Covering Spaces

Some pictures to keep in mind when it comes to covers and path lifting:

Picture to keep in mind

A more complicated situation

Useful Facts

remark:

When covering spaces are involved in any way, try computing Euler characteristics - this sometimes yields nice numerical constraints.

fact (Euler characteristics are multiplicative on covering spaces):

For $p: A \xrightarrow{} B$ an $n{\hbox{-}}$ fold cover, $\begin{align*} \chi(A) = n\, \chi(B) .\end{align*}$

fact:

Covering spaces of orientable manifolds are orientable.

fact:

The preimage of a boundary point under a covering map must also be a boundary point

fact:

Normal subgroups correspond to normal/regular coverings, where automorphisms act freely/transitively. These are “maximally symmetric”.

Universal Covers

definition (Galois/normal/regular covers):

A covering $\tilde X \xrightarrow{p} X$ is Galois (or normal/regular) over $(X, x_0)$ iff $\mathop{\mathrm{Deck}}(\tilde X)$ acts transitively on the fibers: for any two lifts $\tilde x_1, \tilde x_2\in \tilde X$ of $x_0 \in X$ , there is a $\psi\in \mathop{\mathrm{Deck}}(\tilde X)$ with $\psi(\tilde x_1) = \tilde x_2$ .

proposition (Fundamental theorem of covering spaces, Hatcher 1.39):

If $X$ is

Path-connected,
Locally path-connected, and
Semilocally simply connected,

then $X$ admits a universal cover $\widehat{X} \to X$ : if $C \xrightarrow{q} X$ is any other covering map with $C$ connected, then there exists a covering map $\tilde p: \widehat{X} \to C$ making the following diagram commute:

Link to diagram

That is, any other cover $C$ of $X$ is itself covered by $\widehat{X}$ . Note that by the universal property, $\widehat{X}$ is unique up to homeomorphism when it exists.

Moreover, letting $\tilde X \to X$ be an arbitrary path-connected cover and $H\coloneqq p_* \pi_1(\tilde X; \tilde x_0), G\coloneqq\pi_1(X; x_0)$ ,

The $\tilde X\to X$ is Galois iff $H{~\trianglelefteq~}G$ ,
$\mathop{\mathrm{Deck}}(\tilde X\to X) \cong N_{G}(H)/H$ where $N$ is the normalizer in $G$ of $H$ ,
$\mathop{\mathrm{Deck}}(\tilde X\to X) \cong G/H$ if $\tilde X\to X$ is Galois,
$\mathop{\mathrm{Deck}}(\widehat{X} \to X) \cong G$ .

remark:

A Galois cover $\tilde X\to X$ with $\mathop{\mathrm{Deck}}(\tilde X) = G$ is equivalently a principal $G{\hbox{-}}$ bundle for a discrete $G$ :

remark:

Covering spaces of $X$ are classified by subgroups of $\pi_1(X)$ : $\begin{align*} \left\{{\substack{ \text{Covering spaces $\tilde X\to X$} }}\right\}/\text{Isomorphisms over }X &\rightleftharpoons \left\{{\substack{ \text{Subgroups } H\leq \pi_1(X) }}\right\}/\text{Conjugacy} \\ \\ \left\{{\substack{ \text{Galois covering spaces $\tilde X\to X$} }}\right\} &\rightleftharpoons \left\{{\substack{ \text{Normal subgroups } H {~\trianglelefteq~}\pi_1(X) }}\right\} \end{align*}$

theorem (Homotopy lifting property for covers, Hatcher 1.30):

Let $p:\tilde X \to X$ be any covering space, $F: Y \times I \to X$ be any homotopy, and $\tilde F_0: Y\to \tilde X$ be any lift of $F_0$ . Then there exists a unique homotopy $\tilde F:Y\to \tilde X$ of $\tilde F_0$ that lifts $F$ :

Link to diagram

theorem (Lifting criterion for covers, Hatcher 1.33):

If $f: Y\to X$ with $Y$ path-connected and locally path-connected, then there exists a unique lift $\tilde f: Y\to \tilde X$ if and only if $f_*(\pi_1(Y)) \subset \pi_*(\pi_1 (\tilde X))$ :

Link to diagram

Moreover, lifts are unique if they agree at a single point.

remark (Automatic lifts):

Note that if $Y$ is simply connected, then $\pi_1(Y) = 0$ and this holds automatically!

proposition (Covering spaces induce injections on

$\pi_1$ , Hatcher 1.31):

Given a covering space $\tilde X \xrightarrow{p} X$ , the induced map $p^*: \pi_1(\tilde X) \to \pi_1(X)$ is injective. The image consists of classes $[\gamma]$ whose lifts to $\tilde X$ are again loops.

theorem (Fundamental theorem of covering spaces, Hatcher 1.39):

