\classheader{2012-10-22}

The midterm will be on November 2nd in class; it will be closed book, and cover everything up to the class before that.

\begin{theorem}[Homology of quotients]
Let $X$ be a topological space, and $A\subseteq X$ a reasonable subspace. Then the quotient map $p:(X,A)\to(X/A,A/A)$ induces for all $i\geq 0$ an isomorphism
\[p_*:H_i(X,A)\to H_i(X/A,A/A)\;\;\underset{\text{\raisebox{-0.5ex}{$\mathclap{\text{\emph{from homework}}}$}}}{\cong}\;\; \widetilde{H}_i(X/A).\]
\end{theorem}

This theorem is not as powerful as it may seem, because it only allows us to study smashing the subspace $A$ to a point; we often want to glue two subspaces together without turning them into a single point.

\begin{proof}
We will do the case that $A$ is open.

As always, we get the induced map $p_*:H_n(X,A)\to H_n(X/A,A/A)$.
\[\xymatrix{H_n(X,A) \ar[d]_{p_*} \ar[r]^(0.27){\cong}_(0.27){\text{excision}} & H_n(X-A,A-A) = H-n(X-A) \ar[d]^{(p|_{X-A})_*} \\
H_n(X/A,A/A) \ar[r]_{\cong} & H_n(X/A-A/A)}\]

The key is that $(p|_{X-A})$ is a homeomorphism onto its image.
\end{proof}

We want to be able compute the homology of $\R P^n$, $\C P^n$, a knot $\S^3-K$, and other interesting spaces.

Two applications of what we've done so far include the Euler characteristic and the Lefschetz Fixed Point Formula.

\begin{definition}
Let $X$ be a finite $\Delta$-complex. Let $c_n(X)$ denote the rank of $C_n^\Delta(X)$. The Euler characteristic of $X$ is
\[\chi(X)=\sum_{n\geq 0}(-1)^n c_n.\]
\end{definition}

Let's look at some $\Delta$-complex structures on the disk:

\begin{center}
\begin{tikzpicture}[thick,scale=1.7]
\draw (0,0) -- (0.5,0.86) -- (1,0) -- cycle;
\begin{scope}[every node/.style={fill,circle,inner sep=0pt,outer sep=0pt,minimum size=5pt}]
\node at (0,0) {};
\node at (0.5,0.86) {};
\node at (1,0) {};
\end{scope}
\end{tikzpicture}
\qquad
\begin{tikzpicture}[thick]
\draw (0,0) circle (1);
\draw (135:1) -- (315:1);
\draw (45:1) -- (225:1);
\begin{scope}[every node/.style={fill,circle,inner sep=0pt,outer sep=0pt,minimum size=5pt}]
\node at (0,0) {};
\node at (45:1) {};
\node at (135:1) {};
\node at (225:1) {};
\node at (315:1) {};
\end{scope}
\end{tikzpicture}\qquad
\begin{tikzpicture}[thick,scale=1.7]
\draw (0,0) -- (0.5,0.86) -- (1,0) -- cycle;
\draw (0.5,0.86) -- (1.5, 0.86) -- (1,0);
\begin{scope}[every node/.style={fill,circle,inner sep=0pt,outer sep=0pt,minimum size=5pt}]
\node at (0,0) {};
\node at (0.5,0.86) {};
\node at (1,0) {};
\node at (1.5,0.86) {};
\end{scope}
\end{tikzpicture}
\end{center}

\begin{center}
\begin{tabular}{ccc|c}
0-simplices & 1-simplices & 2-simplices & $\#0-\#1+\#2$\\\hline
3 & 3 & 1 & 1\\
5 & 8 & 4 & 1\\
4 & 5 & 2 & 1\\
\end{tabular}
\end{center}


\begin{theorem}[Topological invariance of $\chi$]
Let $X$ be a finite $\Delta$-complex. Then
\[\chi(X)=\sum_{n\geq 0}(-1)^nb_n(X)\]
where $b_n(X)$ is the $n$th Betti number of $X$,
\[b_n(X)=\operatorname{rank}(H_n(X)/T_n(X)).\]
\end{theorem}
In particular, because $X\simeq Y \implies b_n(X)=b_n(Y)$ for all $n\geq 0$, the right side, and hence the left side, depends only on the homotopy type of $X$.

The essential idea is that (letting lowercase letters denote the ranks of the groups typically denoted by the corresponding uppercase letters):
\[h_n=z_n-b_n,\quad c_n=z_n+b_{n-1}\]
and using this in the alternating sum makes the proof pop out.


\subsection*{Hopf Trace Formula}
Let $\cal{C}=\{C_n,\partial_n\}$ be a finite chain complex of finitely generated free abelian groups. Let $f:\cal{C}\to\cal{C}$ be a chain map. We know that $C_n\cong\Z^d$ for some $d$, so picking a basis for $C_n$, we can think of each  $f_n\in\M_{d}(\Z)$. Then $\tr(f_n)$ is the trace of this matrix. Recall that
\[\tr(ABA^{-1})=\tr(B),\]
so $\tr(f_n)$ doesn't depend on choice of basis.

We also have that $(f_n)_*$ acts on $H_n(\cal{C})/T_n(\cal{C})$, where $T_n$ denotes the torsion subgroup of $H_n(\cal{C})$. Thus, we can also consider $\tr((f_n)_*)$.

\begin{theorem}[Hopf Trace Formula]
We have
\[\sum_{n\geq 0} (-1)^n\tr(f_n)=\sum_{n\geq 0}(-1)^n\tr((f_n)_*)\]
\end{theorem}
\begin{remark}
Let $C_n=C_n^\Delta(X)$, and $f=(\id_X)_\#$. Then we get that
\[\sum_{n\geq 0}(-1)^n\operatorname{rank}(C_n^\Delta(X))=\sum_{n\geq 0}(-1)^nb_n(X).\]
\end{remark}
\begin{proof}
We have $B_n\subseteq Z_n\subseteq C_n$, all free abelian groups. First, pick a basis $\partial\sigma_1,\ldots,\partial\sigma_r$ for $B_n$; extend to (\textbf{not a basis; a }\textbf{maximal $\Z$-linearly independent set}) $z_1,\ldots,z_s\in Z_n$; then extend to a basis of $C_n$. Thus, we have a basis for $C_n$ that we've broken up into boundaries, plain old chains, and cycles.

We want to compute, for each element of the basis above, the coefficient of $v$ in $f_n(v)$. Let's call this $\lambda(f_n(v))$. 

Because $f_n$ is a chain map,
\[\lambda(f_n(\partial_{n+1}\sigma_j))=\lambda(f_{n+1}(\sigma_j))\]
Then
\[\sum_{n\geq 0}(-1)^n\tr(f_n)=\sum_{n\geq 0}(-1)^n\sum_{k=0}^n\lambda(f(z_k))\]
because all other terms cancel in pairs. Because the action of $f_*$ on $[z_i]\in H_i$ is just sending it to $[f(z_i)]$, the coefficents of the action of $f_*$ and of $f$ itself are the same. Thus, the left side is equal to
\[\sum_{n\geq 0}(-1)^n\tr((f_n)_*).\qedhere\]
\end{proof}