Munkres §53: Covering Spaces

◄ Munkres §53: The Fundamental Group of the Circle | Munkres §52: The Fundamental Group ►

Claim: If $Y$ has the discrete topology and $p:X\times Y\to X$ is projection on the first coordinate, then $p$ is a covering map.

Proof: Clearly, $p$ is continuous and surjective, and if $U_x$ is a neighborhood of $x\in X$, then

$$p^{-1}\left(U_x\right)=U_x\times Y=\underset{y\in Y}{⨆}\underset{V_y}{\underbrace{U_x\times \left\{y\right\}}},$$

where ${p|}_{V_y}$ is a homeomorphism of $V_y$ onto $U_x$.

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Claim: If $U$ is evenly covered by $p$ and connected, then the partition of $p^{-1}\left(U\right)$ into slices is unique.

Proof: If there exist two distinct slices $\left\{V_{\alpha }\right\}$ and $\left\{W_{\beta }\right\}$, then there exists ${\alpha }^{\prime }$ with $V_{{\alpha }^{\prime }}\ne W_{\beta }$ for all $\beta$. If ${\beta }^{\prime }$ is such that $A:=V_{{\alpha }^{\prime }}\cap W_{{\beta }^{\prime }}\ne \varnothing$, then

$$\bigcup _{\begin{array}{c}\alpha \ne {\alpha }^{\prime }\\ \beta \ne {\beta }^{\prime }\end{array}}{V}_{\alpha }\cap W_{\beta }=A^c.$$$$\begin{array}{}↯& \end{array}$$

Claim: If $p:E\to B$ is a covering map, $B$ is connected, and $\left|p^{-1}\left(b_0\right)\right|=k$, then $\left|p^{-1}\left(b\right)\right|=k$ for all $b$, so $E$ is a $k$-fold covering of $B$.

Proof: Let $C:=\{b\in B:\left|p^{-1}\left(b\right)\right|=k\}$, $D:=\{b\in B:\left|p^{-1}\left(b\right)\right|\ne k\}$, and $b\in C$. Then there is a neighborhood $U\ni b$ such that there is a partition of $p^{-1}\left(U\right)$ into $k$ slices because each slice is homeomorphic to $U$. Therefore, for all $x\in U$, $\left|p^{-1}\left(x\right)\right|=k$, implying that $C$ is open. Similarly, $D$ is open. Finally, note that $C\cap D=\varnothing$ and $C\cup D=B$ by definition. $$\begin{array}{}↯& \end{array}$$

Claim: If $q:X\to Y$ and $r:Y\to Z$ are covering maps and $r^{-1}\left(z\right)$ is finite for all $z\in Z$, then $p=r\circ q$ is a covering map.

Proof: Let $z\in Z$. Then there is a neighborhood $U$ of $z$ such that $\left\{U_i\right\}$ is a finite partition of $r^{-1}\left(U\right)$ into slices. Let $v_i$ be the only element in $r^{-1}\left(z\right)\cap U_i$. Then there is a neighborhood $V_i$ of $v_i$ such that $\left\{V_{i_{\alpha }}\right\}$ is a partition of $q^{-1}\left(V_i\right)$ into slices. Let$$C:=\bigcap _{i}r(U_i\cap V_i).$$Then $C$ is open and evenly covered by $r$. Let $W_{i_{\alpha }}:=q^{-1}(U_i\cap V_i)\cap V_{i_{\alpha }}$. Then $C$ is a neighborhood of $z$ such that $\left\{W_{i_{\alpha }}\right\}$ is a partition of $p^{-1}\left(C\right)$ into slices.$$\begin{array}{}◼& \end{array}$$

Claim: If $p:S^1\to S^1$ is defined by $z\mapsto z^n$, where $S^1:=\{z\in \mathbb{C}:\left|z\right|=1\}$, then $p$ is a covering map.

Proof: Let $z\in S^1$. Then $z=e^{i{\theta }^{\prime }}$ for some ${\theta }^{\prime }\in [0,2\pi )$. Define $\varphi :\mathbb{R}\to S^1$ by $\theta \mapsto e^{i\theta }$, let $U:=\varphi \left(({\theta }^{\prime }-\pi /2,{\theta }^{\prime }+\pi /2)\right)$, and let $n=2$. Then$$p^{-1}\left(U\right)=\varphi \left((\frac{1}{2}{\theta }^{\prime }-\frac{1}{4}\pi ,\frac{1}{2}{\theta }^{\prime }+\frac{1}{4}\pi )\right)\bigsqcup \varphi \left((\frac{1}{2}{\theta }^{\prime }+\frac{3}{4}\pi ,\frac{1}{2}{\theta }^{\prime }+\frac{5}{4}\pi )\right).$$ This generalizes analogously to arbitrary $n$.$$\begin{array}{}◼& \end{array}$$

Claim: If $p:E\to B$ is a covering map and $B$ is Hausdorff, then $E$ is Hausdorff.

Proof: Let $x,y\in E$ be distinct. If $p(x)=p(y)$, then $x$ and $y$ are in different slices. Otherwise, there are neighborhoods $U$ and $V$ of $p\left(x\right)$ and $p\left(y\right)$ with slices $\left\{U_i\right\}$ and $\left\{V_i\right\}$. If $U$ and $V$ are disjoint, then $x$ and $y$ are in disjoint slices. Otherwise, since $B$ is Hausdorff, there are disjoint neighborhoods $U^{\prime }$ and $V^{\prime }$ of $p\left(x\right)$ and $p\left(y\right)$. Therefore, $p^{-1}\left(U^{\prime }\right)\cap U_i$ and $p^{-1}\left(V^{\prime }\right)\cap V_i$ are disjoint neighborhoods of $x$ and $y$. Analogous arguments work for $B$ regular, completely regular, and locally compact Hausdorff.$$\begin{array}{}◼& \end{array}$$

Claim: If $p:E\to B$ is a covering map, $B$ is compact, and $p^{-1}\left(b\right)$ is finite for all $b$, then $E$ is compact.

Proof: Let $\left\{U_i\right\}$ be an open cover of $E$ and let $b\in B$. Then there is a neighborhood $V$ of $b$ such that $\left\{V_j\right\}$ is a finite partition of $p^{-1}\left(V\right)$ into slices. Let $U_{i_j}$ be such that $p^{-1}\left(b\right)\cap U_{i_j}\cap V_j\ne \varnothing$. Then $W_b:=\bigcap _{j}p(U_{i_j}\cap V_j)$ is a neighborhood of $b$ such that $p^{-1}\left(W_b\right)$ is covered by finitely-many $U_i$. Since $B$ is compact, so finitely-many $W_b$ cover $B$, implying that finitely-many $p^{-1}\left(W_b\right)$ cover $E$. Therefore, finitely-many $U_i$ cover $E$.$$\begin{array}{}◼& \end{array}$$