For $\tilde X \xrightarrow{p} X$ a covering space with

$\tilde X$ path-connected,
$X$ path-connected and locally path-connected,

letting $H$ be the image of $\pi_1(\tilde X)$ in $\pi_1(X)$ , we have

$\tilde X$ is normal if and only if $H{~\trianglelefteq~}\pi_1(X)$ ,
For the normalizer $N_G(H)$ where $G\coloneqq\pi_1(X)$ , $\begin{align*} G(\tilde X) \coloneqq\mathop{\mathrm{Aut}}_{\mathrm{Cov}(X) }(\tilde X) \cong {N_G(H) \over H} .\end{align*}$

In particular, $\begin{align*} \tilde X \text{ normal} &\implies G(\tilde X) \cong \pi_1(X) / H \\ \widehat{X} \text{ universal} &\implies G(\widehat{X}) \cong \pi_1(X) .\end{align*}$

fact:

There is a contravariant bijective correspondence $\begin{align*} \left\{{\substack{ \text{Connected covering spaces} \\ p: \tilde X \xrightarrow{} X }}\right\}_{/\sim} &\rightleftharpoons \left\{{\substack{ \text{Conjugacy classes of subgroups} \\ \text{of } \pi_1(X) }}\right\} .\end{align*}$ If one fixes $\tilde x_0$ as a basepoint for $\pi_1(\tilde X)$ , this yields $\begin{align*} \left\{{\substack{ \text{Connected covering spaces} \\ p: \tilde X \xrightarrow{} X }}\right\}_{/\sim} &\rightleftharpoons \left\{{\substack{ \text{Subgroups of } \pi_1(X) }}\right\} .\end{align*}$

proposition (Number of sheets in a covering space, Hatcher 1.32):

For $X, \tilde X$ both path-connected, the number of sheets of a covering space is equal to the index $\begin{align*} [p^*(\pi_1(\tilde X)): \pi_1(X)] .\end{align*}$

Note that the number of sheets is always equal to the cardinality of $p ^{-1} (x_0)$ .

Examples

fact (Some common covering spaces):

Any subgroup $H \leq \pi_1(S^1; 1) = {\mathbf{Z}}$ is of the form $H = n{\mathbf{Z}}$ , so intermediate covers are obtained from $\widehat{S^1}/n{\mathbf{Z}}= {\mathbf{R}}/n{\mathbf{Z}}\cong S^1$ by pulling back the universal bundle:

Link to Diagram

given by the $n{\hbox{-}}$ valent Cayley graph covering a wedge of circles.

For n=2

$T^2 \xrightarrow{\times 2} {\mathbb{K}}$

example (The circle

$S^1$ ):

Identify $S^1 \subset {\mathbf{C}}$ , then every map $p_n: S^1 \to S^1$ given by $z\mapsto z^n$ a yields a covering space $\tilde X_n$ . The induced map can be described on generators as $\begin{align*} p_n^*: \pi_1(S^1) &\to \pi_1(S^1) \\ [\omega_1] &\mapsto [\omega_n] = n[\omega_1] \end{align*}$ and so the image is isomorphic to $n{\mathbf{Z}}$ and thus $\begin{align*} p_n^*(\pi_1(S^1)) = \mathop{\mathrm{Aut}}_{\mathrm{Cov} }(\tilde X_n) = {\mathbf{Z}}/n{\mathbf{Z}} .\end{align*}$ where the deck transformations are rotations of the circle by $2\pi/n$ . The universal cover of $S^1$ is ${\mathbf{R}}$ ; this is an infinitely sheeted cover, and the fiber above $x_0$ has cardinality ${\left\lvert {{\mathbf{Z}}} \right\rvert}$ .

example (Projective

$n\dash$ space

$\RP^n$ ):

The universal cover of ${\mathbf{RP}}^n$ is $S^n$ ; this is a two-sheeted cover. The fiber above $x_0$ contains the two antipodal points.

example (The torus

$S^1 \cross S^1$ ):

The universal cover of $T = S^1 \times S^1$ is $\tilde X ={\mathbf{R}}\times{\mathbf{R}}$ . The fiber above the base point contains every point on the integer lattice ${\mathbf{Z}}\times{\mathbf{Z}}= \pi_1(T) = \text{Aut}(\tilde X)$

proposition (General construction for wedge products):

For a wedge product $X = \bigvee_i^n \tilde X_i$ , the covering space $\tilde X$ is constructed as a infinite tree with $n{\hbox{-}}$ colored vertices:

Each vertex corresponds to one of the universal covers $\tilde X_i$ ,
The color corresponds to which summand $\tilde X_i$ appears,
T The neighborhood of each colored vertex has edges corresponding (not bijectively) to generators of $\pi_1(X_i)$ .

example (Wedge of circles):

The fundamental group of $S^1 \vee S^1$ is ${\mathbf{Z}}\ast {\mathbf{Z}}$ by van Kampen, and the universal cover is the following 4-valent Cayley graph:

The universal cover of \S^1 \vee S^1

See Hatcher p.58 for other covers.

corollary (Every subgroup of a free group is free):

