\classheader{2012-10-31}

The midterm will be Friday, November 2 from 11am to 12:30pm, in Ryerson 352 (the Barn).

\subsection*{Cellular Homology}

\begin{proposition}
Let $X$ be a CW-complex. Then for all $n\geq 0$,
\begin{enumerate}
\item $\widetilde{H}_i(X^{(n)},X^{(n-1)})=\left.\begin{cases}
\Z^d &\text{ if }i=n\!\\ 0 & \text{ if }i\neq n\!
\end{cases}\right\}$ where $d$ is the number of $n$-cells in $X$.
\item $\widetilde{H}_i(X^{(n)})=0$ for all $i>n$.
\item The inclusion $X^{(n)}\hookrightarrow X$ induces an isomorphism $i_*:H_k(X^{(n)})\to H_k(X)$ for all $k<n$.
\end{enumerate}
\end{proposition}
\begin{proof}
Consider the LES of the pair $(X^{(n)},X^{(n-1)})$. Also, observe that $\widetilde{H}_i(X^{(n)},X^{(n-1)})\cong \widetilde{H}_i(X^{(n)}/X^{(n-1)})$, and $X^{(n)}/X^{(n-1)}$ is a wedge of spheres. Lastly, we have a commutative diagram
\begin{center}
\begin{tikzcd}
\D_\alpha^n \ar{r} \ar{rd} \ar{d} & X^{(n)} =\dfrac{X^{(n-1)}\sqcup \coprod_\alpha \D_\alpha^n}{x\sim \phi_\alpha(x)\text{ for all }x\in\partial \D_\alpha^n}\ar{d}\\
\D_\alpha^n/\partial \D_\alpha^n \ar{r} & X^{(n)}/X^{(n-1)}\cong\bigvee \S^n
\end{tikzcd}
\end{center}
\end{proof}

\begin{definition}
The group of cellular $n$-chains on $X$ is defined to be $C_n^{\CW}(X):= H_n(X^{(n)},X^{(n-1)})$. Thus, $C_n^{\CW}(X)\cong\Z^d$ where $d$ is the number of $n$-cells in $X$.
\end{definition}

Our goal is to define the boundary homomorphisms $d_n:C_n^{\CW}(X)\to C_{n-1}^{\CW}(X)$ and show that they satisfy $d_{n-1}\circ d_n=0$. 

Thinking about the definition of $C_n^{\CW}$, we recall that in the LES of $(X^{(n)},X^{(n-1)})$, we had a boundary map $\partial_n:H_n(X^{(n)},X^{(n-1)})\to H_{n-1}(X^{(n-1)})$.

We can also consider the map $(i_{n-1})_*:H_{n-1}(X^{(n-1)})\to H_{n-1}(X^{(n)},X^{(n-1)})$.

Now, we define $d_n=i_{n-1}\circ \partial_n$, which has the right domain and range: $d_n$ goes from $H_n(X^{n},X^{(n-1)})=C_n^{\CW}(X)$ to $H_{n-1}(X^{(n-1)},X^{(n-2)})=C_{n-1}^{\CW}(X)$.

You should check on your own $d_{n-1}\circ d_n=0$ for all $n\geq 0$. Thus, $\cal{C}^{\CW}(X)=\{C_n^{\CW}(X),d_n\}$ is a chain complex.

