\classheader{2012-11-09}


\subsection*{The Universal Coefficient Theorem}

Fix an abelian group $G$, and let $\cal{C}=\{C_n,\partial_n\}$ be a chain complex. As we discussed last time, this gives us a cochain complex $\cal{C}^*=\{C_n^*,\delta^n=\partial_{n-1}^*\}$, and then the cohomology groups $H^n(\cal{C};G)$ with coefficients in $G$.

Last time, we defined a map $\Psi:H^n(\cal{C};G)\to\Hom(H_n(\cal{C}),G)$. It is easy to show that $\Psi$ is onto, so we get a short exact sequence
\begin{center}
\begin{tikzcd}
1\ar{r} & \ker(\Psi) \ar{r} & H^n(\cal{C};G) \ar{r} & \Hom(H_n(\cal{C}),G) \ar{r} & 1
\end{tikzcd}
\end{center}
We want to compute $\ker(\Psi)$, since it represents the difference between homology and cohomology.

You should read in Hatcher about the functor $\Ext(-,G)$, which goes from $\Ab$ to $\Ab$, and also read about free resolutions. The key step will be showing that $\ker(\Psi)=\Ext(H_{n-1}(\cal{C}),G)$. This then imples the universal coefficient theorem, because we get a short exact sequence
\begin{center}
\begin{tikzcd}
1\ar{r} & \Ext(H_{n-1}(\cal{C}),G) \ar{r} & H^n(\cal{C};G) \ar{r} & \Hom(H_n(\cal{C}),G) \ar{r} & 1
\end{tikzcd}
\end{center}
The following proposition explains how to compute $\Ext(-,G)$.
\begin{proposition}
$\text{}$

\begin{enumerate}
\item $\Ext(A\oplus B,G)=\Ext(A,G)\oplus \Ext(B,G)$.
\item $\Ext(A,G)=0$ if $A$ is a free abelian group.
\item $\Ext(\Z/n\Z,G)=G/nG$, where $nG=\{ng\mid g\in G\}$.
\end{enumerate}
\end{proposition}

In fact, when $G=\Z$, we have
\[\Ext(H_{n-1}(\cal{C},G))\cong \frac{H_n(\cal{C})}{\text{torsion}(H_n(\cal{C}))}\oplus\text{torsion}(H_{n-1}(\cal{C})).\]

\begin{remark}
The functor $\Ext(A,G)$ is called that because it has to do with extensions of $G$ by $A$, i.e. ways of picking a group $\Gamma$ so that
\begin{center}
\begin{tikzcd}
1 \ar{r} & A \ar{r} & \Gamma \ar{r} & G \ar{r} & 1
\end{tikzcd}
\end{center}
Thus, cohomology is a key tool for the classification of groups.
\end{remark}


\subsection*{Cohomology of Spaces}

\providecommand{\CW}{\ensuremath{\text{\small CW}}}

For any topological space $X$ (respectively, any $\Delta$-complex, CW-complex), we define
\begin{align*}
H^n(X;G)&:=H^n(\{C_n^{\text{sing}}(X)\})\\
H_{\CW}^n(X;G)&:=H^n(\{C_n^{\CW}(X)\})\\
H_\Delta^n(X;G)&:=H^n(\{C_n^\Delta(X)\})
\end{align*}
Then because the universal coefficient theorem is natural in $\cal{C}$, and the isomorphisms between the various homology theories are natural, we can conclude that
\[H^n(X;G)\cong H_\Delta^n(X;G)\cong H_{\CW}^n(X;G)\]
when they are all defined (i.e. when $X$ is a $\Delta$-complex, CW-complex). Moreover, naturality together with the corresponding fact about homology implies that $f\sim g$ we have $f^*=g^*$ (so that $H^i$ is a homotopy functor), and we even get the long exact sequence of a pair, excision, and Mayer-Vietoris for cohomology, all from naturality.

\subsection*{Cup Product}

Fix a ring $R$ (e.g. $R=\Z,\,\Q,\,\Z/d\Z,\,\R,\,\Q_p,\,\ldots$). Let $X$ be a space. Considering the underlying additive abelian group of $R$, we get the abelian groups $H^n(X;R)$ for all $n\geq 0$, and we let
\[H^*(X;R)=\bigoplus_{n\geq 0}H^n(X;R)\]
which is a (graded) abelian group. Our goal is to make $H^*(X;R)$ into a ring. This additional ring structure on cohomology will let us distinguish $\S^1\vee\S^1\vee\S^2$ and $\T^2$, even though\\
\[H_i(\S^1\vee\S^1\vee\S^2)=\left\{\begin{array}{cc}
\Z &\;\; i=0\\
\Z^2 &\;\; i=1\\
\Z & \;\;i=2 
\end{array}\right\}=H_i(\T^2).\]
For $a\in C^i(X;R)=\Hom(C_i(X);R)$ and $b\in C^j(X;R)$, for any $(\sigma:[v_0v_1\cdots v_{i+j}]\to X)\in C_{i+j}(X)$ we define $a\cupprod b\in C^{i+j}(X;R)$ by
\[(a\cupprod b)(\sigma)=a(\sigma|_{[v_0v_1\cdots v_i]})\;\;\mathclap{\underset{\substack{\uparrow\\ \text{mult. in }R}}{\cdot}}\;\;b(\sigma|_{[v_i\cdots v_{i+j}]}).\]
We have the ``graded product rule for $\delta$'':
\[\delta(a\cupprod b)=\delta a\cupprod b+(-1)^ia\cupprod \delta b\]
This implies that if $\delta a=\delta b=0$, then $\delta(a\cupprod b)=0$ and $\delta a\cupprod b=a\cupprod \delta b=0$, so that the cup product map \[\cupprod:C^i(X;R)\times C^j(X;R)\to C^{i+j}(X;R)\] induces a map in homology \[\cupprod:H^i(X;R)\times H^j(X;R)\to H^{i+j}(X;R),\] defined by $[a]\cupprod [b]=[a\cupprod b]$. This is a bilinear map, so passing to the tensor product and putting all degrees together, we get a homomorphism
\[\cupprod:H^*(X;R)\otimes H^*(X;R)\to H^*(X;R),\]
In summary, $H^*(X;R)$ is a graded commutative ring $a\cupprod b=(-1)^{ij}b\cupprod a$, where addition is just addition of cocycles, and multiplication is $\cupprod$. For any map $f:X\to Y$, we get an induced ring homomorphism $f^*(a\cupprod b)=f^*(a)\cupprod f^*(b)$.

The additive and multiplicative identity are the constant functions $0,1\in H^0(X;R)=\Hom(C_0(X);R)$.




