22
$\begingroup$

Given a set $S$, let $Top(S)$ denote the partially ordered set (poset) of topologies on $S$, ordered by fineness, so the discrete topology, $Disc(S)$, is maximal.

Suppose that $Q$ is a presheaf on $Disc(S)$, that $\tau$ is a topology on $S$, and that $\epsilon\colon Disc(S)\to (S,\tau)$ is the canonical map. I'll say that $Q$ is a sheaf with respect to $\tau$ if the direct image presheaf $\epsilon_*(Q)$ is a sheaf.

For a given set $S$ and presheaf $Q$, let $$Top_{Q}(S)\subseteq Top(S)$$ denote the subposet of topologies on $S$ for which $Q$ is a sheaf.

For example, if $1$ is the terminal presheaf then $Top_{1}(S)=Top(S)$, i.e., every topology makes $1$ a sheaf. In general $Top_{Q}(S)$ may be empty, e.g., if $Q$ assignes a non-terminal set $Q(\emptyset)\not\cong\{*\}$ to the emptyset $\emptyset\subseteq S$.

Question 1: In general, what can one say about the poset $Top_Q(S)$? For example, is it closed under binary meets or joins in $Top(S)$?

Let $Top^{sep}_Q(S)$ denote the poset of $S$-topologies on which $Q$ is a separated presheaf. Recall that a presheaf on a space $X$ is separated if every matching family of sections on a cover extends to at most one section on their union. This condition is less stringent than the sheaf condition, which replaces at most one with exactly one. In general, we have $$Top_Q(S)\subseteq Top^{sep}_Q(S).$$ For example the initial presheaf $0$ on $Disc(S)$ is a separated presheaf but not a sheaf, so $Top_0(S)\subsetneq Top^{sep}_0(S)$.

Question 2: What can one say about the poset $Top^{sep}_Q(S)$?


Edit provenance: I added the second paragraph to address a question of Zhen Lin, which pointed out an ambiguity in the phrase "topologies on $S$ for which $Q$ is a sheaf." Later I added even more detail about that notion, to coincide with clarifications made in the comments. Even later, I realized that the notion I had originally chosen (that $Q$ was the inverse image presheaf $\epsilon^{-1}(Q')$ of some sheaf $Q'$ on $(S,\tau)$) was a bit pathological, so I weakened the condition to say that $\epsilon_*Q$ is a sheaf.

$\endgroup$
  • $\begingroup$ There seems to be a subtlety in the predicate "$Q$ is a sheaf". For instance, if $Q (\emptyset) = 1$, then is $Q$ automatically a sheaf for the indiscrete topology? Or do we also require that $Q (V) \to Q (U)$ be a bijection for all $\emptyset \subsetneq U \subseteq V \subseteq S$? $\endgroup$ – Zhen Lin Apr 14 '14 at 10:19
  • $\begingroup$ @ZhenLin, I don't understand. Are you asking whether $\emptyset\to X$ covers $X$ in the indiscrete topology on $X\neq \emptyset$? I think this definition en.wikipedia.org/wiki/… would say "no". $\endgroup$ – David Spivak Apr 14 '14 at 21:27
  • $\begingroup$ That's not what I'm asking. Rather, there are two reasonable interpretations of your condition: one only considers $Q (V)$ for open $V \subseteq S$, and another considers all $V \subseteq S$. $\endgroup$ – Zhen Lin Apr 14 '14 at 22:57
  • 1
    $\begingroup$ I think Zhen's point is that for two different topologies on the same set, the notions of "presheaf" are different since they are functors on a different poset of opens. So when you talk about topologies S "for which Q is a sheaf", you should really say topologies S "for which the restriction of Q to S is a sheaf" (assuming that that's what you mean). $\endgroup$ – Mike Shulman Apr 16 '14 at 4:37
  • 3
    $\begingroup$ Have you worked this out in the case when $S$ has two elements? $\endgroup$ – Tom Goodwillie Apr 28 '14 at 2:53

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.