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.

So you're done showing openness if the image of a "sheet" under an étale map is open.

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.

  • $\begingroup$ @CyrylL. I reverted to my original thought which was correct in this respect. Didn't you have another question on the global section? It seems to be gone. $\endgroup$ – Henno Brandsma Aug 26 '18 at 8:50
  • $\begingroup$ @CyrylL. the both must go. And $\varphi$ is a map between the covering spaces, so that's consistent. $\endgroup$ – Henno Brandsma Aug 26 '18 at 9:46
  • $\begingroup$ sorry to bother you, but i really don't think your argument is true, may you spell out the details? $\endgroup$ – Cy L Shih Aug 27 '18 at 8:33

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