\classheader{2013-01-29}

This week, we'll cover the connection with algebraic topology.

\begin{theorem}
If $X$ is a compact connected (non-empty) orientable $n$-manifold, then
\begin{itemize}
\item[(a)] $H_n(X)\cong \Z$,
\item[(b)] for all $x\in X$, there is a natural map $H_n(X)\xrightarrow{\;\cong\;} H_n(X,X-x)\cong\Z$.
\end{itemize}
\end{theorem}
\begin{definition}
An orientation of $X$ is a generator of $H_n(X)$, and an oriented manifold is a pair $(X,\theta)$ where $\theta$ is an orientation of $X$.
\end{definition}
\begin{theorem}
If we remove the hypothesis that $X$ is orientable, then it is still true that
\begin{itemize}
\item[($a'$)] $H_n(X;\Z/2\Z)\cong\Z/2\Z$,
\item[($b'$)] there is a natural map $H_n(X;\Z/2\Z)\xrightarrow{\;\cong\;} H_n(X,X-x;\Z/2\Z)$.
\end{itemize}
\end{theorem}
\begin{definition}
Let $f:X\to Y$ be continuous, where both $(X,\theta_X)$ and $(Y,\theta_Y)$ are compact connected oriented $n$-manifolds. The degree of $f$ is the unique integer $\deg(f)$ such that $f_*(\theta_X)=\deg(f)\cdot\theta_Y$ where $f_*:H_n(X)\to H_n(Y)$ is the induced map on homology. This does not depend on the choice of $\theta_X$ or $\theta_Y$.

When $X$ and $Y$ are not assumed to be orientable, we can still define the degree of $f$ in $\Z/2\Z$.
\end{definition}

\begin{definition}
Let $X$ and $Y$ be oriented $n$-manifolds, and let $x\in X$ be an isolated point of $f^{-1}(f(x))$. Let $y=f(x)$. Thus, we have a neighborhood $U\subseteq X$ of $x$ such that $f^{-1}(f(x))\cap U=\{x\}$. In fact, we can get a neighborhood $U\subseteq X$ of $x$ and $U'\subseteq Y$ of $y$ such that $f(U)\subseteq U'$ and $f(U\setminus\{x\})\subseteq U'\setminus\{y\}$. The map $f$ induces
\begin{center}
\begin{tikzcd}
{} & H_n(U,U-x) \ar{d}{\text{excision}}[swap]{\cong} \ar{r}{\text{`` $f_*$ ''}} & H_n(U',U'-y) \ar{d}{\text{excision}}[swap]{\cong} & \\
 H_n(X) \ar{r}{\cong} & H_n(X,X-x) &  H_n(Y,Y-y) & H_n(Y)\ar{l}[swap]{\cong}
\end{tikzcd}

\begin{tikzcd}
\theta_X\in H_n(X) \ar[mapsto]{r} & \theta_X(x)\in H_n(U,U-x) \ar[mapsto]{r} & \theta_Y(y)\in H_n(U',U'-y) \ar[mapsto]{r} & \theta_Y\in H_n(Y)
\end{tikzcd}
\end{center}
We then define the local degree of $f$ at $x$ to be the unique integer $\deg_x(f)$ such that
\[\deg_x(f)\cdot \theta_Y(y)=\text{``$f_*$''}\theta_X(x).\]
\end{definition}
\begin{theorem}
Let $X$ and $Y$ both be compact connected oriented (non-empty) $n$-manifolds, and let $f:X\to Y$ be continuous. Let $y\in Y$, and assume that $f^{-1}(y)$ is finite. Then
\[\deg(f)=\sum_{x\in f^{-1}(y)}\deg_x(f).\]
\end{theorem}
\begin{proof}
We will use singular homology. We see that there is a neighborhood $U_y'$ of $y$ in $Y$ such that $f^{-1}(U_y)=\coprod_{x\in f^{-1}(y)}U_x$ where $U_x$ is a neighborhood of $x$. We get an open cover of $X$, consisting of
\[\{U_x\mid x\in f^{-1}(y)\}\cup\{X\setminus f^{-1}(y)\}\]
\pagebreak 

{\color{myred}{\rule{\textwidth}{0.02in}}\vspace{-0.1in}\bf
\begin{center}
$\blacktriangledown$\quad Why is this true?\quad $\blacktriangledown$
\end{center}}
The homology class $\theta_X$ is representable by a cycle
\[\xi=\sum_{x\in f^{-1}(y)} \sigma_x+\eta,\]
where $\sigma_x\in C_n(U_x)$ and $\eta\in C_n(X-f^{-1}(y))$.

