\classheader{2012-11-19}


\subsection*{The Mayer-Vietoris Argument}

\begin{definition}
An open cover $\{U_i\}$ of an $n$-manifold $M$ is a good cover if each $U_i\cong\R^n$ and for all $i_1,\ldots,i_r$, $U_{i_1}\cap\cdots\cap U_{i_r}\cong\R^n$ or is empty.
\end{definition}

\begin{theorem}
Every smooth manifold $M$ has a good cover. Of course, if $M$ compact, then we can choose the good cover to be finite.
\end{theorem}
\begin{proof}
First, we will prove that any smooth manifold $M$ admits a Riemannian metric. Take a cover of $M$ by coordinate patches $U_i$, and on each coordinate patch, let $\langle\,\cdot\,,\,\cdot\,\rangle_i$ be the pullback of the standard Riemannian metric on $\R^n$. Let $P_i$ be a partition of unity subordinate to the cover. Then $\langle\,\cdot\,,\,\cdot\,\rangle=\sum P_i\langle\,\cdot\,,\,\cdot\,\rangle_i$ is a Riemannian metric on $M$.

Now we appeal to a theorem of Whitehead (or Alexander): if $M$ is any Riemannian manifold, then every $m\in M$ has a convex neighborhood $U_m$, where convex means that there is a unique geodesic in $U_m$ between any two points. Because the intersection of convex sets is convex, and convex $\cong\R^n$, we are done.
\end{proof}

\begin{remark}
Good covers are cofinal in the set of covers ordered by refinement.
\end{remark}

\begin{theorem}
If a manifold $M$ has a finite good cover, then the $i$th homology $H_i(M)$ of $M$ is finitely generated for all $i\geq 0$.
\end{theorem}
\begin{proof}
Mayer-Vietoris implies that for open $U,V\subseteq M$, we have an exact sequence
\begin{center}
\begin{tikzcd}
\cdots\ar{r} & H_i(U)\oplus H_i(V)\ar{r} & H_i(U\cup V) \ar{r}  & H_{i-1}(U\cap V)\ar{r} & \cdots
\end{tikzcd}
\end{center}
Thus, by the rank-nullity theorem,
\[\dim(H_i(U\cup V))\leq\dim(H_{i-1}(U\cap V))+\dim(H_i(U))+\dim(H_i(V)),\]
where $\dim$ here means the minimial number of generators.

If $U_1,\ldots,U_r$ is a good cover, then $U=U_1\cup\cdots\cup U_{r-1}$ is an open submanifold of $M$, and we can let $V=U_r$. By induction on the size of a minimal open cover, we have that $H_i(U)$ and $H_i(V)$ are finitely generated, and thus we are done by our observation above.

The base case is just $M=\R^n$ which is trivial.
\end{proof}

\subsection*{K\"unneth theorem via a Mayer-Vietoris argument}

\begin{theorem}
Let $H^*(X)=H^*(X;\Q)$. Then for any manifolds $M_1,M_2$ with finite good covers (in fact we only need one of them to have this), we have
\[H^*(M_1\times M_2)\cong H^*(M_1)\otimes H^*(M_2).\]
\end{theorem}
\begin{proof}
We will show the map $\psi:H^*(M_1)\otimes H^*(M_2)\to H^*(M_1\times M_2)$ defined by sending $a\smalltensor b$ to $a\times b$ is an isomorphism.

Let $U_1,\ldots,U_r$ be a good cover of $M$. We induct on $r$.

For $r=1$, since $\R^n\times M_2\simeq M_2$, we have $H^*(M_2)\cong H^*(\R^n\times M_2)\cong H^*(\R)\otimes H^*(M_2)$.

