Munkres §54: The Fundamental Group of the Circle

Munkres §53: Covering Spaces ►

Even though it has been four years since I left academia, I still miss solving math problems. They always leave my brain completely fried. 🫠 Math is fairly non-intuitive.

Topological spaces are fascinating mathematical objects. What is most fascinating is that their definition is simple: If a set $X$ and a subset $\tau$ of its power set $2^X$ are such that $\varnothing ,X\in \tau$, arbitrary unions of elements of $\tau$ are elements of $\tau$, and finite intersections of elements of $\tau$ are elements of $\tau$, then $(X,\tau )$ is a topological space. A variety of other objects emerge from this definition, and one of them is the homeomorphism, which is a continuous bijection from one space to another with a continuous inverse. If two spaces are homeomorphic, then we consider them to be the same space, but finding homeomorphisms is hard. We created a new object—called the fundamental group—to represent a property of topological spaces in terms of algebraic groups, which enriches our study of them, and in this post, we take a look at the fundamental group of the circle $S^1$. However, in order to do so, we must first take a look at a handful of new definitions and lemmas.

Definition. If $p:E\to B$ is continuous and surjective, $U\subset B$ is open, and $p^{-1}(U)$ is the union of disjoint, open sets $V_{\alpha }\stackrel{p}{\simeq }U$, then $U$ is said to be evenly covered by $p$.

Definition. If $p:E\to B$ is continuous and surjective, and for all $b\in B$, there exists an open neighborhood $U$ of $b$ that is evenly covered by $p$, then $p$ is said to be a covering map.

Definition. If $p:E\to B$, and $f:X\to B$, then $\tilde{f}:X\to E$ such that $p\circ \tilde{f}=f$ is said to be a lifting of $f$.

Lemma 1. If $p:E\to B$ is a covering map, $p(e_0)=b_0$, and $f$ is a path in $B$ beginning at $b_0$, then there exists a unique continuous lifting of $f$ beginning at $e_0$.

Proof. We spare the reader the pain of this intricate proof. $$\begin{array}{}◼& \end{array}$$

Lemma 2. If $p:E\to B$ is a covering map, $p(e_0)=b_0$, and $F:I\times I\to B$ is continuous such that $F(0,0)=b_0$, then there exists a unique continuous lifting of $F$ such that $\tilde{F}(0,0)=e_0$. If $F$ is a path homotopy, then $\tilde{F}$ is a path homotopy.

Proof. We once again spare the reader the pain of this intricate proof. $$\begin{array}{}◼& \end{array}$$

Lemma 3. If $p:E\to B$ is a covering map, $p(e_0)=b_0$, and $f$ and $g$ are path homotopic paths in $B$ beginning at $b_0$, then the unique continuous liftings of $f$ and $g$ beginning at $e_0$ are path homotopic.

Proof. If $F$ is the path homotopy between $f$ and $g$, then $F(0,0)=b_0$. Therefore—by Lemma 2—the unique continuous lifting of $F$ such that $\tilde{F}(0,0)=e_0$ is a path homotopy. Note that $\tilde{F}(x,0)$ is a lifting of $F(x,0)=f(x)$. Hence, $\tilde{F}(x,0)=\tilde{f}(x)$ since $\tilde{f}$ is unique. Similarly, $\tilde{F}(x,1)=\tilde{g}(x)$. $$\begin{array}{}◼& \end{array}$$

Definition. If $p:E\to B$ is a covering map, and $p(e_0)=b_0$, then $\varphi :{\pi }_1(B,b_0)\to p^{-1}(b_0)$ defined by $[f]\mapsto \tilde{f}(1)$, where $\tilde{f}$ is the unique continuous lifting of $f$ beginning at $e_0$, is said to be the lifting correspondence derived from $p$. Note that $\varphi$ depends on $e_0$ and is well-defined—by Lemmas 1 and 3.

Lemma 4. If $p:E\to B$ is a covering map, $p(e_0)=b_0$, and $E$ is path connected, then the lifting correspondence $\varphi$ derived from $p$ is surjective. If $E$ is simply connected, then $\varphi$ is bijective.

Proof. If $E$ is path connected and $e_1\in p^{-1}(b_0)$, then there exists a path $\tilde{f}$ in $E$ beginning at $e_0$ and ending at $e_1$. Therefore, $p\circ \tilde{f}=f$ is a loop in $B$ at $b_0$, and $\varphi ([f])=\tilde{f}(1)=e_1$.

