Lemma 4

The relation “$(g_1, u_1)$ is congruent to $(g_2, u_2)$” is an equivalence relation in Z, compatible with the operation of F.

This relation is reflexive, for one has $x_e=\varepsilon$. Let $(g_1, u_1)$ and $(g_2, u_2)$ be elements of Z such that $(g_1, u_1)$ is congruent to $(g_2, u_2)$. Let $s\in S$ be such that $g_2=g_1x_s$ and $s\cdot u_2=u_1$; since $x_{s^{-1}}=x^{-1}_s$, one has $g_1=g_2x_{s^{-1}}$ and $u_2=s^{-1}\cdot u_1$, hence $(g_2, u_2)$ is congruent to $(g_1, u_1)$; consequently, this relation is symmetric. Let us show finally that it is transitive. Let $(g_1, u_1)$, $(g_2, u_2)$ and $(g_3, u_3)$ be points of Z such that $(g_1, u_1)$ and $(g_2, u_2)$ are congruent, and also $(g_2, u_2)$ and $(g_3, u_3)$. Let $s$ and $t$ in S be such that $g_2=g_1x_s$ and $s\cdot u_2=u_1$ on the one hand, and $g_3=g_2x_t$ and $t\cdot u_3=u_2$ on the other hand. One has $u_1=s\cdot u_2=st\cdot u_3$, hence $u_1\in M\cap s\cdot M\cap st\cdot M$, which shows that $(s, t)$ belongs to T and that $st$ belongs to S. One then has $g_3=g_2x_t=g_1x_sx_t=g_1x_{st}$ and $u_1=st\cdot u_3$; consequently, $(g_1, u_1)$ and $(g_3, u_3)$ are congruent. Thus, the relation "to be congruent" is an equivalence relation in Z. It is compatible with the operation of F in Z.

Let Y be the quotient topological space of Z for the equivalence relation defined above. Let us denote by $\pi : Z\rightarrow Y$ the canonical mapping. Let us also denote by $q: Z\rightarrow X$ the mapping given by $(g, x)\mapsto \varphi (g)\cdot x$; it is surjective (lemma 3). By passing to the quotient, the mapping $q$ induces a continuous and surjective mapping $p: Y\rightarrow X$; by lemma 4, the operation of F in Z induces a continuous operation of F in Y such that $p(g\cdot y) =\varphi (g)\cdot p(y)$ for every $g\in F$ and every $y\in Y$. In particular, the group N = Ker($\varphi$ ) operates continuously on the X-space $(Y, p)$.