Question about hyperbolic space-forms

Question

I have a map between hyperbolic space-forms $\varphi:B^n/\Gamma \longrightarrow B^n/H$ (where $\Gamma, H$ are discrete groups of isometries that act freely), and a lift to a map $\tilde{\varphi}:B^n \longrightarrow B^n$. Then, calling $\pi:B^n \longrightarrow B^n/\Gamma$ and $\eta:B^n \longrightarrow B^n/H$ the projections, I have a commutative diagram such that $\eta \tilde{\varphi} = \varphi \pi$.

Let $g \in \Gamma$. Then I have the equation $\eta \tilde{\varphi} g = \eta \tilde{\varphi}$. I want to prove that, in this situation, I have a unique isometry $\varphi_*(g)$ such that $\tilde{\varphi}g=\varphi_*(g) \tilde{\varphi}$.

What I can get is that for every $x \in B^n$ there is an isometry $h_x \in H$ such that $\tilde{\varphi}g(x) = h_x \tilde{\varphi}(x)$, but I don't know how to prove that this isometry $h_x$ does not depend on the point $x$. How can I prove it?

Thanks in advance.

Answer 1

$\tilde \varphi : B^n \to B^n$ is invertible, so $\varphi_*(g) = \tilde \varphi g \tilde \varphi^{-1}$.