Restriction to a fundamental domain is an isometry

Question

Suppose the group $\Gamma$ acts on a Riemannian manifold $M$ by isometries and properly and discontinuously. Let $\pi : M \to M/\Gamma$ be the quotient map, which is a Riemannian covering map if we endow $M/\Gamma$ with the right metric. For $p \neq q \in M$, define

$$H_{p,q} = \{ x \in M : d(p, x) < d(q,x) \}.$$

The fundamental domain of $\Gamma$ centered at $p \in M$ is

$$\Delta_p = \bigcap_{g \in \Gamma \setminus \{e\} } H_{p, g(p)} = \{ x \in M : d(p, x)< d(g(p), x) \text{ for all } g \in \Gamma \setminus \{e\} \}.$$

Facts about $\Delta_p$ can be found in this paper. Is it true that $\pi$ restricted to $\Delta_p$ is an isometry? It suffices to show that this restriction is injective.

Answer 1

Assume that $x,\ y\in \Delta_p$ and $\pi (x)=\pi(y)$. In further assume that $d(p,y)\geq d(p,x)$.

Then $y\in \pi^{-1}(\pi (x))$ so that there is $g$ s.t. $gx=y$.

So $$ d(p,y)\geq d(p,x)=d(gp,gx)=d(gp,y) $$ That is, $y$ is not in $\Delta_p$. It is a contradiction.