\classheader{2012-10-01}

Today I'll try to present a general idea of algebraic topology.

\begin{definition}
A category $\cat{C}$ is a set $\cat{C}$ (though not always really a set) of objects and a set \[\Mor(\cat{C})=\{f:A\to B\mid A,B\in\cat{C}\}\]
of morphisms between the objects of $\cat{C}$. We can always compose morphisms (whose domains and codomains are appropriately related), this composition is always associative, and for all objects $A\in\cat{C}$, there is a morphism $\id_A:A\to A$ that acts as the identity for composition.
\end{definition}

\begin{examples}
The following are some common examples of categories.
\begin{itemize}
\item $\cat{C}=\{\text{topological spaces}\}$, and $\Mor(\cat{C})=\{\text{continuous maps}\}$
\item $\cat{C}=\{\text{topological spaces}\}$, and $\Mor(\cat{C})=\{\text{homeomorphisms}\}$
\item $\cat{C}=\{\text{vector spaces over a field }K\}$, and $\Mor(\cat{C})=\{\text{$K$-linear maps}\}$
\item $\cat{C}=\{\text{abelian groups}\}$, and $\Mor(\cat{C})=\{\text{homomorphisms}\}$
\end{itemize}
\end{examples}

\begin{definition}
A (covariant) functor $\fun{F}:\cat{C}\to\cat{D}$ is a mapping that takes objects $A\in\cat{C}$ to objects $\fun{F}(A)\in\cat{D}$, and morphisms $(f:A\to B)$ between objects of $\cat{C}$ to morphisms $(\fun{F}(f):\fun{F}(A)\to \fun{F}(B))$ between the corresponding objects of $\cat{D}$. 

We'll often refer to $\fun{F}(f)$ as $f_*$ when the functor $\fun{F}$ is understood.

We can also define a contravariant functor to be a functor that reverses the direction of morphisms, so that $\fun{F}(f):\fun{F}(B)\to \fun{F}(A)$. In this situation, the shorthand for $\fun{F}(f)$ will be $f^*$.

In either case, it must satisfy $\fun{F}(\id_A)=\id_{\fun{F}(A)}$. It also must satisfy functoriality; that is, we must have $\fun{F}(f\circ g)=\fun{F}(f)\circ \fun{F}(g)$, or in shorthand $(f\circ g)_*=f_*\circ g_*$. This goes in the other order for contravariant functors.
\end{definition}

\begin{examples}
The following are some common examples of functors.
\begin{itemize}
\item $\cat{C}=\{\text{finite sets}\}$ with $\Mor(\cat{C})=\{\text{set maps}\}$, $\cat{D}=\{\R\text{-vector spaces}\}$ with $\Mor(\cat{D})=\{\text{linear maps}\}$, and $\fun{F}:\cat{C}\to\cat{D}$ defined on objects by
\[S\mapsto \text{free }\R\text{-vector space on }S\]
and for a morphism $f:S\to T$, $\fun{F}(f)$ is the linear map uniquely determined by sending the basis element $s$ to the basis element $f(s)$.
\item $\cat{C}=\{\R\text{-vector spaces}\}$ with $\Mor(\cat{C})=\{\text{linear maps}\}$, and $\fun{F}:\cat{C}\to\cat{C}$ defined on objects by $V\mapsto V^*$, and on morphisms by $(A:V\to W)\mapsto (A^*:W^*\to V^*)$.
\end{itemize}
\end{examples}
Broadly speaking,
\[\text{algebraic topology}\;\;\simeq\;\; \text{construction of functors }\{\text{spaces}\}\to\{\text{algebraic objects}\}.\]
The functors we'll look at in this class are $H_i$, $H^i$, $\pi_1$, and $\pi_n$.

Here are some demonstrations of the power of functors. Let's suppose we know there exists a functor $\fun{F}_i:\{\text{spaces}\}\to\{\text{abelian groups}\}$ such that $\fun{F}_i(\D^n)=0$ for any $n$, $\fun{F}_i(\S^n)=0$ for $n\neq i$, and $\fun{F}_i(\S^i)\neq 0$. We don't have to know what else it does.

\begin{example}
From this assumption alone, we can prove that $\S^n\isom\S^m\iff n=m$, and as a corollary, prove ``invariance of dimension'', i.e. $\R^n\isom\R^m\iff n=m$.
\end{example}
\begin{proof}
Suppose $h:\S^n\to\S^m$ is a homeomorphism, with inverse $g$. Then applying the functor $\fun{F}_n$,
\[g\circ h=\id_{\S^n}\implies g_*\circ h_*=\id_{\fun{F}_n(\S^n)},\]
\[h\circ g=\id_{\S^m}\implies h_*\circ g_*=\id_{\fun{F}_n(\S^m)}.\]
This implies that $h_*:\fun{F}_n(\S^n)\to \fun{F}_n(\S^m)$ is a bijection, and hence an isomorphism. But $\fun{F}_n(\S^n)\neq 0$ and $\fun{F}_n(\S^m)=0$ unless $n=m$.
\end{proof}

\begin{example}[Brouwer Fixed Point Theorem]
Any continuous $f:\D^n\to\D^n$ has a fixed point ($n\geq 2$).
\end{example}
\begin{proof}
Suppose there are no fixed points. Then we can define $r:\D^n\to\bound\D^n=\S^{n-1}$ by drawing a line from $f(v)$ to $v$ (this is possible only because we are assuming they are different) and extending it until it hits the boundary of $\D^n$, and setting that to be $r(v)$.
\begin{center}
\begin{tikzpicture}[scale=2.4]
\node (center) at (0,0) {};
\begin{scope}[every node/.style={fill,circle,inner sep=0pt,outer sep=1pt,minimum size=0.15cm}]
\node (v) at (135:0cm) [label={45:$v$}] {};
\node (fv) at (315:0.4cm) [label={45:$f(v)$}] {};
\node (rv) at (135:1cm) [label={135:$r(v)$}] {};
\end{scope}
\draw[very thick] (0cm,0cm) circle(1cm);
\draw[thick,arrows={-angle 60}] (fv.center) -- (rv.center);
\end{tikzpicture}
\end{center}
You can prove that this is continuous on your own. Note that for any $v\in\S^{n-1}$, we will have that $r(v)=v$. Thus, letting $i:\S^{n-1}\to\D^n$ be the obvious inclusion map, we have that $r\circ i=\id_{\S^{n-1}}$ (we say that ``$r$ is a retraction of $\D^n$ onto $\S^{n-1}$''). Because $r\circ i=\id_{\S^{n-1}}$, we have that
\[\fun{F}_{n-1}(r)\circ \fun{F}_{n-1}(i)=\id_{\fun{F}_{n-1}(\S^{n-1})}\]
so that $\fun{F}_{n-1}(r)$ must be surjective. But $\fun{F}_{n-1}(r):\fun{F}_{n-1}(\D^n)\to \fun{F}_{n-1}(\S^{n-1})$ can't be surjective because $\fun{F}_{n-1}(\D^n)=0$ and $\fun{F}_{n-1}(\S^{n-1})\neq 0$. This is a contradiction.
\end{proof}
A final comment: a good method for turning invariants for spaces into invariants for maps is to associate to a map $f:X\to Y$ its graph $\Gamma_f\subset X\times Y$.