3

Let $F$ be a presheaf (of sets) on some topological space $X$, and define the etale space of $F$ as $\widehat{F} = \bigsqcup_{x\in X} F_x$, where $F_x$ is the stalk over $x \in X$. There is a canonical projection $\pi\colon \widehat{F} \to X$, $\pi(F_x) = x$, and the topology on $\widehat{F}$ is supposed to be the coarsest one such that $\pi$ is continuous.

I have some notes that say that then the subsets of $\widehat{F}$ of the form \begin{align*} T(s, U) = \{(x, [s]_x) \mid x \in U\} \end{align*} form a subbase, for $U \subset X$ open, $s \in F(U)$ a section, and $[s]_x$ the germ of $s$ at $x$.

Can that be true, though? If the topology on $\widehat{F}$ is the coarsest one such that $\pi$ is continuous, then a subbase is given by sets of the form \begin{align*} T'(U) = \pi^{-1}(U) = \bigsqcup_{x \in U} F_x \end{align*} for $U$ open.

But if define the sets $T(s, U)$ to be open, then they generate a finer topology than the sets $T'(U)$, since $T(s,U)$ is a (proper) subset of $T'(U)$.

In the answer to this question, the topology is immediately described as the one generated by $T(s,U)$, without any mention of its coarseness.

To be more clear: my question is whether the notes are correct that this is the coarsest topology making $\pi$ continuous and if I am overlooking something?

Jo Mo
  • 2,075
  • 2
    It seems like the topology generated by $T(s,U)$ is the correct one. If the topology on etale space is the coarsest, then any section $\tilde s:U\to \bigsqcup_{x\in X} F_x$ is continuous, but this wouldn't be true for general sheaves. – lEm Oct 19 '21 at 10:45
  • @lEm: I don't quite understand that, since I'm new to sheaves, but it would be true if you define the topology in this way haha. But, ok, you're confirming here that the topology generated by $T(s,U)$ is indeed not the coarsest one? – Jo Mo Oct 19 '21 at 13:13
  • 1
    It is not the coarsest. For example, you can look at vector bundle over a manifold $\pi:E\to M$. The sheaf associated to this is the sheaf of continuous sections $F(U)={s:U\to E|,\pi\circ s=\mathrm{id}_U}$. The etale space is the same as $E$ set theoretically. In this case, the coarsest topology would not recover you the vector bundle's topology, but the topology generated by $T(s,U)$ would. In general, you should be able to recover your original sheaf by taking sections of the etale space. As we saw, this isn't true if we pick the wrong topology. – lEm Oct 19 '21 at 13:29

1 Answers1

1

As per the comments, the usual topolgy is the one generated by the $T(s, U)$.

Jo Mo
  • 2,075