Lifting symmetries to the universal cover

Question

If $X$ is a connected topological space with universal cover $p: \tilde{X} \to X$, I believe any homeomorphism $f : X \to X$ can be 'lifted' to a homeomorphism $\tilde{f} : \tilde{X} \to \tilde{X}$.

(I'm using the word 'lift' in a nonstandard way here. What I mean is that we can find a map $\tilde{f} : \tilde{X} \to \tilde{X}$ such that $p \circ \tilde{f} = f \circ p$. I believe, but haven't checked, that any choice of such $\tilde{f}$ is a homeomorphism.)

However, there is in general no canonical way to choose $\tilde{f}$ given $f$, so if we have a discrete group $G$ acting as homeomorphisms of $X$ this action may not give rise to an action of $G$ on $\tilde{X}$.

Is there a functorial way to build a group $G_X$ such that any action $\alpha$ of $G$ on $X$ gives rise to an action $\tilde{\alpha}$ of $G_X$ on $\tilde{X}$?

That question is a bit vague; to make it precise I think I want an onto homomorphism $\rho: G_X \to G$ such that for all $g \in G_X$, $\tilde{\alpha}(g) : \tilde{X} \to \tilde{X}$ 'lifts' $\alpha(\rho(g)) : X \to X$ in the above sense.

If $G$ acts on $X$ in a way that preserves a basepoint $\ast$ it would act on $\pi_1(X,\ast)$, allowing us to define the semidirect product of $G$ and $\pi_1(X,\ast)$, and I would try using this as $G_X$. However, I'm mainly interested in the case where the action of $G$ does not preserve any point of $X$.

Answer 1

Assuming that $G$ is discrete, the homotopy quotient $X/G$ fits into a fiber sequence

$$X \to X/G \to BG$$

and hence, by the long exact sequence in homotopy, its fundamental group $\widetilde{G} = \pi_1(X/G)$ fits into a short exact sequence

$$1 \to \pi_1(X) \to \widetilde{G} \to G \to 1$$

and moreover $X$ and $X/G$ have the same universal cover, hence $\widetilde{G}$ acts on $\widetilde{X}$ and this action extends the action of $G$ on $X$ as expected. (All of this is in a homotopical sense; I haven't thought about what it would take to get a point-set action.)

One way to describe the classification of group extensions is that extensions of $G$ by $N$ correspond to actions of $G$ on the groupoid $BN$, and the extension above comes in this way from the induced action of $G$ on the fundamental groupoid $\Pi_1(X)$. This extension splits as a semidirect product iff the corresponding action has a homotopy fixed point, which is weaker than the condition that some particular point-set model of it has an actual fixed point but still need not be satisfied in general.

Answer 2

Inspired by Qiaochu Yuan's answer, I'll note that we can solve my problem rather tautologously in the universal example where $G$ is the group of all homeomorphisms of $X$ (treated as a discrete group, for simplicity) and $\alpha$ is the identity.

In this case we can let $G_X$ be the group of all homeomorphisms $f : \tilde{X} \to \tilde{X}$ that 'lift' homeomorphisms $f : X \to X$, in the sense that $p \circ \tilde{f} = f \circ p$ . For any such $\tilde{f}$ there is a unique $f$ that makes this equation hold, so there is a homomorphism $\rho: G_X \to G$ with

$$ \rho(\tilde{f}) = f .$$

The kernel of $\rho$ is group of deck transformations of $\tilde{X}$, so we have a short exact sequence

$$1 \to \pi_1(X) \to G_X \stackrel{\rho}{\longrightarrow} G \to 1$$

so $G_X$ is an extension of $G$ by $\pi_1(X)$. When $G$ fixes a point $\ast \in X$, then $G_X$ is a semidirect product of $G$ and $\pi_1(X,\ast)$.

Answer 3

A way to solve this is to consider for every action $\alpha$ of $G$ on $X$ the group of gauge transformations $G(\alpha)$. An element of $G(\alpha)$ is an automorphism $h$ of the universal cover $\tilde X$ such that there exists $g\in G$ such that $p\circ h=\alpha(g)\circ p$. Then you can naturally defines a notion of (iso)morphism between representations $\alpha$. I assume the isomorphim class of these representations is a set $A$. Write $G_X=\Pi_{\alpha\in A}G(\alpha)$. Any action of $G$ on $X$ isomorphic to $\alpha$ yields to the action of $G_X$ on $\tilde X$ defined by the projection $G_X\rightarrow G(\alpha)$.