\classheader{2012-11-28}

Recall that if $X$ is a path-connected topological space and $x_0\in X$, then $\pi_1(X,x_0)$ is the fundamental group of $X$ with basepoint $x_0$. It consists of homotopy classes (rel endpoints) of loops $\alpha:[0,1]\to X$ satisfying $\alpha(0)=\alpha(1)=x_0$. The group operation is $\ast$.

\begin{theorem}
If $f:X\to X$ is a homotopy equivalence, then $\pi_1(X,x_0)\xrightarrow{\;f_*\;}\pi_1(Y,f(x))$ is an isomorphism.
\end{theorem}
\begin{example}
We know that $\R^2\htopequiv\text{pt}$, so $\pi_1(\R^2,(0,0))\cong\pi_1(\text{pt},\text{pt})=1$.
\end{example}

Today, we'll compute $\pi_1(\S^1,1)$, and introduce covering spaces.

Are these two loops equivalent in $\pi_1(\S^1,1)$?

\begin{center}
\begin{tikzpicture}[every path/.style={thick}]
\draw (0,0) circle (1);
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] at (1,0) {};
\draw [red,domain=0:6.28,variable=\t,smooth,samples=75,->,>=angle 60] plot ({\t r}: {1.15+0.02*\t});
\begin{scope}[shift={(3.5,0)}]
\draw (0,0) circle (1);
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] at (1,0) {};
\draw [blue,domain=0:2*6.28,variable=\t,smooth,samples=75,->,>=angle 60] plot ({\t r}: {1.15+0.02*\t});
\end{scope}
\end{tikzpicture}
\end{center}

No; they go around different numbers of times. It's clear intuitively that $\pi_1(\S^1,1)\cong\Z$. But how can we prove this?

The right way to think about $\S^1$ is as $\R/\Z$, under the quotient $p:\R\to\S^1$ defined by $p(t)=e^{2\pi it}$.

\begin{center}
\begin{tikzpicture}
\draw[thick] (0,0) circle (1);
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] at (1,0) {};
\draw[thick] (-3.5,2.1) to (3.5,2.1);
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] at (-3,2.1) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] at (-2,2.1) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] at (-1,2.1) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] at (0,2.1) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] at (1,2.1) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] at (2,2.1) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] at (3,2.1) {};
\draw[shorten >=5pt,shorten <=5pt,->,>=angle 60,thick] (0,2) to node[midway,auto] {$p$} (0,1);
\node at (1.5,0) {$\S^1$};
\node at (4,2.1) {$\R$};
\end{tikzpicture}
\end{center}

Note that we can ``lift'' paths in $\S^1$ up to $\R$, as long as we have specified where to start.

\begin{theorem}
For any $\alpha:[0,1]\to\S^1$ with $\alpha(0)=x_0$, and any choice of $\widetilde{x_0}\in p^{-1}(x_0)$, then there is a unique $\widetilde{\alpha}:[0,1]\to\R$ with $p\circ\widetilde{\alpha}=\alpha$ and $\widetilde{\alpha}(0)=\widetilde{x_0}$.
\begin{center}
\begin{tikzcd}
& \R \ar{d}{p}\\ [0,1] \ar{ur}{\widetilde{\alpha}} \ar{r}[swap]{\alpha} & \S^1
\end{tikzcd}
\end{center}
\end{theorem}
\begin{proof}
We can cover $\S^1$ by open $U$, $V$ like this:

\begin{center}
\begin{tikzpicture}
\draw[thick] (0,0) circle (1);
\draw[thick,red,(-)] (260:1.2) arc (260:460:1.2);
\draw[thick,blue,(-)] (80:1.4) arc (80:280:1.4);
\node[red] at (1.5,0) {$V$};
\node[blue] at (-1.7,0) {$U$};
\end{tikzpicture}
\end{center}

Let $\{U_i\},\{V_i\}$ be the connected components of $p^{-1}(U)$ and $p^{-1}(V)$, respectively.

\begin{center}
\begin{tikzpicture}
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] at (1,0) {};
\draw[thick] (-3,0) to (3,0) node [label=right:$\R$] {};
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] at (-2,0) {};
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\draw[thick,red,(-)] (-2.4,0.5) -- (-1.6,0.5);
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\begin{scope}[shift={(1,0)}]
\draw[thick,red,(-)] (-2.4,0.5) -- (-1.6,0.5);
\draw[thick,blue,(-)] (-1.9,-0.5) -- (-1.1,-0.5);
\end{scope}
\begin{scope}[shift={(2,0)}]
\draw[thick,red,(-)] (-2.4,0.5) -- (-1.6,0.5);
\draw[thick,blue,(-)] (-1.9,-0.5) -- (-1.1,-0.5);
\end{scope}
\begin{scope}[shift={(3,0)}]
\draw[thick,red,(-)] (-2.4,0.5) -- (-1.6,0.5);
\draw[thick,blue,(-)] (-1.9,-0.5) -- (-1.1,-0.5);
\end{scope}
\begin{scope}[shift={(4,0)}]
\draw[thick,red,(-)] (-2.4,0.5) -- (-1.6,0.5);
\end{scope}
\end{tikzpicture}
\end{center}