Idea for a particular case: use the fact that $\pi_1\qty{\bigvee^k S^1} = {\mathbf{Z}}^{\ast k}$ , so if $G \leq {\mathbf{Z}}^{\ast k}$ then there is a covering space $X \twoheadrightarrow\bigvee^k S^1$ such that $\pi_1(X) = G$ . Since $X$ can be explicitly constructed as a graph, i.e. a CW complex with only a 1-skeleton, $\pi_1(X)$ is free on the edges in the complement of a maximal tree.

example (Wedge of projective spaces):

The fundamental group of ${\mathbf{RP}}^2 \vee {\mathbf{RP}}^2$ is ${\mathbf{Z}}_2 \ast {\mathbf{Z}}_2$ , corresponding to an infinite string of copies of 2-valent $S^2$ s:

Another universal cover.

example (

$\RP^2 \wedgeprod T^2$ ):

The fundamental group of ${\mathbf{RP}}^2 \vee T^2$ is ${\mathbf{Z}}_2 \ast {\mathbf{Z}}$ , and the universal cover is shown in the following image. Each red vertex corresponds to a copy of $S^2$ covering ${\mathbf{RP}}^2$ (having exactly 2 neighbors each), and each blue vertex corresponds to ${\mathbf{R}}^2$ cover ${\mathbb{T}}^2$ , with ${\left\lvert {{\mathbf{Z}}^2} \right\rvert}$ many vertices as neighbors.

Universal cover of {\mathbb{T}}^2 \vee {\mathbf{RP}}^2

Applications

theorem (Maps into contractible spaces are always nullhomotopic):

If $X$ is contractible, every map $f: Y \to X$ is nullhomotopic.

proof (?):

If $X$ is contractible, there is a homotopy $H: X\times I \to X$ between $\operatorname{id}_X$ and a constant map $c: x \mapsto x_0$ . So construct $\begin{align*} H': Y\times I &\to X \\ H'(y, t) &\coloneqq \begin{cases} H(f(y), 0) = (\operatorname{id}_X \circ f)(y) = f(y) & t=0 \\ H(f(y), 1) = (c \circ f)(y) = c(y) = x_0 & t=1 \\ H(f(y), t) & \text{else}. \end{cases} \end{align*}$ Then $H'$ is a homotopy between $f$ and a constant map, and $f$ is nullhomotopic.

corollary (Factoring through a contractible space implies nullhomotopic):

Any map $f:X\to Y$ that factors through a contractible space $Z$ is nullhomotopic.

proof (?):

We have the following situation where $f = p \circ \tilde f$ :

Link to diagram

Since every map into a contractible space is nullhomotopic, there is a homotopy $\tilde H: Y\times I \to Z$ from $\tilde f$ to a constant map $c: Y\to Z$ , say $c(y) = z_0$ for all $y$ . But then $p\circ \tilde H: X \times I \to Y$ is also a homotopy from $f$ to the map $p\circ c$ , which satisfies $(p\circ c)(y) = p(z_0) = x_0$ for some $x_0 \in X$ , and is in particular a constant map.

proposition (Showing one space can not cover another):

There is no covering map $p: {\mathbf{RP}}^2 \to {\mathbb{T}}^2$ .

proof (?):

Use the fact that $\pi_1({\mathbb{T}}^2) \cong {\mathbf{Z}}^2$ and $\pi_1({\mathbf{RP}}^2) = {\mathbf{Z}}/2{\mathbf{Z}}$ are known.
The universal cover of ${\mathbb{T}}^2$ is ${\mathbf{R}}^2$ , which is contractible.
Using the following two facts, $p_*$ is the trivial map:
- By the previous results, $p$ is thus nullhomotopic.
- Since $p$ is a covering map, $p_*: {\mathbf{Z}}/2{\mathbf{Z}}\hookrightarrow{\mathbf{Z}}^2$ is injective.
Since $p$ was supposed a cover, this can be used to imply that $\operatorname{id}_{{\mathbb{T}}^2}$ is nullhomotopic.
Covering maps induce injections on $\pi_1$ , and the only way the trivial map can be injective is if $\pi_1(T^2) = 0$ , a contradiction.

theorem (When actions yield covering maps onto their quotients, Hatcher 1.40):

If $G\curvearrowright X$ is a free and properly discontinuous action, then

The quotient map $p:X \to X/G$ given by $p(y) = Gy$ is a normal covering space,
If $X$ is path-connected, then $G = \mathop{\mathrm{Aut}}_{\mathrm{Cov}} (X)$ is the group of deck transformations for the cover $p$ ,
If $X$ is path-connected and locally path-connected, then $G\cong \pi_1(X/G) / p_*(\pi_1(X))$ .

fact:

If $f:X\to Y$ is a covering map of degree 1, then $f$ is necessarily a homeomorphism.

Exercises

See https://www.math.utah.edu/~patrikis/6520Spring2018/6520hw4.pdf