{\color{myred}{\rule{\textwidth}{0.02in}}}

Because $\xi$ is a cycle, we have $\partial \xi=0\in C_{n-1}(X)$. Thus, $f_*(\theta_X)$ is represented by
\begin{center}
\qquad\qquad\qquad\qquad\begin{tikzcd}
\mathllap{\sum f(\sigma_x)+f(\eta)\in }\,Z_n(Y)\, \ar[hook]{r}\ar{d} & C_n(Y) \ar{d}\\
Z_n(Y,Y-y) \ar[hook]{r} & C_n(Y,Y-y)
\end{tikzcd}
\end{center}
We have $\eta\in C_n(X-f^{-1}(y))$, and $f(X-f^{-1}(y))\subseteq Y-\{y\}$, so $f(\eta)\in C_n(Y-y)$. Thus, when we send $\theta_X\in H_n(X)$ via
\begin{center}
\begin{tikzcd}
H_n(X)\ar{r}{f_*} & H_n(Y) \ar{r}{\cong} & H_n(Y,Y-y)
\end{tikzcd}
\end{center}
the image can be represented by $\sum_{x\in f^{-1}(y)}f(\sigma_x)$, because $f(\eta)$ is supported on $Y-y$.

{\color{myred}{\rule{\textwidth}{0.02in}}\vspace{-0.1in}\bf
\begin{center}
$\blacktriangledown$\quad Missing content\quad $\blacktriangledown$
\end{center}}

It follows that $\partial \sigma_x\in C_{n-1}(U_x-\{x\})$ for all $x\in f^{-1}(y)$.

We claim that $\sigma_x\in C_n(U_x,U_x-\{x\})$ is a generator of $H_n(U_x,U_x-x)$.

\begin{center}
\begin{tikzcd}
H_n(X) \ar{r}{\cong} & H_n(X,X-x)\\
& H_n(U_x,U_x-x) \ar{u}[swap]{\text{excision}}
\end{tikzcd}
\end{center}

$\sigma_x\in Z_n(U_x,U_x-x)$
\[f_*(\overline{\sigma_x})=\deg_x(f)\cdot\theta_Y\]
where $\overline{\sigma_x}$ denotes the corresponding homology class.

{\color{myred}{\rule{\textwidth}{0.02in}}}
\end{proof}

\begin{corollary}
Let $X$ and $Y$ be as before, and now assume that $f$ is $\Cinfty$. Suppose that $y\in Y$ is a regular value of $f$. Then $f^{-1}(y)$ is finite and $\deg_x(f)=\pm 1$ for all $x\in f^{-1}(y)$.
\end{corollary}
\begin{proof}
For each $x\in f^{-1}(y)$, the restriction $(U_x,U_x-x)\to (U_y',U_y'-y)$ is a diffeomorphism, which induces $H_n(U_x,U_x-x)\xrightarrow{\cong} H_n(U_y',U_y'-y)$, so $\deg_x(f)=\pm 1$.
\end{proof}
I haven't shown you how to determine whether the local degree is $+1$ or $-1$, but it is still useful to compute modulo 2. For $f$ as above (we can drop the hypothesis that $X$ and $Y$ are orientable), then if $y$ is a regular value of $f$ and $\# f^{-1}(y)\notin 2\Z$, then $f$ is not null-homotopic.

\begin{example}
Let $P\in \C[z]$ be a polynomial with $\deg(P)=d>0$. Because the map $P:\C\to\C$ is proper, it extends to one-point-compactifications, so we have
\begin{center}
\begin{tikzcd}
{}&\C \ar{r}{P} \ar[hook]{d} & \C\ar[hook]{d}\\
\S^2\ar{r}{\cong} & \C\cup\{\infty\} \ar{r}[swap]{P^*} & \C\cup\{\infty\} \ar{r}{\cong} & \S^2
\end{tikzcd}
\end{center}
and $\deg(P^*)=d$. 

{\color{myred}{\rule{\textwidth}{0.02in}}\vspace{-0.1in}\bf
\begin{center}
$\blacktriangledown$\quad Why did I write this?\quad $\blacktriangledown$
\end{center}}

For all $a\in\C$, we have $\# P^{-1}(a)=d$.  $P(z)-P(a)$, $P'(z)\neq 0$ (?)

