# 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

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

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

Covering spaces of orientable manifolds are orientable.

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

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

## Universal 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 $$\psi(\tilde x_1) = \tilde x_2$$.

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: 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,
• $$\mathop{\mathrm{Deck}}(\widehat{X} \to X) \cong G$$.

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

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 $$F$$: 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 $$f_*(\pi_1(Y)) \subset \pi_*(\pi_1 (\tilde X))$$: Moreover, lifts are unique if they agree at a single point.

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

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.

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*}

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*}

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

For n=2  • $$T^2 \xrightarrow{\times 2} {\mathbb{K}}$$

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 $${\left\lvert {{\mathbf{Z}}} \right\rvert}$$.

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.

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 $${\mathbf{Z}}\times{\mathbf{Z}}= \pi_1(T) = \text{Aut}(\tilde X)$$

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 $$\pi_1(X_i)$$.

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.

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.

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.

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

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

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.

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

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

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

• 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.

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 $$G\cong \pi_1(X/G) / p_*(\pi_1(X))$$.

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