Now suppose that $\{U,V\}$ is a cover of $M_1$. Mayer-Vietoris implies that we have a long exact sequence
\begin{center}
\begin{tikzcd}
\cdots\ar{r} & H^i(U\cap V) \ar{r} & H^i(U)\oplus H^i(V) \ar{r} & H^i(U\cap V) \ar{r} & \cdots
\end{tikzcd}
\end{center}
Tensoring with $H^{n-i}(M_2)$ throughout (?), we get\vspace{-0.1in}
\begin{center}
\begin{tikzcd}[column sep=12pt]
\cdots\ar{r} & H^i(U\cap V) \otimes H^{n-i}(M_2)\ar{r} & (H^i(U)\oplus H^i(V) )\otimes H^{n-i}(M_2)\ar{r} & H^i(U\cap V) \otimes H^{n-i}(M_2)\ar{r} & \cdots
\end{tikzcd}
\end{center}
which is still exact because tensoring with a vector space preserves exactness.

Summing the above exact sequence over all $i$, we get an exact sequence
\begin{center}
\begin{tikzcd}
\cdots \ar{r} & \underbrace{\bigoplus_{i=1}^n[H^i(U\cup V)\otimes H^{n-i}(M-2)]}_{\mathclap{n\text{th degree of }H^*(U\cup V)\otimes H^*(M_2)}} \ar{r} & \cdots
\end{tikzcd}
\end{center}
which $\psi$ maps to
\begin{center}
\begin{tikzcd}
\cdots \ar{r} & H^n((U\cup V)\times M_2) \ar{r} & \cdots
\end{tikzcd}
\end{center}
Check that $\psi$ forms a chain map between these two exact sequences, and use the 5 lemma. (?)
\end{proof}


\subsection*{Orientation Revisited}

Let $x\in\R^n$, and let $B$ be a neighborhood of $x$. We have
\begin{align*}
H_i(\R^n,\R^n-\{x\}) & \underset{\substack{\uparrow\\ \mathclap{\text{excision}}}}{\cong} H_i(\R^n,\R^n-\overline{B})\\
&\cong \widetilde{H}_i(\R^n/(\R^n-\overline{B}))\\
&\cong \widetilde{H}_i(\overline{B}/\partial B)\\
&\cong \widetilde{H}_i(\S^n)\\
&\cong\begin{cases}
\Z & \text{ if }i=n,\\
0 & \text{ otherwise.}
\end{cases}
\end{align*}
Now let $M$ be a manifold. For any $x\in M$, we can find a neighborhood $B$ of $x$ lying in a coordinate chart. By the same reasoning,
\begin{align*}
H_i(M,M-\{x_i\})&\cong H_i(B,B-\{x\})\\
&\cong\begin{cases}
\Z & \text{ if }i=n,\\
0 & \text{ otherwise.}
\end{cases}
\end{align*}
A choice of generator of $H_n(M,M-\{x\})\cong \Z$ is called a local orientation at $x$. 

Now let $M$ be a triangulated, closed, and connected $n$-manifold for which there exists a compatible orientation on each simplex of the triangulation. Let $\{\sigma_i\}$ be the set of $n$-simplices, and let $\sigma=\sum \sigma_i\in C_n(M)$; this sum makes sense because $M$ is compact, so there are finitely many terms. Because $M$ is a manifold without boundary, and the orientations on $\sigma_i\cap \sigma_j$ cancel, we have that $\partial\sigma=\sum \partial \sigma_i=0$. Thus, $\sigma\in Z_n(M)$. But $C_{n+1}(M)=0$ because $M$ is $n$-dimensional; therefore, $[\sigma]\in H_n(M)$ is of infinite order. This homology class is called the fundamental class of $M$, and is denoted $[M]$.

\begin{corollary}
The Klein bottle $K$ is not orientable since $H_2(K)=0$.
\end{corollary}

Check for yourself that for any $x\in M$, the inclusion map $i:(M,\varnothing)\hookrightarrow (M,M-x)$ induces a map $i_*:H_n(M)\to H_n(M,M-x)$ that takes a generator to a generator.



