Group action on topological spaces. 
Let $X$ be a simply connected topological space, and $G$ a group that acts faithfully on $X$ by homeomorphism (i.e. the map $x \mapsto g \cdot x$ is a homeomorphism for each $g \in G)$. Let $X / G$ be the orbit space, i.e. the quotient $X / \sim$ where $x \sim y$ iff there is a $g \in G$ such that $x=g \cdot y$, and assume that the projection $p: X \rightarrow X / G$ is a covering map. Show that then, $G \cong \pi_{1}\left(X / G,\left[x_{0}\right]\right)$.

Below is my attempt to prove the statement, in which I seek an isomorphism between $G$ and $\pi_{1}(X / G,\left[x_{0}\right])$. I have some questions for both the surjectivity and injectivity parts.

For $g \in G$, let $\tilde{\alpha}_g$ be a path in $X$ with $\tilde{\alpha}_g(0) = x_0$ and $\tilde{\alpha}_g(1) = g \cdot x_0$. Since $X$ is simply connected, such path exists. Also let $\alpha_g = p \circ \tilde{\alpha}_g$. We observe that, since the images of $x_0$ and $g \cdot x_0$ under $p$ are in the same orbit by definition, $\alpha_g$ is a loop in $X/G$ based at $[x_0]$. Thus $\alpha_g \in \pi_1(X/G, [x_0])$.
Now we define $\varphi: G \to \pi_1(X/G, [x_0])$ to be $\varphi(g) = [\alpha_g]$. We claim that $\varphi$ is an isomorphism.

To show that $\varphi$ is surjective, I let $[\gamma] \in \pi_1(X/G, [x_0])$. So $\gamma(0) = \gamma(1) = [x_0]$. This implies $p$ send the pre-images of both $\gamma(0)$ and $\gamma(1)$ to the same orbit. So $p^{-1} \circ \gamma(0) = g \cdot x_0$, and $p^{-1} \circ \gamma(1) = h \cdot x_0$, for some $g$ and $h$ in $G$. It is here that I’m stuck, as I can’t say which element of $G$ would give the path between $g \cdot x_0$ and $h \cdot x_0$. Another approach I could think of is to say we choose a lift $\tilde{\gamma}$ of $\gamma$ that starts at $x_0$ (instead of $g \cdot x_0$). But am I allowed to do that?
For injectivity, I sort of running into the same problem. Let $g \in G$ be such that $\varphi(g) = [x_0]$. So I need to show that $g$ is the identity in $G$. This would be true if $\tilde{\alpha_g}$ is the constant path at $x_0$ in $X$. But why can’t we have the case where $\tilde{\alpha_g}$ is a path from $x_0$ to $(g \cdot x_0) \neq x_0$, and each point on the path is $g^\prime \cdot x_0$ for some $g^\prime \in G$?
 A: In equations like $p^{-1} \circ \gamma(0) = g \cdot x_0$, you seem to be treating $p^{-1}$ as if it were a function from $X/G$ to $X$. But it's not: the function $p : X \to X/G$ is not injective so $p^{-1} : X/G \to X$ is not defined.
Of course $p^{-1}$ is defined as a set function: for any set $A \subset X$ we have
$$p^{-1}(A) = \{x \in X \mid p(x) \in A\}
$$
So for example $p^{-1}\{\gamma(0)\}$ are $p^{-1}\{\gamma(1)\}$ are both subsets of $X$, but because $\gamma(0)=\gamma(1)=[x_0]$ it follows that they are the same subset. And that subset is $G \cdot x_0$, the entire $G$-orbit of $x_0$.
What you seem to be missing for purposes of proving surjectivity is the path lifting property. And for purposes of proving injectivity you are missing the homotopy lifting property. Both of those properties are stated briefly here. And then you need the "path lifting lemma" and the "homotopy lifting lemma" which says that both of those properties are true in the case of a covering map. This is a situation where it is worth cracking open a good topology book such as Munkres "Topology".
A: In general if you have a covering map $p: X\to Y$, then there exists a natural map
$f: \pi (Y,y_0)\to p^{-1}(y_0) , y_0:=p(x_0)$
that is the so called monodromy map. This map is defined in the following way:
If $\gamma$ is a loop of base point $y_0$, then $f(\gamma):=\gamma^\sim(1)$ is the unique lifting of $\gamma$ in $X$ started at $x_0$ and computed at the time $t=1$.
At this point one can observe $f$ is surjective, when $X$ is path connected. Conversely one can observe $f(\gamma)$ is zero if and only if $\gamma=p\circ \gamma’$ , where $\gamma’$ is a loop of $\pi(X,x_0)$.
By first theorem of isomorphism (of sets) one get
$p^{-1}(y_0)\cong \pi(Y,y_0)/ \sim_f=\pi(Y,y_0)/p^*(\pi(X,x_0))$
In your case $X$ is path and simply connected, so $\pi(X,x_0)=0$ and we get
$G\cong p^{-1}(y_0) \cong \pi(Y,y_0)$
This equivalence is only an equivalence of sets, so you must to prove that in your case the monodromy map $f$ is also an omomorphism of groups.
Can you continue?
