Proposition 1

The mapping $p: E_{\mathscr{F}}\rightarrow$ B is etale. For every open set $U\in \mathscr{B}$ and every $s\in \mathscr{F}(U)$, the mapping $\sigma_{\mathscr{F}}(U, s)$ is therefore a continuous section of $p$ over U.

Let $\lambda = (U, s)$ and $\mu= (U', s')$ be elements of L. By definition of the relation $R_{\mathscr{F}}$, the set $A_{\lambda \mu}$ of points $x$ of $U\cap U'$ at which $\sigma_{\mathscr{F}}(U, s)$ and $\sigma_{\mathscr{F}}(U', s')$ coincide is the interior in B, of the set of $x\in U\cap U'$ such that $s(x) =s'(x)$. It follows that $A_{\mu\lambda}= A_{\lambda \mu}$. We then denote by $h_{\mu\lambda}: A_{\lambda \mu}\rightarrow A_{\mu\lambda}$ the mapping Id$_{A_{\lambda \mu}}$. The set $E_{\mathscr{F}}$ is obtained by gluing together the open sets U along the $A_{\lambda \mu}$ by means of the bijections $h_{\mu\lambda}$ (TG, I, p. 16). By prop. 9 of TG, I, p. 17, the mapping $\sigma_{\mathscr{F}}(U, s)$ induces a homeomorphism of U onto an open subset of $E_{\mathscr{F}}$. This proves that the mapping $p$ is etale (I, p. 33, prop 9).

For every open set $U\in \mathscr{B}$ and every $s\in \mathscr{F}(U)$, one has $\sigma_{\mathscr{F}}(U, s)(x) =$ $[U, s, x]$ for every $x\in U$. The second assertion therefore follows from the definition of $p$.