\classheader{2012-12-03}

\begin{theorem}[Gordon-Luecke, 1980's]
Let $K_1,K_2$ be knots in $\R^3$. Then $K_1$ is equivalent to $K_2$, i.e. there is a homeomorphism $h:\R^3\to\R^3$ with $h(K_1)=K_2$, if and only if $\pi_1(\R^3\setminus K_1)\cong\pi_1(\R^3\setminus K_2)$.
\end{theorem}

(warning: this is an incorrect statement of the theorem)


\subsection*{Comparing $H_1$ and $\pi_1$}

Let $\Gamma$ be any group. The commutator subgroup of $\Gamma$ is defined to be
\[[\Gamma,\Gamma]=\langle \{ghg^{-1}h^{-1}\mid g,h\in\Gamma\}\rangle.\]
Then $\Gamma/[\Gamma,\Gamma]$ is abelian, and $\Gamma/[\Gamma,\Gamma]\cong \Gamma^{\ab}$, where $\Gamma^{\ab}$ is by definition the unique group satisfying the following universal mapping property: for any abelian group $A$ and homomorphism $\phi:\Gamma\to A$, there is a unique $\overline{\phi}:\Gamma^{\ab}\to A$ such that
\begin{center}
\begin{tikzcd}
\Gamma\ar{r}{\phi} \ar{d}[swap]{\pi} & A\\
\Gamma^{\ab} \ar[dashed]{ru}[swap]{\overline{\phi}}
\end{tikzcd}
\end{center}
You proved the following theorem on the most recent homework:
\begin{theorem}
Let $X$ be any path-connected space. Then the map $\phi:\pi_1(X)\to H_1(X,\Z)$ defined by $\phi([\gamma])=\gamma_*([\S^1])$ induces an isomorphism
\[\overline{\phi}:\pi_1(X)/[\pi_1(X),\pi_1(X)]\to H_1(X,\Z).\]
\end{theorem}


\subsection*{Covering Spaces}
Throughout, let $X$ and $Y$ be path-connected and locally path-connected spaces.
\begin{definition}
A map $p:Y\to X$ is a covering space when any $x\in X$ has a neighborhood $U_x$ such that, for each $x_\alpha\in p^{-1}(x)$, there is a neighborhood $V_\alpha$ of $x_\alpha$ such that $p|_{V_\alpha}:V_\alpha\to U_x$ is a homeomorphism, and $V_\alpha\cap V_\beta$ for $\alpha\neq \beta$.
\end{definition}
\begin{example}
A group action on $X$ is just a subgroup $\Gamma\subset \text{Homeo}(X)$. The action is discrete if, for all $x\in X$, there is a neighborhood $U_x$ of $x$ such that for all $g\in \Gamma\setminus\{\id\}$, $gU_x\cap U_x=\varnothing$. You can check that if $\Gamma$ acts discretely on $X$, then the quotient $X\to X/\Gamma$ is a covering map.

To give explicit examples of this, we can take $\Gamma\subset\text{Homeo}(\R^2)$ to be
\[\Gamma=\langle (x,y)\mapsto (x+1,y),\;(x,y)\mapsto(x,y+1)\rangle,\]
in which case we get a covering map $\R^2\to \R^2/\Gamma\cong\T^2$. We could also consider
\[\Lambda=\langle (x,y)\mapsto (x,y+1)\rangle,\]
in which case we get a covering map $\R^2\to\R^2/\Gamma\cong \S^1\times\R$. Quotienting the rest of the way produces a covering map $\S^1\times \R\to\T^2$. In fact, $\T^2$ covers itself; for any $a,b\in\Z^2$, the map which acts as a degree $a$ map on the first factor of $\T^2=\S^1\times\S^1$ and as a degree $b$ map on the second factor produces a covering map $\T^2\to\T^2$.
\end{example}

Note that there are lots of different covering maps of $\T^2$, and some of them cover others:

\begin{center}
\qquad\qquad\begin{tikzpicture}
\begin{scope}[scale=0.25]
\draw[help lines] (-4,-4) grid (4,4) node[label={[black]right:$\R^2$}] {};
\draw[thick] (-4,0) -- (4,0);
\draw[thick] (0,-4) -- (0,4);
\end{scope}
\draw[thick,->,>=angle 90] (0,-1.5) -- (0,-3);

\begin{scope}[scale=0.9,shift={(0,-4.8)}]
\node at (1.3,1.1) {$\T^2$};
\draw [,fill=gray!20] (0,0) ellipse (1.2 and 1);
\begin{scope}
\clip (0,0) ellipse (0.6 and 0.4);
\draw[,fill=white] (0,-0.1) ellipse (0.52 and 0.42);
\end{scope}
\begin{scope}
\clip (-1,0) rectangle (1,-1);
\draw[,postaction={decorate,decoration={markings,mark=at position 0.65 with {\node (a) {};},mark=at position 0.7 with {\node (b) {};}}}] (0,0) ellipse (0.6 and 0.4);
\end{scope}
\draw[,postaction={decorate,decoration={markings,mark=at position 0.65 with {\node (c) {};},mark=at position 0.7 with {\node (d) {};}}}] (0,0) ellipse (1.2 and 1);
\end{scope}

