Lemma 5

If M is connected, the space Y is connected.

Let $g\in F$ and $s\in S$ be given. Let $u$ and $v$ be elements of M such that $v=s\cdot u$; we have therefore $\pi (g, v) =\pi (gx_s, u)$, and the sets $\pi (\{g\} \times M)$ and $\pi (\{gx_s\} \times M)$ of Y have a common point. Since they are connected, they are contained in the same connected component of Y. Since S is a symmetric subset of the group G, and since $x_{s^{-1}}=x^{-1}_s$ for every $s\in S$, every element $g$ of F is of the form $x_{s_1}. . . x_{s_n}$, where $n\in \mathbf{N}$ and $s_1, . . . , s_n$ are elements of S. By induction on $n$, the sets $\pi (\{e\} \times M)$ and $\pi (\{g\} \times M)$ are contained in the same connected component of Y, for every $g\in F$. It follows that Y is connected.