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 $V_\alpha \overset{p}{\simeq} U$, 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. $$\tag*{$\blacksquare$}$$
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. $$\tag*{$\blacksquare$}$$
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, $\tilde{F}(x,1) = \tilde{g}(x)$. $$\tag*{$\blacksquare$}$$
Definition. If $p: E \to B$ is a covering map, and $p(e_0) = b_0$, then $\phi: \pi_1(B, b_0) \to p^{-1}(b_0)$ 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 $\phi$ 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 $\phi$ derived from $p$ is surjective. If $E$ is simply connected, then $\phi$ 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 $\phi([f]) = \tilde{f}(1) = e_1$.
If $E$ is simply connected, then suppose that $\phi([f]) = \phi([g])$ and $[f] \neq [g]$. 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. $$\tag*{$\blacksquare$}$$
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 $V_n = (n - 1/4, n + 1/4)$. $p|_{\bar{V}_n}$ is bijective onto $\bar{U}$ since it is clearly surjective and $\sin 2\pi x$ is strictly monotonic. $\bar{V}_n \overset{p}{\simeq} \bar{U}$ since $\bar{V}_n$ is compact and $\bar{U}$ is Hausdorff. Therefore, $V_n \overset{p}{\simeq} U$. Repeat this argument with the remaining $90^\circ$ rotations of $U$. $$\tag*{$\blacksquare$}$$
Claim. 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 $\phi: \pi_1(S^1, b_0) \to \mathbb{Z}$ 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 $\tilde{\tilde{g}} = n + \tilde{g}$ is a lifting of $g$ beginning at $n$ since $p(n + x) = p(x)$. Moreover, $\tilde{f} * \tilde{\tilde{g}}$ is the lifting of $f * g$ beginning at $e_0$, and $\tilde{\tilde{g}}(1) = n + m$. Therefore, $\phi([f] * [g]) = n + m = \phi([f]) + \phi([g])$.$$\tag*{$\blacksquare$}$$
Bonus 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 $\tilde\alpha*\tilde\beta$ is a lifting of $\alpha*\beta$.
Proof.
$$\begin{align}p\circ\tilde\alpha*\tilde\beta(x)&=\begin{cases}p\circ\tilde\alpha(2x)=\alpha(2x)&0\leq x\leq1/2\\p\circ\tilde\beta(2x-1)=\beta(2x-1)&1/2<x\leq1\end{cases}\\&=\alpha*\beta(x).\tag*{$\blacksquare$}\end{align}$$
Bonus 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 $\phi: \pi_1(B, b_0) \to p^{-1}(b_0)$ 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.$$\tag*{$\blacksquare$}$$