Is there a natural homorphism $\pi_1(X/G) \rightarrow G$?

Question

In Basic Topology (Armstrong, M.A., 1983) we have the following:

(5.13) Theorem. If $G$ acts as a group of homeomorphisms on a simply connected space $X$, and if each point $x \in X$ has a neighbourhood $U$ which satisfies $U \cap g(U) = \emptyset$ for all $g \in G - \{e\}$, then $\pi_1(X/G)$ is isomorphic to $G$.

Probably a very naive question, but is there a version of this without the assumption of simply-connectedness that gives us a (mere) homomorphism $\pi_1(X/G) \rightarrow G$? If so, how is this homomorphism defined?

I think that deck transformations are relevant here, but my understanding of them is a bit sketchy and I'm having trouble pulling my intuitions together into a coherent thought.

This question is related, but seems to go the other way.

Answer 1

To get a morphism you would first need a "concrete group" not just an isomorphism class, that is,you'd need to fix a point in $X/G$, say $x_0$.

For the morphism you'll also need a specific preimage of $x_0$, say $a_0\in X$

Then what you can see is that $p:X\to X/G$ is a covering map. So given $[\alpha]\in \pi_1(X/G, x_0)$, you can see it as a loop $\alpha$ and lift this to a path in $X$, with beginning and end points in the same orbit, say $a_0, a_1, a_1 = g\cdot a_0$.

Then you can use the hypothesis and see that $g$ is actually the only member of the group that sends $a_0$ to $a_1$, and actually it's the only (provided $X$ is connected - though not necessarily simply connected) automorphism of the covering sending $a_0$ to $a_1$. So you can define the image of $[\alpha]$ as $g$; see that this is well-defined and a morphism $\pi_1(X/G, x_0) \to G$.

In the general case, this won't be injective though of course if $X$ is path-connected (still no assumption of simple-connectedness though) it will be surjective.

In this case the kernel of this map is $p_\star(\pi_1(X, a_0))$, and so you get an isomorphism $\pi_1(X/G, x_0)/p_\star(\pi_1(X, a_0)) \simeq G$. This is exactly what you have when $X$ is simply connected, but here it's a bit more general.

Answer 2

I'm assuming that $G$ is a discrete group. Then given the existence of neighbourhoods $U$ at each point $x\in X$ satisfying $U\cap g(U)=\emptyset$, a standard result is that $X\rightarrow X/G$ is a principal $G$-covering (this is given as Proposition 14.1.12 in tom Dieck's "Algebraic Topology", for instance).

This means that $G\hookrightarrow X\xrightarrow{p}X/G$ is a fibration sequence, where $p$ is the quotient map. This gives a long exact homotopy sequence

$$0\rightarrow\pi_1X\xrightarrow{p_*} \pi_1(X/G)\xrightarrow{\Delta} \pi_0G\rightarrow \pi_0X\rightarrow\dots$$

where $\Delta$ is the fibration connecting map.

Since we assumed that $G$ was discrete we have that $\pi_0G\cong G$, not only as a set, but also as a group. This follows since the fibration is principal, which means that it classifies by some map $\delta:X/G\rightarrow BG\simeq K(G,1)$ which in turn allows us to identify $\pi_0G\cong \pi_1BG\cong G$ and the map $\Delta$ as the homomorphism induced by $\delta_*:\pi_1(X/G)\rightarrow \pi_1BG$.

Hence, identifying $\pi_0G\cong G$ we have a homomorphism

$$\Delta:\pi_1(X/G)\rightarrow G.$$

Of course, to get this homomorphism we may equally well construct the map $\delta:X/G\rightarrow BG$, identify $\pi_1BG\cong G$ and then identify the homomorphism $\delta_*:\pi_1(X/G)\rightarrow G$.

Of course in either case, the kernel of the homomorphism is simply $\pi_1X$.