If $E$ is simply connected, then suppose that $\varphi ([f])=\varphi ([g])$ and $[f]\ne [g]$. If $\tilde{f}$ and $\tilde{g}$ are the unique continuous liftings of $f$ and $g$ beginning at $e_0$, then $\tilde{f}(1)=\tilde{g}(1)$. Therefore, there exists a path homotopy $\tilde{F}$ between $\tilde{f}$ and $\tilde{g}$, and $p\circ \tilde{F}$ is a path homotopy between $f$ and $g$, which is a contradiction. $$\begin{array}{}◼& \end{array}$$

Lemma 5. $p:\mathbb{R}\to S^1$ defined by $x\mapsto (\cos 2\pi x,\sin 2\pi x)$ is a covering map.

Proof. $p$ is clearly continuous and surjective. $U=\{(x,y)\in S^1:x>0\}$ is open. $p^{-1}(U)$ is the union of disjoint intervals $V_n=(n-1/4,n+1/4)$. $p{|}_{{\overline{V}}_n}$ is bijective onto $\overline{U}$ since it is clearly surjective and $\sin 2\pi x$ is strictly monotonic. ${\overline{V}}_n\stackrel{p}{\simeq }\overline{U}$ since ${\overline{V}}_n$ is compact and $\overline{U}$ is Hausdorff. Therefore, $V_n\stackrel{p}{\simeq }U$. Repeat this argument with the remaining ${90}^{\circ }$ rotations of $U$. $$\begin{array}{}◼& \end{array}$$

Theorem. The fundamental group of $S^1$ is isomorphic to the additive group of integers.

Proof. If $p$ is the covering map of Lemma 5, $e_0=0$, and $b_0=p(e_0)$, then $p^{-1}(b_0)=\mathbb{Z}$, and—by Lemma 4—the lifting correspondence $\varphi :{\pi }_1(S^1,b_0)\to \mathbb{Z}$ derived from $p$ is bijective since $\mathbb{R}$ is simply connected.

If $\tilde{f}$ and $\tilde{g}$ are the unique continuous liftings of $f$ and $g$ beginning at $e_0$, $\tilde{f}(1)=n$, and $\tilde{g}(1)=m$, then $\widetilde{\stackrel{~}{g}}=n+\tilde{g}$ is a lifting of $g$ beginning at $n$ since $p(n+x)=p(x)$. Moreover, $\tilde{f}\ast \widetilde{\stackrel{~}{g}}$ is the lifting of $f\ast g$ beginning at $e_0$, and $\widetilde{\stackrel{~}{g}}(1)=n+m$. Therefore, $\varphi ([f]\ast [g])=n+m=\varphi ([f])+\varphi ([g])$.$$\begin{array}{}◼& \end{array}$$

Claim. If $p:E\to B$ is a covering map, $\alpha$ and $\beta$ are paths in $B$ such that $\alpha (1)=\beta (0)$, $\tilde{\alpha }$ and $\tilde{\beta }$ are liftings of $\alpha$ and $\beta$ such that $\tilde{\alpha }(1)=\tilde{\beta }(0)$, then $\tilde{\alpha }\ast \tilde{\beta }$ is a lifting of $\alpha \ast \beta$.

Proof.

$$\begin{array}{rl}p\circ \tilde{\alpha }\ast \tilde{\beta }(x)& =\{\begin{array}{ll}p\circ \tilde{\alpha }(2x)=\alpha (2x)& 0\le x\le 1/2\\ p\circ \tilde{\beta }(2x-1)=\beta (2x-1)& 1/2<x\le 1\end{array}\\ ◼& & =\alpha \ast \beta (x).\end{array}$$

Claim. If $p:E\to B$ is a covering map, $E$ is path connected, and $B$ is simply connected, then $p$ is a homeomorphism.

Proof. The lifting correspondence $\varphi :{\pi }_1(B,b_0)\to p^{-1}(b_0)$ derived from $p$ is surjective—by Lemma 4—since $E$ is path connected, and its domain is trivial since $B$ is simply connected. Therefore, $p$ is injective.

If $V\subset E$ is open and $b\in p(V)$, then there exists an open neighborhood $U$ of $b$ that is evenly covered by $p$. If, $e\in V$, $p(e)=b$, and $V_{\alpha }$ is the slice containing $e$, then $p(V\cap V_{\alpha })\subset p(V)$ is open since $p{|}_{V_{\alpha }}$ is a homeomorphism. Therefore, $p(V)$ is a union of open sets. Hence, $p$ is open.$$\begin{array}{}◼& \end{array}$$