Note that, for each $U_i$ and $V_i$, the maps $p|_{U_i}:U_i\to U$ and $p|_{V_i}:V_i\to V$ are all homeomorphisms, so at least \textit{locally}, there are lifts; indeed for any space $Z$ and any map $f:Z\to U$, there is a unique lift $\widetilde{f}$, i.e. a map such that $p\circ \widetilde{f}=f$.
\begin{center}
\begin{tikzcd}
& U_i \ar{d}{p}\\ Z\ar[dashed]{ur}{\widetilde{f}} \ar{r}[swap]{f} & U
\end{tikzcd}
\end{center}
Now pull back $U$ and $V$ under $\alpha$ to cover $[0,1]$ (which has a finite subcover because it is compact).
\begin{center}
\begin{tikzpicture}
\draw[thick] (-3,0) to (3,0);
\draw[thick] (-2.96,-0.15) -- (-3,-0.15) -- (-3,0.15) -- (-2.96,0.15);
\draw[thick] (2.96,-0.15) -- (3,-0.15) -- (3,0.15) -- (2.96,0.15);
\begin{scope}[shift={(-0.7,0)}]
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] at (-2,0) {};
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\end{scope}
\begin{scope}[shift={(-0.25,0)}]
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] at (-2,0) {};
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\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] at (3,0) {};
\end{scope}
\draw[thick,red,(-)] (-2.4,0.5) -- (-1.6,0.5);
\draw[thick,blue,(-)] (-1.9,-0.5) -- (-1.1,-0.5);
\begin{scope}[shift={(1,0)}]
\draw[thick,red,(-)] (-2.4,0.5) -- (-1.6,0.5);
\draw[thick,blue,(-)] (-1.9,-0.5) -- (-1.1,-0.5);
\end{scope}
\begin{scope}[shift={(2,0)}]
\draw[thick,red,(-)] (-2.4,0.5) -- (-1.6,0.5);
\draw[thick,blue,(-)] (-1.9,-0.5) -- (-1.1,-0.5);
\end{scope}
\begin{scope}[shift={(3,0)}]
\draw[thick,red,(-)] (-2.4,0.5) -- (-1.6,0.5);
\draw[thick,blue,(-)] (-1.9,-0.5) -- (-1.1,-0.5);
\end{scope}
\begin{scope}[shift={(4,0)}]
\draw[thick,red,(-)] (-2.4,0.5) -- (-1.6,0.5);
\draw[thick,blue,(-)] (-1.9,-0.5) -- (-1.1,-0.5);
\end{scope}
\begin{scope}
\clip (-3,-1) rectangle (3,1);
\begin{scope}[shift={(5,0)}]
\draw[thick,red,(-)] (-2.4,0.5) -- (-1.6,0.5);
\draw[thick,blue,(-)] (-1.9,-0.5) -- (-1.1,-0.5);
\end{scope}
\begin{scope}[shift={(1,0)}]
\draw[thick,red,(-)] (-2.4,0.5) -- (-1.6,0.5);
\draw[thick,blue,(-)] (-1.9,-0.5) -- (-1.1,-0.5);
\end{scope}
\begin{scope}[shift={(0,0)}]
\draw[thick,red,(-)] (-2.4,0.5) -- (-1.6,0.5);
\draw[thick,blue,(-)] (-1.9,-0.5) -- (-1.1,-0.5);
\end{scope}
\begin{scope}[shift={(-1,0)}]
\draw[thick,red,(-)] (-2.4,0.5) -- (-1.6,0.5);
\draw[thick,blue,(-)] (-1.9,-0.5) -- (-1.1,-0.5);
\end{scope}
\end{scope}
\begin{scope}[shift={(0,0.5)}]
\draw[thick,red] (-2.96,-0.15) -- (-3,-0.15) -- (-3,0.15) -- (-2.96,0.15);
\draw[thick,red] (2.96,-0.15) -- (3,-0.15) -- (3,0.15) -- (2.96,0.15);
\end{scope}
\end{tikzpicture}
\end{center}
Thus, $\alpha$ breaks into subpaths $\alpha_i$, with $\alpha_i:[v_{i-1},v_i]\to U$ (or $V$). Therefore, there is a unique lift $\widetilde{\alpha_1}:[0,v_1]\to U_i\subseteq\R$, where this is the unique $U_i$ containing $\widetilde{x_0}$. Then there is a unique lift $\widetilde{\alpha_2}:[v_1,v_2]\to V_j\subseteq\R$ such that $\widetilde{\alpha_2}(v_1)=\widetilde{\alpha_1}(v_1)$. We then proceed inductively.
\end{proof}