{\color{myred}{\rule{\textwidth}{0.02in}}}
\end{example}

\subsection*{Vector bundles}

\begin{definition}
A map $p:E\to B$ is a fiber bundle if there is an open covering $\{U_\alpha\}_{\alpha\in A}$ of $B$, homeomorphisms $\psi_\alpha:p^{-1}(U_\alpha)\to U_\alpha\times F$ where $F$ is a (possibly varying) topological space, such that for each $\alpha\in A$, the following diagram commutes
\begin{center}
\begin{tikzcd}[column sep=0.1in]
p^{-1}(U_\alpha) \ar{rr}{\psi_\alpha} \ar{rd}[swap]{p|_{p^{-1}(U_\alpha)}} && U_\alpha\times F \ar{dl}{p_1}\\
& U_\alpha
\end{tikzcd}
\end{center}
If $B$ is connected, then all of the spaces $F$ must be homeomorphic, and we say that $p$ is a fiber bundle with fiber $F$.
\end{definition}
%{\color{myred}{\rule{\textwidth}{0.02in}}\vspace{-0.1in}\bf
%\begin{center}
%$\blacktriangledown$\quad I missed the details here / possible mistakes\quad $\blacktriangledown$
%\end{center}}

Let $U_\alpha,U_\beta$ be two elements of this open cover of $B$. We can construct a commutative diagram:
\begin{center}
\begin{tikzcd}[column sep=0.8in]
(U_\alpha\cap U_\beta)\times F \ar{rd} & p^{-1}(U_\alpha \cap U_\beta)\ar{l}[swap]{\psi_\beta|_{p^{-1}(U_\alpha\cap U_\beta)}} \ar{r}{\psi_\alpha|_{p^{-1}(U_\alpha\cap U_\beta}} \ar{d} & (U_\alpha \cap U_\beta)\times F\ar{ld}\\
&U_\alpha\cap U_\beta
\end{tikzcd}
\end{center}

Let $\phi_{\alpha\beta}=\psi_\alpha|_{p^{-1}(U_\alpha\cap U_\beta)}\circ\psi_\beta|_{p^{-1}(U_\alpha\cap U_\beta)}^{-1}$, which is a homeomorphism making the following diagram commute:
\begin{center}
\begin{tikzcd}[column sep=0.04in]
(U_\alpha\cap U_\beta)\times F \ar{rd} \ar{rr}{\phi_{\alpha\beta}} && (U_\alpha\cap U_\beta)\times F \ar{ld}\\
& U_\alpha\cap U_\beta
\end{tikzcd}
\end{center}
When it is the case that that $\phi_{\alpha\beta}$ is a \emph{diffeomorphism} for all $\alpha$, $\beta$, we will say that ``$U_\alpha\cap U_\beta\to\mathrm{Diffeo}(F)$''.

Note that for all $x\in U_\alpha\cap U_\beta\cap U_\gamma$, we have \[\phi_{\alpha\beta}(x)\phi_{\beta\gamma}(x)=\phi_{\alpha\gamma}(x).\] This is called the cocycle condition.

%{\color{myred}{\rule{\textwidth}{0.02in}}}

Conversely, given an open cover $\{U_\alpha\}$, and maps $\phi_{\alpha\beta}$ satisfying the cocycle condition, we can construct a fiber bundle $E\to B$ by taking $\coprod_{\alpha\in A}(U_\alpha\times F)$, and quotienting by the equivalence relation $(x,v)\in U_\alpha\times F \sim (x,\phi_{\alpha\beta}(x)v)\in U_\beta\times F$ for all $x\in U_\alpha\cap U_\beta$ and $v\in F$.


Now we will define the notion of a vector bundle. We let $F=\R^k$ (or $\C^k$ for complex vector bundles), and we require that $\phi_{\alpha\beta}:U_\alpha\cap U_\beta\to \GL_k(\R)$ is a diffeomorphism for all $\alpha,\beta\in A$, so that $(x,v)\mapsto (x,\phi_{\alpha\beta}(x)v)$ is a diffeomorphism from $(U_\alpha\cap U_\beta)\times \R^k$ to itself. The resulting fiber bundle on $B$ is then a vector bundle. The number $k$ is referred to as the rank of the vector bundle. 

Given a $\Cinfty$ fiber bundle $p:V\to B$, then $p$ is a vector bundle if
\begin{itemize}
\item[(a)] there is an ``addition'' map
\begin{center}
\begin{tikzcd}[column sep=0.1in]
V\times_B V \ar{rr} \ar{rd} && V \ar{dl}\\
 & B
\end{tikzcd}
\end{center}
\item[(b)] and there is a ``scalar multiplication'' map
\begin{center}
\begin{tikzcd}[column sep=0.1in]
\R\times V \ar{rr} \ar{rd} && V \ar{dl}\\
 & B
\end{tikzcd}
\end{center}
that gives $p^{-1}(b)$ the structure of a vector space for all $b\in B$.
\end{itemize}
Basically any construction you can do with vector spaces, you can do with vector bundles: direct sum, tensor product, exterior algebra, etc.