\classheader{2012-10-19}

Last time we computed that $\widetilde{H}_i(\S^n)=\begin{cases}
\Z & \text{ if }i=n,\\[-0.06in] 0 & \text{ if }i\neq n,
\end{cases}$, and that $\widetilde{H}_i(\D^n)=0$ for all $i$. Thus, as a corollary we can now conclude the Brouwer fixed point theorem.

This theorem is a big deal. Thurston used the Brouwer fixed point theorem to classify homeomorphisms of hyperbolic 3-manifolds up to homotopy by taking the space of hyperbolic metrics, which any homotopy class of homeomorphisms acts on, and looking at the various places a fixed point could be.

Now we'll finally prove that $H_i^\Delta(X)\cong H_i^S(X)$.

Let $X$ be a $\Delta$-complex and $A$ a subcomplex of $X$. There exists a chain map $\psi:C_i^\Delta(X,A)\to C_n^S(X,A)$ sending $\sigma$ to itself. This induces a homomorphism $\psi_*:H_n^\Delta(X,A)\to H_n^S(X,A)$.

\begin{theorem}
For all $(X,A)$ and $n\geq 0$, $\psi_*$ is an isomorphism.
\end{theorem}
\begin{corollary}[``Amazing'']
$H_n^\Delta(X,A)$ doesn't depend on which $\Delta$-complex structure you choose!
\end{corollary}
\begin{proof}
We'll do this for finite-dimensional $X$ and $A=\varnothing$. Read about other cases in Hatcher.

Let $X^k$ = ``$k$-skeleton of $X$'' = $\{\text{the }k\text{-simplices }\sigma_i:\Delta^k\to X\}$.

We have a LES of the pair $(X^k,X^{k-1})$ in each homology theory, and a commutative diagram comparing them because $\psi$ is a chain map:
\[\xymatrix{ H_{n+1}^\Delta(X^k,X^{k-1}) \ar[r] \ar[d]^{\psi_*}& H_n^\Delta(X^{k-1}) \ar[r]  \ar[d]^{\psi_*}& H_n^\Delta(X^k) \ar[r] \ar[d]^{\psi_*} & H_n^\Delta(X^k,X^{k-1}) \ar[r]  \ar[d]^{\psi_*}& H_{n-1}^\Delta(X^{k-1}) \ar[d]^{\psi_*} \\
 H_{n+1}^S(X^k,X^{k-1}) \ar[r] & H_n^S(X^{k-1}) \ar[r] & H_n^S(X^k) \ar[r] & H_n^S(X^k,X^{k-1}) \ar[r] & H_{n-1}^S(X^{k-1}) }\]
We want to induct on $k$ to prove that $H_n^\Delta(X^k)\xrightarrow{\;\cong\;} H_n^S(X^k)$. If this is true for all $k$, then we're done, because (by our assumption that $X$ is finite-dimensional), we have $X=X^k$ for some $k$. The claim is true for $k=0$ because we know how to look at points. By our inductive hypothesis, the 2nd and 5th vertical arrows in our diagram are isomorphisms.

Now we will prove that the 1st and 4th vertical arrows are isomorphisms. Intuitively, taking the $k$-skeleton of $X$ (which is just a gluing of $k$-simplices) and quotienting by the $(k-1)$-skeleton is just crushing the boundaries of all of the $k$-simplices, turning the $k$-skeleton into a union and/or wedge of spheres.

We have that
\[C_n^\Delta(X^k,X^{k-1})=\begin{cases}
0 & \text{ if }n\neq k,\\
\Z^d & \text{ if }n=k.
\end{cases}\]
where $d=\#(X^k\setminus X^{k-1})$. Thus, $H_n^\Delta(X^k,X^{k-1})$ is the same. Now we want to find $H_i^S(X^k,X^{k-1})$. 

Recall that $X^k=\{\sigma_i:\Delta^k\to X^k\}$. Define the map $\Phi:\coprod_i(\Delta^k,\partial\Delta^k)\xrightarrow{\;\sigma_i\;}(X^k,X^{k-1})$. Then $\Phi$ induces a \textbf{homeomorphism}
\[\textstyle\coprod\Delta^k / \coprod \partial\Delta^k \xrightarrow{\;\cong\;} X^k/X^{k-1}\]
essentially by the definition of $X$ being a $\Delta$-complex. This is the same as our proof that \[(\D^n,\partial\D^n)\to(\D^n/\partial\D^n,\partial\D^n/\partial\D^n)=(\D^n/\partial\D^n,\text{pt})\]
induces an isomorphism in homology. This is the same map that we're using in our big diagram, so \textbf{that} map is the isomorphism (a priori, just knowing the domain and codomain are isomorphic doesn't tell us that our map is an isomorphism).

\begin{lemma}[Five Lemma]
Given a map between exact sequences
\[\xymatrix{A \ar[r] \ar[d]^\alpha & B \ar[r] \ar[d]^\beta & C \ar[r] \ar[d]^\gamma & D \ar[r] \ar[d]^\delta & E \ar[d]^{\eta} \\
A \ar[r] & B \ar[r] & C \ar[r] & D \ar[r] & E
}\]
where $\alpha,\beta,\delta,\eta$ are isomorphisms, then $\gamma$ is an isomorphism.
\end{lemma}

Now we're done, as long as we can prove formally 

\begin{theorem}[Excision]
Given a topological space $X$ and $A,U\subseteq X$ subspace with $\overline{U}\subset\operatorname{int}(A)$, the inclusion $i:(X/U,A/U)\hookrightarrow (X,A)$ induces an isomorphism on homology.
\end{theorem}


\begin{proof}
We'll do the case when $\Delta$-complex and each of $A$, $X-U$, $A-U$ is a subcomplex of $X$. 

We get chain maps
\[C_n(X-U)\to C_n(X)\to C_n(X,A)\]
and call the composition $\phi$. The map $\phi$ is surjective, which is clear - any chain in $X$ without any part in $A$ certainly is a chain in $X$ without any part in $U$. Then $\phi$ induces an isomorphism.
\[C_n(X-U,A-U)\;\stackrel{\text{def}}{=}\;\frac{C_n(X-U)}{C_n(A-U)}\cong C_n(X,A).\qedhere\]
\end{proof}


Thus, we have proven that $\Delta$-complex homology and singular homology are isomorphic.
\end{proof}






