\classheader{2012-10-17}

\begin{theorem}[Fundamental Theorem of Homological Algebra]
Let $\cal{A}$, $\cal{B}$, and $\cal{C}$ be chain complexes, and suppose that
\[0\to\cal{A}\to\cal{B}\to\cal{C}\to0\]
is a SES of chain complexes. Then there exists a connecting homomorphism $\delta:H_n(\cal{C})\to H_{n-1}(\cal{A})$ and a long exact sequence of abelian groups
\[\xymatrix{\cdots\ar[r] & H_n(\cal{A})\ar[r]^{\phi_*} & H_n(\cal{B})\ar[r]^{\psi_*} & H_n(\cal{C})\ar[r]^{\delta} & H_{n-1}(\cal{A}) \ar[r]^{\phi_*} & H_{n-1}(\cal{B})\ar[r] & \cdots}\]
\end{theorem}
\begin{proof}
Our SES of chain complexes gives us the following diagram:
\[\xymatrix@M=0.13in{
0 \ar[r] & A_{n+1}\ar[r]^{\phi_{n+1}} \ar[d]^{\partial_{n+1}}& B_{n+1}\ar[r]^{\psi_{n+1}}\ar[d]^{\partial_{n+1}'} & C_{n+1}\ar[r] \ar[d]^{\partial_{n+1}''} & 0\\
0 \ar[r] & A_{n}\ar[r]^{\phi_{n}} \ar[d]^{\partial_{n}}& B_{n}\ar[r]^{\psi_{n}}\ar[d]^{\partial_{n}'} & C_{n}\ar[r] \ar[d]^{\partial_{n}''} & 0\\
0 \ar[r] & A_{n-1}\ar[r]^{\phi_{n-1}}& B_{n-1}\ar[r]^{\psi_{n-1}}& C_{n-1}\ar[r] & 0}\]
We want to define $\delta: H_n(\cal{C})\to H_{n-1}(\cal{A})$, mapping a homology class $[c]$ to a homology class $[a]$. Note that $H_n(\cal{C})=Z_n(\cal{C})/B_n(\cal{C})$ and $H_{n-1}(\cal{A})=Z_{n-1}(\cal{A})/B_{n-1}(\cal{A})$.

There are three key rules to use in any ``diagram chase'':
\begin{enumerate}
\item $\partial^2=0$
\item exactness
\item chain maps
\end{enumerate}

Given an $n$-cycle $c\in C_n$, so that $\partial c=0$, the fact that $\psi$ is onto implies that there is a $b\in B_n$ such that $\psi(b)=c$. Note that
\[0=\partial c=\partial\psi(b)=\psi(\partial b)\implies \partial b\in \ker(\psi)=\im(\phi).\]
Thus, there is an $a\in A_{n-1}$ such that $\phi(a)=\partial b$. Check that $a\in Z_{n-1}(\cal{A})$:
\[\phi(\partial a)=\partial\phi(a)=\partial\partial b=0.\]
But $\phi$ is injective, so $\partial a=0$.

Thus, given $[c]\in H_n(\cal{C})$, we can choose a representative $c$ in $Z_n(\cal{C})$ and produce an $a\in Z_{n-1}(\cal{A})$. Now we need to check that regardless of the representative we choose, the cycle $a$ represents the same class in homology. In other words, we want to check that the proposed map $\delta:H_n(\cal{C})\to H_{n-1}(\cal{A})$ is well-defined.

Thus, suppose instead of $c$ we'd chosen $c+\partial\theta$ as a representative of $[c]$, where $\theta\in C_{n+1}$. We want to show that $\delta$ will map it to the same $[a]$. 

Because $\psi$ is surjective, there is some $b'$ such that $\psi(b+b')=c+\partial\theta$. Thus, in particular $\psi(b')=\partial\theta$, and hence \[\psi(\partial b')=\partial \psi(b')=\partial\partial\theta=0.\]
This implies that $\partial b'\in \ker(\psi)=\im(\phi)$, hence $\partial b'=\phi(a')$ for some $a'$. Our goal is to show that there is an $a''$ such that $a'=\partial a''$.

Note that $\partial(b+b')=\partial b+\partial b'=\partial b+\partial\partial\theta=\partial b$, hence $\partial(b')=0$.

To finish, note that
\[\phi((a+a')-a)=\phi(a+a')-\phi(a)=\phi(a')=\partial b'=0.\]


Now we need to prove the exactness of the long exact sequence (LES). Let's just prove exactness at $H_n(\cal{B})$ for example. Given $[b]\in H_n(\cal{B})$ such that $\psi_*([b])=0$, we want to show that there is some $[a]\in H_n(\cal{A})$ such that $\phi_*([a])=[b]$.

We know that $\partial b=0$, and because $\psi_*([b])=0\in H_n(\cal{C})$, we have that $\psi(b)=\partial c$ for some $c\in C_{n+1}$. Thus $\psi(\partial b)=\partial\psi(b)=\partial\partial c=0$, so that $\partial b\in\ker(\psi)=\im(\phi)$, and hence there is some $a'$ such that $\phi(a')=\partial b$.

Finish as an exercise.
\end{proof}

Now let's apply this theorem to a pair of spaces $(X,A)$. The SES of chain complexes
\[0\to C_n(A)\to C_n(X)\to C_n(X,A)\to 0\]
produces a LES
\[\cdots\to H_{n+1}(X,A)\to H_n(A)\to H_n(X)\to H_n(X,A)\to H_{n-1}(A)\to\cdots\]
We'll prove later using this tool that if $(X,A)$ is reasonable, then $H_n(X,A)\cong H_n(X/A)$.

\begin{example}
Let $(X,A)=(\D^n,\partial\D^n)$ for $n\geq 1$. Note that $\partial \D^n=\S^{n-1}$.

Let $k>1$. The LES says that
\[H_{k+1}(\D^n,\S^{n-1})\to H_k(\S^{n-1})\to H_k(\D^n)\to H_k(\D^n,\S^{n-1})\to H_{k-1}(\S^{n-1})\to H_{k-1}(\D^n)\]
Any homology of the disk for $k>1$ is 0, because it is contractible. We can even show that directly in singular homology, because homology is a homotopy functor and the disk is homotopy equivalent to a point, whose singular homology we can compute. Thus, we get for any $k>1$
\[0\to H_k(\S^n)\to H_{k-1}(\S^{n-1})\to 0\]
and this just says that $H_k(\S^n)\cong H_{k-1}(\S^{n-1})$.

Let's look at what happens at the end.
\[\xymatrix@R=0.1in{H_1(\D^n,\S^{n-1})\ar[r]& H_0(\S^{n-1})\ar[r]& H_0(\D^n)\ar[r]& H_0(\D^n,\S^{n-1})\ar[r] & 0\\
\genfrac{}{}{0pt}{}{0\text{ if } n>1}{\Z\text{ if }n=1} & \genfrac{}{}{0pt}{}{\Z\text{ if } n>1}{\Z^2\text{ if }n=1} & \Z & 0}\]

Our inductive claim is that for all $n\geq 1$ and $k>1$,
\[H_k(\S^n)=\begin{cases}
0&\text{ if }k\neq 0,n\\
\Z&\text{ if }k=0,n.
\end{cases}\]
Here is a table:
\renewcommand{\arraystretch}{1.5}
\[\begin{array}{c|ccc}
& H_0 & H_1 & H_2 \\\hline
\S^0 & \Z^2 & 0 & 0\\
\S^1 & \Z & \Z & 0 \\
\S^2 & \Z & 0 & \Z\end{array}\]
\end{example}
\renewcommand{\arraystretch}{1}