Let $X$ be any compact, connected surface of constant negative curvature $-1$. It is known that there is a covering map $\rho: \mathbb H \to X$ that is a local isometry, where $\mathbb H$ is the hyperbolic upper half plane.
Let
$$\Gamma:=\{\gamma\in SL(2,\mathbb R): \rho(\gamma z)=\rho(z), \forall z\in \mathbb H \}$$
I wonder how to prove that $\Gamma$ is discrete and that $\Gamma \backslash SL(2,\mathbb R)$ is compact.
For the discreteness, I feel like we need to use the local homeomorphism property of $\rho$ somehow but I can't formula it into a rigorous proof.
For the compactness, I know usually one wants to embed $\Gamma \backslash SL(2,\mathbb R)$ into a compact space as a closed subspace. One way to do this is by
$$f: \Gamma \backslash SL(2,\mathbb R) \to X, \Gamma x \mapsto \rho(x).$$
This map is well-defined, continuous and injective. As long as I could show that its image is closed in $X$, we are done (but I don't know how to show it rigorously).
You can do this with covering space theory.
First of all, the deck transformation group $D(\rho)$ of the covering map $\rho : \mathbb H \to X$ is defined to be the group of all homeomorphism $h : \mathbb H \to \mathbb H$ such that $\rho(hz)=\rho(z)$ for all $z \in \mathbb H$.
Obviously $\Gamma$ is a subgroup of $D(\rho)$. The key is to prove that $\Gamma = D(\rho)$, and to do this one has to prove that for each $h \in D(\rho)$, the homeomorphism $h : \mathbb H \to \mathbb H$ is an isometry. The first step is to use the fact that $\rho$ is a locally isometric covering map. Using this, one proves that $h$ is also a local isometry, because for each $x \in \mathbb H$ with $y = h(x) \in \mathbb H$, there exist open subsets $U \subset X$, $V_x \subset \mathbb H$, and $V_y \subset \mathbb H$ such that
The next step is to prove that a bijective local isometry of $\mathbb H$ is an isometry of $\mathbb H$, but I'll leave that for you to ponder.
So, knowing that $\Gamma=D(\rho)$, discreteness of $\Gamma$ follows from a general theorem that the deck transformation group of any covering map acts properly discontinuously,: to see why this follows, if $\Gamma$ is not discrete then it has a non-identity subsequence converging to the identity in $SL(2,\mathbb R)$, and one can show that this contradicts proper discontinuity using the fact that the Lie group topology on $SL(2,\mathbb R)$ is the same as the compact open topology with respect to the action of $SL(2,\mathbb R)$ on $\mathbb H$.
Finally, cocompactness of $\Gamma=D(\rho)$ follows from the fact that the orbit space is homeomorphic to $X$ which is cocompact.