Lemma 7

Endowed with the action of N, the X-space $(Y, p)$ is a principal left covering. It is trivializable over M.

Let $x\in X$; fix an element $g\in E_x$ and an element $\overline{g}\in F$ such that $\varphi (\overline{g}) =g$. We denote by $\mu_x: E_x\times M\rightarrow X$ the mapping given by $(h, u)\mapsto h\cdot u$.

Let $n\in N$, let $h, k\in E_x$ and let $u, v\in M$ be such that $h\cdot u=k\cdot v$; put $s=g^{-1}h,t=h^{-1}k$ and $r=st=g^{-1}k$. Since $x$ belongs to $g\cdot M\cap h\cdot M\cap k\cdot M$, the pair $(s, t)$ belongs to T, whence $x_sx_t=x_r$. Hence

There thus exists a unique mapping $\theta : N\times (E_x\cdot M)\rightarrow Y$ such that $\theta (n, h\cdot u) =\pi (ngx_{g^{-1}h}, u)$ for every $n\in N$, every $h\in E_x$ and every $u\in M$. We have

Moreover, for $n, n'\in N$ and $y\in E_x\cdot M$, one has $\theta (n'n, y) =n'\cdot \theta (n, y)$. The mapping $\theta ': N\times (E_x\times M)\rightarrow Y$ defined by $\theta '(n,(h, u)) =\theta (n, \mu_x(h, u))$ is continuous; since the mapping $\mu_x$ is universally strict (I, p. 133, Lemma 2), the mapping $\theta$ is continuous. It is a bijection of $N\times (E_x\cdot M)$ onto the subspace $Y\times_X(E_x\cdot M)$ of Y (Lemma 6).

Let $z= (k, v)\in Z$ and $(h, u)\in E_x\times M$ be such that $q(k, v) =\mu_x(h, u)$. Put $s$ = $h^{-1}\varphi (\overline{k})$; since $\varphi (\overline{k})\cdot v$ = $h\cdot u$, one has $s\in$ S. Then put $\lambda '(z,(h, u)) =kx^{-1}_sx_{h^{-1}g}\overline{g}^{-1}$; one has $\varphi (\lambda '(z,(h, u))) =$ $\varphi (\overline{k})s^{-1}h^{-1}gg^{-1}$ = $e$, hence $\lambda '(z,(h, u))\in$ N. Thus a continuous mapping $\lambda ': Z\times_X(E_x\times M)\rightarrow N$ is defined. Moreover, for every $z$ and $(h, u)$ as above, one has

and $\lambda '(z,(h, u))$ is the unique element $n$ of N such that $\theta (n, h\cdot u) =\pi (z)$. In particular there exists a unique mapping

such that $\lambda (\pi (z), h\cdot u) =\lambda '(z,(h, u))$ for every $z\in Z$ and every $(h, u)\in$ $E_x\times M$ such that $q(z) =h\cdot u$. It is continuous, since the mapping $\mu_x$ is universally strict. It follows that the $E_x\cdot M$-space $Y_{E_x\cdot M}$ deduced from Y by base change, endowed with the operation of N, is a principal fibre space with group N, trivializable.

The lemma is thus proved.