Let $\Gamma$ be a group os isometries of $\mathbb{S}^n$ acting properly and discontinuously. For $p \neq q \in \mathbb{S}^n$, define
$$H_{p,q} = \{ x \in \mathbb{S}^n : d(p, x) < d(q, x) \},$$
where $d$ is the distance function on $\mathbb{S}^n$. The fundamental domain of $\Gamma$ centered at $p$ is then
$$\Delta_p = \bigcap_{g \in G \setminus \{e\}} H_{p, g(p)}.$$
Facts about $\Delta_p$ can be found on this paper. For example, $\Delta_p$ is open and star-like with respect to $p$. My question is: does it always happen that the antipodal point of $p$ does not lie in $\overline{\Delta}_p$ when $\Gamma$ is not trivial?
Based on the following facts, found on the above cited paper, I prove that $-p \not \in \overline{\Delta}_p$.
Fact 1: $\partial \Delta_p = \bigcup_{g \neq e} \partial \Delta_p \cap \partial \Delta_{g(p)}$
Fact 2: $\overline{\Delta}_p \cap \overline{\Delta}_{g(p)} \subset A_{p, g(p)}$, where $A_{p,g(p)} = \{ x \in \mathbb{S}^n : d(p, x) = d(g(p), x)\}$.
Suppose $-p \in \overline{\Delta}_p$. We divide in two cases
Case 1: $-p \in \Delta_p$. In this case, we would have, for each $g\neq e$,
$$\pi = d(p, -p) < d(g(p), -p),$$
absurd.
Case 2: $-p \in \partial \Delta_p$. In this case, from the facts above, we conclude that there exists $g_0 \in \Gamma, g_0 \neq e$, such that $-p \in \partial \Delta_p \cap \partial \Delta_{g_0(p)} \subset A_{p,g_0(p)}$. Thus, from
$$\pi = d(p, -p) = d(g_0(p), -p)$$
we conclude that $g_0(p) = p$, which cannot occur, since no element of $\Gamma$ has a fixed point.