\begin{definition}
We define the CW homology of $X$ to be $H_n^{\CW}(X):=H_n(\cal{C}^{\CW}(X))$.
\end{definition}

\begin{theorem}
Let $X$ be any CW-complex. Then $H_n^{\CW}(X)\cong H_n(X)$ for all $n\geq 0$.
\end{theorem}
\begin{corollary}
We then have that
\begin{itemize}
\item $H_n(X)=0$ for all $n>\dim(X)$.
\item $\rank(H_n(X))\leq\#$ of $n$-cells of $X$.
\item $H_n^{\CW}(X)$ is a homotopy invariant.
\end{itemize}
\end{corollary}
\begin{remark}
By the topological invariance of $H_n^{\CW}$ and the Hopf trace formula, we can compute
\[\chi(X)=\sum_{n\geq 0}(-1)^n\rank(C_n^{\CW}(X)).\]
For example,
\[\chi(\S^n)=1+(-1)^n=\begin{cases}
0 & \text{ if }n\text{ odd},\\
2 & \text{ if }n\text{ even}.
\end{cases}\]
\[\chi(\R P^n)=1-1+1-\cdots+(-1)^n=\begin{cases}
0 & \text{ if }n\text{ odd},\\
1 & \text{ if }n\text{ even}.
\end{cases}\]
\[\chi(\Sigma_g)=1-2g+1=2-2g.\]
\end{remark}

To compute $H_n^{\CW}(X)$, we need to understand the boundary maps $d_n$. We have
\[C_n^{\CW}(X)=\text{$\Z$-span of }\{\sigma_\alpha:\D_\alpha^n\to X^{(n)}\}\]
\[C_{n-1}^{\CW}(X)=\text{$\Z$-span of }\{\tau_\beta:\D_\beta^{n-1}\to X^{(n-1)}\}\]
Consider the composition of the attaching map of an $n$-cell $\sigma_\alpha$ with the quotient map sending $X^{(n-1)}$ to $X^{(n-1)}/X^{(n-2)}$:
\begin{center}
\begin{tikzpicture}
\node at (-1,2.3) {$\partial\D^n$};
\draw[thick] (-1,1) circle (1);
\node[inner sep=0pt,outer sep=0pt,minimum size=5pt,fill,circle] (a) at (3,-1) {};
\path[thick] (a.center) edge[out=0,in=100,loop,distance=2cm] node[above right] {$\tau_1(\D^{n-1})$} (a.center);
\path[thick] (a.center) edge[out=120,in=220,loop,distance=2cm] node[above left] {$\tau_2(\D^{n-1})$} (a.center);
\node[rotate=20] (b) at ($(3,-1)+(300:0.5)$) {$\cdots$};
\path[thick,->,>=stealth] (0.1,1) edge[out=0,in=120,shorten >=5pt] node [above right] {$\sigma_\alpha$} (2.8,-0.5);
\node at (6,-1) {$=X^{(n-1)}/X^{(n-2)}$};
\end{tikzpicture}\vspace{-0.2in}
\end{center}
Then, for all $\beta$, consider the composition of that with the quotient map from $X^{(n-1)}/X^{(n-2)}$ to just one of the $(n-1)$-cells $\tau_\beta$:
\begin{center}
\begin{tikzcd}[column sep=0.9in]
\S^{n-1}=\partial\D_\alpha^n \ar{r}{\sigma_\alpha|_{\partial\D_\alpha^n}} & X^{(n-1)}\ar{r}{\text{quotient}} & X^{(n-1)}/X^{(n-2)} \ar{r}{\substack{\text{more}\\\text{quotient}}} & \tau_\beta(\D^{n-1})=\S^{n-1}
\end{tikzcd}
\end{center}
This is a map $\psi_{\alpha,\beta}=\S^{n-1}\to\S^{n-1}$.
\begin{lemma}[Hatcher, p.141]
For all $\alpha,\beta$, we have
\[d_n(\sigma_\alpha)=\sum_{\beta}d_{\alpha,\beta}\tau_\beta\]
where $d_{\alpha,\beta}=\deg(\psi_{\alpha,\beta})$.
\end{lemma}

