# How to prove that $\phi:G\rightarrow \pi_1(X/G,p(x_0))$ is a homomorphism of groups?

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 $p\circ\alpha$, that is $\langle p\circ\alpha\rangle$. This map is independent of the choice of path as $X$ is simply connected.

I am having trouble proving that it is a group homomorphism.

If $g_1,g_2\in G$ then let $\gamma_1$ be a path from $x_0$ to $g_1(x_0)$ and $\gamma_2$ be a path from $x_0$ to $g_2(x_0)$ then $\gamma_1*g_1\circ\gamma_2$ is a path from $x_0$ to $g_1\circ g_2(x_0)$. So ,$$\phi(g_1\circ g_2)=\langle p \circ (\gamma_1*g_1\circ\gamma_2)\rangle=\langle (p \circ \gamma_1)*(p\circ g_1\circ\gamma_2)\rangle$$

But I am not sure where to go from here. The next step should be to say this is equal to $\langle p\circ\gamma_1\rangle\langle p\circ\gamma_2\rangle$. Now obviously $g_1\circ\gamma_2$ is homotopic to $\gamma_2$ and both project to loops based at $p(x_0)$. I know that these loops are homotopic in $X/G$ however why are they path homotopic?

Thanks!

They are not only path homotopic, they are the same curve as $p\circ g_1 = p$ by definition of $X/G$. So $$p\circ g_1\circ\gamma_2 = p\circ\gamma_2 = \phi(g_2).$$