I have been working on this classic exercise from my topology class.
Let $X$ be a simply connected topological space, and $G$ a group that acts faithfully on $X$ by homeomorphism. Let $X/G$ be the orbit space and assume that the projection $p : X \to X/G$ is a covering map. Show: $G \cong \pi_1(X/G, [x_0])$.
I have an idea of the solution, the wanted homeomorphism being $f: G \to \pi_1(X/G, [x_0])$ sending $g\in G$ to $[p\circ\gamma_g]$, where $\gamma_g$ is a path in $X$ from $x_0$ to $g\cdot x_0$. But there is a certain step in the proof I can't get my head around: When proving injectivity, we suppose $p\circ \gamma_g \simeq p\circ \gamma_h$ and lift that homotopy to $\gamma_g \simeq \gamma_h$ to conclude $g\cdot x_0 = h\cdot x_0$.
At this point in any source I found it was then directly concluded that $g = h$, which proves injectivity. But this isn't a general property of group actions, on Wikipedia I found that a free action is needed for that step, which in general is a much stronger condition.
Is the group action in this case automatically free? Does this follow from it acting by homeomorphism (which I didn't use in the proof yet)? Has it to do with $X$ being simply connected?
Thanks in advance for any answer :)
For each $g \in G$, define $$\text{Fix}(g) = \{ x \in X : g(x) = x \} .$$
Claim 1: $\text{Fix}(g)$ is open. To see this, suppose that $x \in \text{Fix}(g)$. Since $p : X \to X / G$ is a covering space, there exists an open neighbourhood $U \subset X / G $ with $p(x) \in U$ such that $p^{-1}(U) = \coprod_i \tilde U_i$, where each open set $\tilde U_i$ in this disjoint union is mapped homeomorphically to $U$ by $p$. Now $x$ is contained in precisely one of these $U_i$'s; to fix our notation, let's say that $x \in \tilde U_{i_1}$. Since $x \in \text{Fix}(g)$, $g(x)$ is equal to $x$, so $g(x) \in \tilde U_{i_1}$ too. Since $g : X \to X$ is a continuous map, there must exist an open neighbourhood $V \subset X$ with $x \in V$ such that $g(V) \subset \tilde U_{i_1}$. For any point $x' \in V \cap \tilde U_i$, we have $x' \in \tilde U_{i_1}$ and $g(x') \in \tilde U_{i_1}$. But $x'$ and $g(x')$ are in the same orbit under the $G$ action, so $p(x') = p(g(x'))$, and since $p$ is injective on $\tilde U_{i_1}$, we have that $x' = g(x')$. Thus $V \cap \tilde U_i \subset \text{Fix}(g)$. This proves that $\text{Fix}(g)$ is open.
Claim 2: $\text{Fix}(g)$ is closed. Proved by a similar argument. Suppose $ x \in X \ \backslash \ \text{Fix}(g)$. Then $x \in \tilde U_{i_1}$ (say) and $g(x) \in \tilde U_{i_2}$ (say), where $\tilde U_{i_1}$ and $\tilde U_{i_2}$ are different. By continuity, there exists an open neighbourhood $V$ of $x$ such that $g(V) \subset \tilde U_{i_2}$. This implies that $V \cap \tilde U_i \subset X \ \backslash \ \text{Fix}(g)$. Hence $X \ \backslash \ \text{Fix}(g)$ is open, i.e. $\text{Fix}(g)$ is closed.
Claim 3: $\text{Fix}(g)$ is either $X$ or $\emptyset$. This follows from Claims 1 and 2, and from the fact that $X$ is connected. (It is connected because it is simply connected.)
Claim 4: $$\text{Fix}(g) =\begin{cases} X & \text{if } \ g = e \\ \emptyset & \text{if } \ g \neq e \end{cases}$$ Here $e$ is the identity element in $G$. This follows from Claim 3, and from the fact that $G$ acts faithfully on $X$ (which is to say that $\text{Fix}(g) \neq X$ if $g \neq e$).
Claim 5: If $g_1(x) = g_2(x)$ for some $x \in X$, then $g_1 = g_2$. This follows from Claim 4. Indeed, if $g_1(x) = g_2(x)$, then $(g_2^{-1}g_1)(x) = x$, so $x \in \text{Fix}(g_2^{-1}g_1)$, which would imply that $g_2^{-1}g_1 = e$.