Let's work out the homology of the torus this way. Let $X=\T^2$. The CW chain complex is
\begin{center}
\begin{tikzcd}[row sep=0.05in]
C_2 & C_1 & C_0 & \\
\Z\ar{r}{d_2} & \Z^2\ar{r}{d_1} & \Z\ar{r} & 0
\end{tikzcd}
\end{center}
We obviously have $d_1=0$. Now let's think about $\psi_{\alpha a}$. It may be helpful to note that the map sending $\partial\D_\alpha^2$ to $\tau_a(\D^1)$ factors through the quotient collapsing the $b$ arcs:
\begin{center}
\begin{tikzpicture}[scale=1]
\node at (-3.8,2.6) {$\partial\D_\alpha^2$};
\begin{scope}[outer sep=2pt,decoration={markings,mark=at position 0.55 with {\arrow[scale=1]{angle 90}}}]
\draw[thick,red,postaction={decorate}] (-1,2) arc (0:90:1);
\node at ($(-2,2)+(45:1.3)$) {$a$};
\draw[thick,blue,postaction={decorate}] (-2,3) arc (90:180:1);
\node at ($(-2,2)+(135:1.3)$) {$b$};
\draw[thick,red,postaction={decorate}] (-2,1) arc (270:180:1);
\node at ($(-2,2)+(235:1.3)$) {$a$};
\draw[thick,blue,postaction={decorate}] (-1,2) arc (360:270:1);
\node at ($(-2,2)+(315:1.3)$) {$b$};
\draw[thick] ($(-2,2)+(0:0.93)$) -- ($(-2,2)+(0:1.07)$);
\draw[thick] ($(-2,2)+(90:0.93)$) -- ($(-2,2)+(90:1.07)$);
\draw[thick] ($(-2,2)+(180:0.93)$) -- ($(-2,2)+(180:1.07)$);
\draw[thick] ($(-2,2)+(270:0.93)$) -- ($(-2,2)+(270:1.07)$);
\end{scope}
\begin{scope}[decoration={snake,amplitude=1pt}]
\draw[thick,->,>=stealth,decorate] (-2,0.6) -- (-2,-0.67);
\end{scope}
\begin{scope}[outer sep=2pt,decoration={markings,mark=at position 0.5 with {\arrow[scale=1]{angle 90}}}]
\draw[thick,red,postaction={decorate}] ($(-2,-2)+(315:1)$) arc (315:135:1);
\node at ($(-2,-2)+(45:1.3)$) {$a$};
\draw[thick,red,postaction={decorate}] ($(-2,-2)+(315:1)$) arc (315:495:1);
\node at ($(-2,-2)+(235:1.3)$) {$a$};
\draw[thick] ($(-2,-2)+(315:0.93)$) -- ($(-2,-2)+(315:1.07)$);
\draw[thick] ($(-2,-2)+(135:0.93)$) -- ($(-2,-2)+(135:1.07)$);
\end{scope}
\begin{scope}[decoration={markings,mark=at position 0.5 with {\arrow[>=angle 90]{>}}}]
\node[inner sep=0pt,outer sep=0pt,minimum size=5pt,fill,circle] (a) at (4,0) {};
\path[thick,dashed] (a) edge[out=320,in=40,loop,distance=2cm,postaction={decorate}] node [label=right:$b$] {} (a);
\path[thick] (a) edge[out=220,in=140,loop,distance=2cm,postaction={decorate}] node [label=left:$a$] {} (a);
\end{scope}
\path[thick,->,>=stealth] (-0.7,2) edge[out=0,in=140] (3.1,0.6);
\path[thick,->,>=stealth] (-0.7,-2) edge[out=0,in=220] (3.1,-0.6);
\end{tikzpicture}
\end{center}
We clearly have that $\deg(\psi_{\alpha a})=1-1=0$, and similarly, $\deg(\psi_{\alpha b})=0$. Therefore $d_2= 0$, and 
\[H_i^{\CW}(\T^2)\cong C_i^{\CW}(\T^2)=\begin{cases}
\Z & \text{ if }i=0,2\\
\Z^2 & \text{ if }i=1\\
0 & \text{ otherwise.}
\end{cases}\]