Thus, we've shown that we can lift paths. We can also lift homotopies:

\begin{theorem}
For any $H_s(t):[0,1]^2\to\S^1$ such that $H_0(0)=x_0$, and any choice of $\widetilde{x}_0\in p^{-1}(x_0)$, there is a unique lift $\widetilde{H}_s(t):[0,1]^2\to\R$ such that $\widetilde{H}_0(0)=\widetilde{x_0}$ and $p\circ\widetilde{H}=H$.
\end{theorem}
\begin{proof}
Break $[0,1]^2$ into squares small enough to guarantee that each maps into either $U$ or $V$, and lift each square inductively in the same way as our earlier proof.
\begin{center}
\begin{tikzpicture}
\draw[thick] (0,0) circle (1);
\foreach \x in {0.2,0.4,...,1.8} {
\draw[gray] (-4.5+\x,-1) -- (-4.5+\x,1);
\draw[gray] (-4.5,-1+\x) -- (-2.5,-1+\x);}
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] at (1,0) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] at (0,2.1) {};
\draw[thick] (-2.5,2.1) to (2.5,2.1);
\draw[shorten >=5pt,shorten <=5pt,->,>=angle 60,thick] (0,2) to node[midway,auto] {$p$} (0,1);
\node at (1.5,0) {$\S^1$};
\node at (3,2.1) {$\R$};
\draw[thick] (-4.5,-1) rectangle (-2.5,1);
\node at (-3.5,-1.4) {$[0,1]^2$};
\draw[thick,-angle 60] (-2.3,0) -- (-1.3,0);
\end{tikzpicture}
\end{center}
\end{proof}

Now let's get back to $\pi_1(\S^1,1)$. Define $\gamma_n:[0,1]\to\S^1$ by $\gamma_n(t)=e^{2\pi i nt}$.
\begin{center}
\begin{tikzpicture}[every path/.style={thick}]
\draw (0,0) circle (1);
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] at (1,0) {};
\draw [red,domain=0:6.28,variable=\t,smooth,samples=75,->,>=angle 60] plot ({\t r}: {1.15+0.02*\t});
\begin{scope}[shift={(3.5,0)}]
\draw (0,0) circle (1);
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] at (1,0) {};
\draw [red,domain=0:2*6.28,variable=\t,smooth,samples=75,->,>=angle 60] plot ({\t r}: {1.15+0.02*\t});
\end{scope}
\begin{scope}[shift={(-3.5,0)},yscale=-1]
\draw (0,0) circle (1);
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] at (1,0) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,red,circle] at (1.15,0) {};
\end{scope}
\begin{scope}[shift={(-7,0)},yscale=-1]
\draw (0,0) circle (1);
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] at (1,0) {};
\draw [red,domain=0:6.28,variable=\t,smooth,postaction={decorate,decoration={markings,mark=at position 0.999 with {\arrow[>=angle 60]{>}}}}] plot ({\t r}: {1.15+0.02*\t});
\end{scope}
\node at (-9,0) {$\cdots$};
\node at (6,0) {$\cdots$};
\end{tikzpicture}
\end{center}
Note that the unique lift $\widetilde{\gamma_n}$ with $\widetilde{\gamma_n}(0)=0$ is
\begin{center}
\begin{tikzpicture}[every path/.style={thick}]
\draw[thick] (-1,0) -- (1,0);
\node[inner sep=0pt,outer sep=0pt,minimum size=3pt,fill,circle] at (0,0) {};
\node[inner sep=0pt,outer sep=0pt] at (0,-0.25) {0};
\node[inner sep=0pt,outer sep=0pt,minimum size=3pt,fill,circle] at (0.5,0) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=3pt,fill,circle] at (1,0) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=3pt,fill,circle] at (-0.5,0) {};
\draw [red,->,>=angle 60] (0,0.15) -- (0.5,0.15);
\begin{scope}[shift={(3.5,0)}]
\draw[thick] (-1,0) -- (1,0);
\node[inner sep=0pt,outer sep=0pt,minimum size=3pt,fill,circle] at (0,0) {};
\node[inner sep=0pt,outer sep=0pt] at (0,-0.25) {0};
\node[inner sep=0pt,outer sep=0pt,minimum size=3pt,fill,circle] at (0.5,0) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=3pt,fill,circle] at (1,0) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=3pt,fill,circle] at (-0.5,0) {};
\draw [red,->,>=angle 60] (0,0.15) -- (1,0.15);
\end{scope}
\begin{scope}[shift={(-3.5,0)},yscale=1]
\draw[thick] (-1,0) -- (1,0);
\node[inner sep=0pt,outer sep=0pt,minimum size=3pt,fill,circle] at (0,0) {};
\node[inner sep=0pt,outer sep=0pt] at (0,-0.25) {0};
\node[inner sep=0pt,outer sep=0pt,minimum size=3pt,fill,circle] at (0.5,0) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=3pt,fill,circle] at (1,0) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=3pt,fill,circle] at (-0.5,0) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=3pt,fill,red,circle] at (0,0.15) {};
\end{scope}
\begin{scope}[shift={(-7,0)},yscale=1]
\draw[thick] (-1,0) -- (1,0);
\node[inner sep=0pt,outer sep=0pt,minimum size=3pt,fill,circle] at (0,0) {};
\node[inner sep=0pt,outer sep=0pt] at (0,-0.25) {0};
\node[inner sep=0pt,outer sep=0pt,minimum size=3pt,fill,circle] at (0.5,0) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=3pt,fill,circle] at (1,0) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=3pt,fill,circle] at (-0.5,0) {};
\draw [red,->,>=angle 60] (0,0.15) -- (-0.5,0.15);
\end{scope}
\node at (-9,0) {$\cdots$};
\node at (6,0) {$\cdots$};
\end{tikzpicture}
\end{center}
Define $\Phi:\Z\to\pi_1(\S^1,1)$ by $\Phi(n)=[\gamma_n]$.

