Let $G$ be a discontinuous group (this means that it acts discontinuously with finite stabilizers) of homeomorphisms of a simply connected, locally compact metric space $X$. Let $p:X\rightarrow X/G$ be the orbit map. Choose a base point $x_0\in X$ and let $\pi_1(X/G,p(x_0))$ be the fundamental group of $X/G$.

Define $\phi:G\rightarrow \pi_1(X/G,p(x_0))$ as follows - for any $g\in G$ let $\alpha$ be a path from $x_0$ to $g(x_0)$. Send $g$ to the homotopy class of loops $p\circ\alpha$. This map is independent of the choice of path as $X$ is simply connected. It is also easy to check that it is a group homomorphism.

We can show that for a discontinuous group of homeomorphisms of a locally compact metric space, $p:X\rightarrow X/G$ has the path lifting property. This helps show that $\phi$ is surjective.

What I want to know is

what is the kernel of $\phi$?

I know that $\pi_1(X/G,p(x_0)) \cong G/N$ where $N$ is the normal subroup generated by all those elements of $G$ that have fixed points. So my guess is that $\ker\phi\cong N$ but I am not sure how to show this. If $g\in G$ has a fixed point say $g(x)=x$ then we can join $x_0$ to $x$ by a path $\gamma$ and use $\gamma*(g\circ\gamma)^{-1}$ to join $x_0$ to $g(x_0)$. Then I have to show $p\circ(\gamma*(g\circ\gamma)^{-1})$ is null homotopic in $X/G$. But what about the other way that is given a null homotopic loop in $X/G$ why should it come from a $g$ with has a fixed point?

Thanks!