\begin{scope}[scale=0.9,shift={(4.5,-3.8)}]
\node at (1.3,1.1) {$\T^2$};
\draw [,fill=gray!20] (0,0) ellipse (1.2 and 1);
\begin{scope}
\clip (0,0) ellipse (0.6 and 0.4);
\draw[,fill=white] (0,-0.1) ellipse (0.52 and 0.42);
\end{scope}
\begin{scope}
\clip (-1,0) rectangle (1,-1);
\draw[,postaction={decorate,decoration={markings,mark=at position 0.65 with {\node (a) {};},mark=at position 0.7 with {\node (b) {};}}}] (0,0) ellipse (0.6 and 0.4);
\end{scope}
\draw[,postaction={decorate,decoration={markings,mark=at position 0.65 with {\node (c) {};},mark=at position 0.7 with {\node (d) {};}}}] (0,0) ellipse (1.2 and 1);
\end{scope}
\draw[thick,-angle 90] (2.8,-3.8) -- (1.3,-4.1);
\begin{scope}[shift={(-0.25,-0.2)}]
\begin{scope}[rotate around={-90:(4.2,0)}]
\node (u) at (3.9,0.8) {};
\node (w) at (3.9,-0.5) {};
\node (p) at (4.5,0.8) {};
\node (q) at (4.5,-0.5) {};
\fill[gray!20] (p.center) arc (0:180:0.3 and 0.1) -- (w.center) -- (q.center) -- cycle;
\draw (p.center) -- (q.center);
\draw (u.center) -- (w.center);
\draw[dashed] (p.center) arc (0:180:0.3 and 0.1) node[label=above right:{$\S^1\times[0,1]$}] {};
\draw[shade,dashed,top color=gray!20] (4.2,-0.5) ellipse (0.3 and 0.1);
\end{scope}
\end{scope}
\draw[thick,-angle 90] (1.3,0.1) -- (3,-0.1);
\draw[thick,-angle 90] (4.1,-0.8) -- (4.1,-2.2);
\end{tikzpicture}
\end{center}

We can build a dictionary between the group theory of $\pi_1(X)$ and the covering space theory of $X$. It is a Galois correspondence. Technically, we need to assume that $X$ is path-connected, locally path-connected, and semi-locally simply connected.

\begin{center}
\begin{tabular}{c@{$\;\;\longleftrightarrow\;\;$}c}
 $\pi_1(X)$ & $X$\\\hline
subgroups $H\subseteq\pi_1(X)$ & covering spaces $p:Y\to X$\\
$\pi_1(X)$ & $\id:X\to X$\\
$\{e\}$ & universal cover $\widetilde{X}\to X$\\
$H\subset \pi_1(X)$ & $\widetilde{X}/H\to X$\\
normal subgroups & $p:Y\to X$ is regular\\
$[\pi_1(X):H]$ & $\#p^{-1}(x)$, i.e. number of sheets\\
conjugacy classes of elements & free homotopy classes of loops
\end{tabular}
\end{center}
\begin{theorem}
If $\Gamma$ acts discretely on $Y$ and $\pi_1(Y)=0$, then $\pi_1(Y/\Gamma)\cong \Gamma$.
\end{theorem}
\begin{example}
We can cover the unit disk model of $\H^2$ with octagons, and act discretely on it by $\Gamma=\langle x,y,z,w\rangle$ for some transformations $x,y,z,w$ which identify the appropriate sides of the octagons, and we get $\H^2/\Gamma\cong\Sigma_2$.
\end{example}
\begin{example}
If $L(p,q)$ is the lens space for $p$ and $q$, we can obtain it as $\S^3/(\Z/p\Z)$.
\end{example}
\begin{non-example}
Let $r\in\R\setminus\Q$, and let $\Z$ act on $\S^1$ by $e^{2\pi i\theta}\mapsto e^{2\pi i (\theta+r)}$. This action is not discrete, and $\S^1/\Z$ is not even Hausdorff.
\end{non-example}
\begin{non-example}
Let $\Z/3\Z$ act on $\D^2$ by rotations by $2\pi/3$. This action is not discrete, though it almost is - it would be discrete if we threw away the origin. This kind of situation is called a branched cover.
\end{non-example}
\begin{proposition}
Let $p:Y\to X$ be a covering map. Then $p_*:\pi_1(Y,y)\to \pi_1(X,p(y))$ is injective.
\end{proposition}
\begin{proof}
The idea is clear: if a loop $p_*([\gamma])$ is trivial in $\pi_1(X,p(y))$, we can lift the homotopy to one demonstrating that $[\gamma]$ is trivial.
\end{proof}