I was reading Serre´s FAC and I didn´t understand why the restrictions to a point are continuous. Let $F$ be a presheaf and $p: \coprod F_x \longrightarrow X $ it´s étale space with topology generated by $[t, U] = \{ \varphi_{x}^{U}(t) ; x \in U \}$. Now, let $i : F_U \longrightarrow \Gamma (U, F)$ such that $i(t)(x) = (\varphi_{x}^{U}(t)) $, then why $i(t)$ is continuous?
Thanks in advance.
Let $x\in U$ and $[s,V]$ be a basis neighbourhood of $i(t)(x)$. $i(t)(x) \in [s,V]$ means the two germs conicide, $\varphi_x^U(t) = \varphi_x^V(s)$. That means there is an open $x \in W \subset U\cap V$ with $\rho_W^U(t) = \rho_W^V(s)$ (where $\rho_B^A \colon F(A)\to F(B)$ is the restriction map). But then $[t,W] = [s,W]$, and
$$i(t)^{-1}([s,V]) \supset i(t)^{-1}([s,W]) = i(t)^{-1}([t,W]) = W$$
is a neighbourhood of $x$, so $i(t)$ is continuous in $x$. $x\in U$ was arbitrary, so $i(t)$ is continuous on $U$.