Dirichlet domain is actually a polygon Let $\Gamma \subset PSL(2,\mathbb{R})$ be a Fuchsian group acting properly discontinuously (without fixed point) on the hyperbolic plane $\mathbb{H}^2$, such that $\mathbb{H}^2/\Gamma$ is compact. Fix a point $z_0 \in \mathbb{H}^2$. We define the Dirichlet domain associated to $z_0$ to be the set
$$
\Delta(z_0)=\left\{z \in \mathbb{H}^2: d(z,z_0)< d(z,\gamma z_0), \forall \gamma \in \Gamma\backslash \left\{id\right\} \right\}
$$
where $d$ denotes the usual hyperbolic metric.

Goal : $\Delta(z_0)$ is a polygon, i.e. a finite intersection of half-spaces.

We take the following approach : 
It is known that if $\mathbb{H}^2/\Gamma$ is a closed surface (which is our case), then for every $z \in \mathbb{H}^2$, we have that 
$$
\mathbb{H}^2=\cup_{\gamma \in \Gamma}B(\gamma z,R)
$$
where $R=diam(\mathbb{H}^2/\Gamma)$. Now set $S=\left\{ \gamma \in \Gamma : d(z_0,\gamma z_0) \leq R\right\}$ : note that $S$ is finite. Define 
$$
\Delta^S(z_0)=\left\{ z \in \mathbb{H}^2 : d(z,z_0) < d(z,\gamma z_0),\forall \gamma \in S \backslash \left\{id\right\} \right\}
$$

Goal : $\Delta(z_0)=\Delta^S(z_0)$

It is readily shown that $\Delta^S(z_0)$ is indeed a polygon, so the approach makes sense. In the details of the proof, I must justify the following : 

Lemma : $\Delta^S(z_0) \cap C(z_0,R)=\emptyset$, where $C(z_0,R)=\left\{z \in \mathbb{H}^2 : d(z,z_0)=R \right\}$.

Somehow I am unable to show this. The inequalities we must prove are very sharp, so basic "algebraic" attempts revealed themselves to be fruitless. Here are two directions I thought of :


*

*Take $z \in C(z_0,R)$. Since the $\gamma$-balls of radius $R$ cover $\mathbb{H^2}$, we can find $\tilde{\gamma} \in \Gamma, \tilde{\gamma} \neq id$ such that $d(z,\tilde{\gamma}z_0) \leq R$; in that case, $\tilde{\gamma} \notin S$. The path from here on out seems difficult.

*Suppose by the absurd that the intersection is nonempty, and let $z$ belong to this intersection. I know that for a sufficiently small $\delta$, the projection map $\pi : \mathbb{H}^2 \to \mathbb{H}^2/\Gamma$ restricts to a bijective isometry $\pi' : B(z,\delta) \to B(\pi(z),\delta)$. Now we can always take a smaller radius such that the ball $B(z,\delta)$ intersects $\Delta^S(z_0)$ (because $\Delta^S(z_0)$ is open) I assume that this isn't possible, but don't know how to show it. With additional information on how $\pi'$ acts on these two sets maybe we can get the desired contradiction.


It is clear that I am missing one good nontrivial argument/construction to prove this and I have hit a dead end. I am also struggling to make use of all our hypotheses (do we need a further interpretation of the compactness of the hyperbolic surface?). Maybe we can work with appropriately chosen geodesics, but my attempts in this direction have also been fruitless. Any help would be greatly appreciated. Thanks for reading ! (I can provide additional details/assumptions on the objects if needed.)
 A: The claim seemed difficult to prove because... it is false as stated. However, we can make a slight adjustment to prove the result : set instead $$S=\left\{ \gamma \in \Gamma : d(z_0,\gamma z_0) \leq 2R\right\}$$ Now we show that $\Delta^S \cap C(z_0,R)=\emptyset$. Take $z \in C(z_0,R)$. We want to show that $z \notin \Delta^S$, i.e. there exists some $\tilde{\gamma} \in S, \tilde{\gamma} \neq id$ such that $$R=d(z,z_0)\geq d(z,\tilde{\gamma}z_0)$$
Since $\cup_{\gamma \in \Gamma}B(\gamma z,R)=\mathbb{H}^2$, there exists $\gamma ' \in \Gamma$ such that $d(\gamma 'z,z_0)<R$. Moreover $\gamma ' \neq id$. Indeed, if that were the case, then we would have $d(z,z_0)<R$ : but we assumed that this distance was exactly $R$. Set now $\tilde{\gamma}=\gamma ' ^{-1}$. It remains only to show that we have $\tilde{\gamma} \in S$, i.e. $d(z_0,\tilde{\gamma}z_0)\leq 2R$. But this is immediate : indeed,
$$
d(z_0,\tilde{\gamma}z_0)\leq d(z_0,z)+d(z,\tilde{\gamma}z_0)\leq 2R$$
by our hypothesis and what precedes.
And we have proved the result.
