Showing that $\pi_1(X/G) \cong G$. Suppose G is a discrete topological group acting freely on a simply connected topological space X. I am trying to show that $\pi_1(X/G) \cong G$.
Here is my progress so far: I have shown that $X \to X/G$ is a universal cover and the fibers are isomorphic to $G$ as $G-$sets. How would I determine where each element of $G$ is sent in the fundamental group?
 A: The idea is the following. We want to define a homomorphism $f$ from the fundamental group to $G$.Take an element of the fundamental group $\gamma: [0,1] \to X/G $ such that $\gamma(0)=\gamma(1)=x_0$ (a representative of the orbit). This lifts to a $\bar \gamma: [0,1] \to X$ such that $\bar \gamma(0)= x_0$, because the last is an universal cover. If $ \bar \gamma(1)= gx_0$, we let $f(\gamma) = g$. Now we should verify:


*

*multiplicativity. This follows from a tricky use of uniqueness of lifting, which I left you as an exercise.

*injectivity. If $\bar \gamma (1) = \bar \gamma'(1)$, then they start and finish in the same points, thus they are homotopically equivalent by simply connectedness of $X$. Then,their image via the projection to $X/G$, namely $\gamma,\gamma'$ are homotopically equivalent.

*surjectivity. $X$ is path connected, thus there exist a $\bar \gamma$ from $x_0$ to every $gx_0$ . Take $\gamma = \pi \bar \gamma $ and you know, by uniqueness of lifting, that $f(\gamma)= g$.
