Let $\mathbb{H}$ be hyperbolic plane, $\Gamma$ is a discrete subgroup of $PSL\_2(\mathbb{R}$) so that $\Gamma \backslash \mathbb{H}$ is a compact hyperbolic surface. Maybe it will be very simple to you but I am very confused when I try to construct a homeomorphism $\phi: \Gamma \backslash T\_1 \mathbb{H} \longrightarrow T\_1(\Gamma \backslash \mathbb{H})$


Suppose $\Gamma (z,v)$ is an element of $\Gamma \backslash T\_1 \mathbb{H}$, so what is $\phi (\Gamma (z,v))$? Actually I read in some book and they always accept this fact obviously, but for me it's not obvious. 


I'm so sorry if it is not a question of researching. 



What is the homeomorphism from $\Gamma \backslash T_1 \mathbb{H}$ to $T_1(\Gamma \backslash \mathbb{H})$
I'll define the map and leave the proof that it is a diffeo to you. Assume $M$ is a Riemannian manifold and $\Gamma$ is a torsion free group that acts properly discontinuously by isometries on $M$. Get the quotient map $M\to \Gamma\backslash M$.
Check that it is differentiable map between Riemannian manifolds which derivative at every point is an isometry and get an associated map $T^1M\to T^1(\Gamma\backslash M)$. Note that this map is $\Gamma$-invariant, thus factors through the orbit space $\Gamma\backslash (T^1M)$. Call the resulting map $\phi:\Gamma\backslash (T^1M)\to T^1(\Gamma\backslash M)$.


