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I'm interested in Proposition 3.5 in Milne's book "Etale Cohomology," which says that a presheaf on a noetherian site is a sheaf if it satisfies the sheaf axiom with respect to finite coverings. By the way, a site is noetherian if every covering has a finite subcovering (e.g. open sets in a noetherian space). Milne proves it by an elementwise calculation, but I'm wondering if there's an abstract nonsense-type argument which works for, say, sheaves with values in an arbitrary cocomplete category. I feel like there's some kind of cofinality argument lurking around which I can't quite formulate.

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First of all, you have to formulate the sheaf axiom correctly, i.e. with sieves. For ease of notation I will work with the standard site of a topological space – it will be clear how to do this for a general site.

Recall that a sieve (of open sets) is a collection $\mathfrak{U}$ of open sets with the following property: if $U \in \mathfrak{U}$ and $U' \subseteq U$, then $U' \in \mathfrak{U}$. A sieve $\mathfrak{U}$ covers $V$ if $V = \bigcup_{U \in \mathfrak{U}} U$. The sheaf condition on a presheaf $F$ can then be stated as follows:

  • If $\mathfrak{U}$ is a sieve that covers $V$, then the diagram $$F (V) \to \prod_{U \in \mathfrak{U}} F (U) \rightrightarrows \prod_{U \in \mathfrak{U}} \prod_{U' \subseteq U} F (U')$$ is an equaliser, where one of the arrows is defined by projection and the other by restriction.

It is straightforward to check that this condition is equivalent to the usual one. The next step is to prove the following:

  • If $F$ satisfies the sheaf condition for the sieve $\mathfrak{U}'$ and $\mathfrak{U}' \subseteq \mathfrak{U}$, then $F$ satisfies the sheaf condition for $\mathfrak{U}$.

Finally, observe that in a noetherian topological space, every covering sieve contains a finitely generated covering sieve. Thus, it suffices to check the sheaf condition on finitely generated covering sieves, or equivalently, finite covers.

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  • $\begingroup$ Actually, I don't see why every sieve is generated by finitely many elements. A generating set is a subcover, but not conversely, right? $\endgroup$ – Justin Campbell May 7 '14 at 2:23
  • $\begingroup$ A subcover is a generating set. $\endgroup$ – Zhen Lin May 7 '14 at 7:12
  • $\begingroup$ I must be using the wrong notion of generating set. I thought it meant that every member of the sieve is contained in some member of the generating set, but of course a subcover need not have this property. What do you mean by generating set and how does it help me prove the claim? $\endgroup$ – Justin Campbell May 7 '14 at 14:51
  • $\begingroup$ Oops – yes, you're right. The noetherian condition only says that every covering sieve contains a finitely generated covering sieve. $\endgroup$ – Zhen Lin May 7 '14 at 16:14
  • $\begingroup$ I see that you've reformulated the problem, but I'm still left with the exact same issue. In the last bullet point, you require that $\mathfrak{U}'$ and $\mathfrak{U}$ cover the same set, correct? The question is, how do I use this to prove that claim without doing some calculation with sections? $\endgroup$ – Justin Campbell May 7 '14 at 17:18

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