Fundamental group of quotient space under a continuous group action Suppose $X$ is a simply connected topological space, and let $\rho$ be a continuous group action of a group $G$ on $X$ that satisfies the following condition: for any $x \in X$, there exists an open set $U$ containing $x$ such that $U \cap \rho(g)(U) = \emptyset$ for all nontrivial $g \in G$ (in some places this is referred to as a properly discontinuous action).
Under the above conditions, I already know that the quotient mapping $X \to X/G$ is a covering map. I want to understand why $\pi_1 (X/G) = G$.
Is there any easy way to see why this is true? I have access to Hatcher's book, but I'm having a bit of a hard time putting the pieces together to understand why this is true. I understand why this is true when $X$ is the Cayley graph of a free group $F$ and $G$ is a subgroup of $F$, as this is a very nice space with a natural group action, but I want to understand why this happens in a more general setting.
 A: Well, since $X$ is simply connected, then by construction $q : X \to X/G$ is the universal cover of $X / G$.
Now, if we have a basepoint $[x_0] \in X / G$, then the fibre $q^{-1}([x_0])$ is $\{ x \mid qx = [x_0] \}$, and (again basically by construction) we see that this is $\{ g x_0 \mid g \in G \}$. That is, we can identify the fibre over $[x_0]$ with $G$. (As an aside, this identification is noncanonical in the sense that it requires us to choose a representative of $[x_0] \in X/G$ in order to make the identification. Properly, then, we should think of the fibre as a $G$ torsor, but this is a fairly subtle distinction that we can gloss over for now).
But now, what is the fundamental group of $X/G$? Well, it's all the paths in $X/G$ which start and end at $[x_0]$. But we can lift one of these paths to $X$, and when we do this we get a path from $x_0$ to some other point in the fibre of $[x_0]$ -- this is because if $\gamma$ is a loop in $X/G$ and $\tilde{\gamma}$ is its lift, then $q\tilde{\gamma} = \gamma$ by the definition of a lift, and so $q \tilde{\gamma}(1) = \gamma(1) = [x_0]$, so that $\tilde{\gamma}(1) \in q^{-1}([x_0])$.

At this point, you should draw pictures of:

*

*$S^1$, with its universal cover $\mathbb{R}$

*$S^1 \times S^1$, with its universal cover $\mathbb{R}^2$ (you'll probably want to draw this as a square with opposite sides identified)

*$S^1 \vee S^1$, with its universal cover "the cayley graph of $F_2$"

and convince yourself that loops in the base space lift to paths in the universal cover whose endpoints
lie in the fibre of the basepoint.

But we know that points in the fibre $q^{-1}([x_0])$ are in bijection with $g x_0$ for group elements $g$! If we're feeling optimistic (and if we know a handful of examples) we might guess that two loops in $X/G$ are homotopic if and only if they lift to the same point in the fibre $q^{-1}([x_0])$. If this is true, then it means we get a bijection $\pi_1(X/G) \to q^{-1}([x_0]) \cong G$, which (with some work) we can show to be an isomorphism of groups).

Now you should look at the pictures you drew before and convince yourself that two loops are homotopic in the base space if and only if they lift to the same endpoints in the universal cover. This will be easiest to see for
$S^1 \vee S^1$, since two loops basically never lift to the same endpoint. It will be very instructive to see this for $S^1 \times S^1$, since now we might get two a priori different loops that lift to the same endpoint, and you'll have to convince yourself that they're actually the same. This will be related to the abelian-ness of $\pi_1(S^1 \times S^1) = \mathbb{Z}^2$.

And importantly, we can show this! If $\gamma_1$ and $\gamma_2$ are loops in $X/G$ whose lifts $\tilde{\gamma_1}$ and $\tilde{\gamma_2}$ both end at the same $gx_0$, then $\tilde{\gamma_1} \tilde{\gamma_2}^{-1}$ is a loop in $X$, and thus there's a homotopy $H$ contracting this loop to the identity (since $X$ is simply connected). Then $qH$ gives the desired homotopy between $q\tilde{\gamma_1} = \gamma_1$ and $q \tilde{\gamma_2} = \gamma_2$.
Now, it takes a little bit of work to show that composing loops downstairs is compatible with the $G$ structure on the fibre. The idea, though, is that even if the identification of the fibre with $G$ is noncanonical, the change $gh^{-1}$ is well defined. But this is all we need, since we're really identifying the loop in $X/G$ with the group element that changes the fibre by $g$. So if we do a loop that moves $x_0$ to $gx_0$, that loop also moves $hx_0$ to $ghx_0$ -- all the changes are the same! Then if we compose two loops, one changing $x_0 \to hx_0$ and one changing $x_0 \to g x_0$, then the composite makes the change $x_0 \to hx_0 \to ghx_0$, so that composing loops is the same thing as composing the $G$ action on the fibre.

At this point you should again consult your pictures. Take two loops downstairs and concatenate the paths.
Where do the paths lift to? Convince yourself by trying a couple examples that if you lift $\gamma_1 \cdot \gamma_2$ then the endpoint of the lift is the group product in the fibre of where we would get by lifting $\gamma_1$ and lifting $\gamma_2$.


To summarize, then:
The fibre of a basepoint $[x_0] \in X/G$ can be identified $G$ after we pick a basepoint $x_0 \in X$ with $q x_0 = [x_0]$. Now given a loop $\gamma : [x_0] \to [x_0]$ in $X/G$, this lifts to a path $\tilde{\gamma} : x_0 \to x_1$ in $X$, where $qx_1 = [x_0]$ so that $x_1 = g x_0$ for some (unique!) $g \in G$.
Now it turns out that the map $\pi_1(X/G) \to G$ sending $\gamma$ to this $g$ is an isomorphism of groups. We sketched an argument above, and you can find a more precise treatment in any algebraic topology book of your choosing. I'm fond of Hatcher's or Rotman's books, personally, but you might also like the treatment in Pierre Albin's recorded lectures (available on youtube). The video relevant to this particular question is probably this one, but you may have to go $\pm 1$ lecture (I didn't rewatch them, I'm going purely off the video title).

I hope this helps ^_^