\begin{theorem}
The map $\Phi$ is an isomorphism.
\end{theorem}
\begin{proof}
First, let's show that $\Phi$ is surjective. Given $[\alpha]\in\pi_1(\S^1,1)$, by our earlier theorem, there is a unique lift $\alpha:[0,1]\to\R$ with $\widetilde{\alpha}(0)=0$. Then $\widetilde{\alpha}(1)\in p^{-1}(1)=\Z$, so $\widetilde{\alpha}(1)=n$ for some $n\in\Z$. Obviously, we are going to want to define a homotopy between $\alpha$ and $\gamma_n$.

Define $H:[0,1]^2\to\R$ by $H_s(t)=snt+(1-s)\widetilde{\alpha}(t)$. Note that
\[H_0(t)=\widetilde{\alpha}(t),\quad H_1(t)=nt=\widetilde{\gamma_n}(t),\quad H_s(0)=0,\quad H_s(1)=n.\]
Thus, $p\circ H$ is a homotopy $\alpha\sim\gamma_n$, so $[\alpha]=[\gamma_n]$.

Now we'll show that $\Phi$ is injective. Suppose that $[\gamma_m]=[\gamma_n]$, so that there is some $H:[0,1]^2\to\S^1$ such that
\[H_0(t)=\gamma_n(t),\quad H_1(t)=\gamma_m(t),\quad H_s(0)=1,\quad H_s(1)=1.\]
\begin{center}
\begin{tikzpicture}[scale=2]
\draw[thick] (0,0) rectangle (1,1);
\node at (0.5,1.15) {$1$};
\node at (0.5,-0.2) {$1$};
\node at (-0.3,0.5) {$\gamma_n$};
\node at (1.3,0.5) {$\gamma_m$};
\end{tikzpicture}
\end{center}
We lift this $H$ to $\widetilde{H}:[0,1]^2\to\R$, with $\widetilde{H}_0(0)=0$. Then we have $\widetilde{H}_0(t)=\widetilde{\gamma_n}(t)$ and $\widetilde{H}_1(t)=\widetilde{\gamma_m}(t)$, and note that $\widetilde{H}_s(0)$ and $\widetilde{H}_s(1)$ have to always map into $p^{-1}(1)=\Z$, and therefore they must be constant. Thus,
\[m=\widetilde{\gamma_m}(1)=\widetilde{H}_1(1)=\widetilde{H}_0(1)=\widetilde{\gamma_n}(1)=n.\qedhere\]
\end{proof}

\begin{definition}
If $\widetilde{X}$ and $X$ are topological spaces, we say that the map $p:\widetilde{X}\to X$ is a covering map when there is an open cover $\{V_i\}$ of $X$ such that $p|_U$ is a homeomorphism for any connected component $U$ of $p^{-1}(V_i)$, for any $V_i$. We say that $\widetilde{X}$ is a covering space of $X$.
\end{definition}