# Why is the etale map an open map?

In Proposition 5.59 (page 276) of his book An Introduction to Homological Algebra, Rotman states that an etale map is always an open map on sheaf space. (5.59iii)

Proposition 5.59 Let $\mathcal{S} = (E, p, X)$ and $\mathcal{S}' = (E', p', X')$ be etale-sheaves over a topological space $X$.

(iii) Every etale-map $\varphi \colon \mathcal{S} \to \mathcal{S}'$ is an open map $E \to E'$.

Proof. The sheets form a base of open sets for $E$.

Why does this follow?

Fact: If $f: X \to Y$ is a function between spaces and $\mathcal{B}$ is a base for the topology of $X$, then $f$ is an open map iff $$\forall B \in \mathcal{B}: f[B] \text{ is open in } Y$$
This is proved simply by observing that for families $A_i \subseteq X, i \in I$ we have that $f[\bigcup_i A_i] = \bigcup_i f[A_i]$ and knowing that open sets are unions of basic open sets and open sets are closed under all unions.
This is clear, as the definition of étale-map implies that for a sheet $S$ of $p$, we have that $\varphi[S] = (p')^{-1}[p[S]]$, where the left to right inclusion follows from $p'\varphi=p$ and the right to left one uses the local injectivity of $p$. So $\varphi[S]$ is open as $p[S]$ is open by the local homeomorphism part of the protosheaf definition and $p'$ is continuous.
• @CyrylL. the both must go. And $\varphi$ is a map between the covering spaces, so that's consistent. – Henno Brandsma Aug 26 '18 at 9:46