Why is the preimage of zero of a global section of a locally constant sheaf closed?

Question

I'm reading Faisceaux Algébriques Cohérents (Serre), no.36 lemma 2. The proof of that lemma states that the preimage of zero of a global section of a sheaf is closed when the sheaf is locally constant.

I know the preimage of zero is open (because it is a sheaf), but I can't understand why locally constant implies it closed.

Answer 1

This is an abstract of the above comments.

Serre uses the "espace étalé" version of sheaves, so let me recall the definition of constant and locally constant sheaves.

A sheaf $F$ on $X$ is said to be constant if there exist a discrete set $E$ such that $F$ is homeomorphic to $X\times E$ with the product topology, and such that the projection $\pi:F\rightarrow X$ corresponds to the first projection. In that case, there is a bijection between sections of $F$ defined on an open subset $U$ and continuous function $U\rightarrow E$.

The bijection is the following : given a section $s:U\rightarrow F$, compose with the homeomorphism $F\simeq X\times E$ and compose with the second projection. This gives a continuous map $f:U\rightarrow E$. Conversely, given a continuous map $f:U\rightarrow E$, takes its graph, that is, consider the map $U\rightarrow X\times E$ where the first coordinate is simply the inclusion, and the second coordinate is $f$. Finally, compose with the homeomorphism $X\times E\simeq F$ to get a section $U\rightarrow F$.

You should check that this correspondence is indeed a bijection.

A sheaf $F$ is said to be locally constant if there exists a open covering $X=\bigcup U_i$ such that the restriction $F_i$ of $F$ to $U_i$ is constant.

Now let us prove that if $s$ is a section of a locally constant sheaf of abelian groups, then the set where $s(x)=0$ is closed.

Let $x\in X$ be a point where $s(x)\neq 0$. There exist an open neighborhood $U$ of $x$ such that $F$ is constant on $U$ with value $E$. Hence the section corresponds to a continuous map $f:U\rightarrow E$. But then, we have $\{x\in U, s(x)\neq 0\}=f^{-1}(E\setminus\{0\})$ which is open. So the complementary of the zero set of $s$ is open.