*$ is an isomorphism.
\end{proof}

\subsection*{How to Prove that $H_i$ is a Homotopy Functor}

\subsubsection*{Method 1: The Simplicial Approximation Theorem}

The idea is that, given $\Delta$-complexes $X$ and $Y$ with $f:X\xrightarrow{\;\cong\;}Y$,

Step 1: We show there exists a simplicial homeomorphism $f':X'\to Y$ (where $X'$ is the barycentric subdivision of $X$)

Step 2: Then show that $H_i(X)\cong H_i(X')$.

Step 3: Then show that if $f''$ is another such map, then $(f'')|_*=(f')_*$.

Step 4: Now declare $f_*=f_*'$.

\subsubsection*{Method 2: Singular Homology}

Step 1: Construct singular homology, $H_i^S$, which is easy to show is a homotopy functor.

Step 2: For any $\Delta$-complex $X$, $H_i^S(X)\cong H_i(X)$.

We'll do it this way.

\subsection*{Singular Homology}

Let $X$ be any topological space, and for any $n\geq 0$, define
\[C_n(X)=\text{free abelian group on }\{\text{continuous maps }\Delta^n\to X\}.\]
Define $\partial_n:C_n(X)\to C_{n-1}(X)$ as the unique linear extension of 
\[\left(\sigma:[v_0\cdots v_n]=\Delta^n\to X\right)\mapsto \partial_n\sigma =\sum_{i=0}^n(-1)^i\sigma|_{[v_0\cdots\widehat{v_i}\cdots v_n]}\]
Check that $\partial_{n-1}\circ\partial_n=0$. This is the complex of singular chains.

\begin{definition}
The singular homology $H_i^S(X)$ of a space $X$ is 
\[H_i^S(X)=H_i(\{C_n(X),\partial_n\}).\]
Induced maps: if $f:X\to Y$ is continuous, then we can send $\sigma:\Delta^n\to X$ to $(f\circ \sigma):\Delta^n\to Y$, and then extend this to $f_\#:C_n(X)\to C_n(Y)$.
\end{definition}

It is trivial to check that $f_\#$ is a chain map, and so induces $f_*:H_n^S(X)\to H_n^S(Y)$, and also that $(f\circ g)_\#=f_\#\circ g_\#$, which then implies that $(f\circ g)_*=f_*\circ g_*$.

Note that we're building a dictionary:

\begin{center}
\begin{tabular}{c|c}
$\{C_n(X),\partial_n\}$ & $\cal{C}=\{C_n,\partial_n\}$\\\hline
continuous maps $f:X\to Y$\rule{0pt}{2.6ex} & chain maps $f:\cal{C}\to\cal{D}$\\[0.03in]
$f\sim g$ homotopy & chain homotopy 
\end{tabular}
\end{center}