Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I need example that associated sheaf functor doesn't preserve arbitrary products. I think that one can provide an example for sheaves over topological space. Thanks for your help.

share|improve this question
    
Homework? And what associated sheaf functor? Do you mean the sheafification functor from presheaves to sheaves? Or something else? Are you allowing sheaves over sites (it seems you must be because you talk about providing an example of sheaves over a topological space)? Finally, if you think you have an example - actually give your example! –  Anon Nov 1 '11 at 14:16
1  
Sorry for my bad english, which leads to some misunderstanding. It's not a homework. Yes I meant sheafification functor. I strongly believe there is example over site of open subsets of a topological space. –  Anonymous Nov 1 '11 at 15:35
1  
Why do you need this example? –  S. Carnahan Nov 1 '11 at 18:59
5  
I don't know why Anon would think this is homework, or why someone is voting to close this. It is a perfectly reasonable question for someone who is not expert in topos theory. –  Todd Trimble Nov 2 '11 at 15:40
add comment

2 Answers 2

An example where sheafification does not preserve arbitrary products is where we take sheaves over a (sober) space $X$ that is not locally connected, for example the space of irrationals or Cantor space.

Recall that a Grothendieck topos $E$ is locally connected if the (essentially unique) geometric morphism $\Gamma = f_\ast: E \to Set$ has a left adjoint $f^\ast$ that in turn has a left adjoint. More generally, a geometric morphism $f_\ast: E \to F$ between toposes is an essential geometric morphism if its left adjoint $f^\ast$ has a left adjoint.

We have the following facts:

  • A presheaf topos $Set^{C^{op}}$ is locally connected. Here the left adjoint to the global sections functor $\Gamma: Set^{C^{op}} \to Set$ is the diagonal functor $\Delta: Set \to Set^{C^{op}}$, which of course has a left adjoint.

  • A geometric morphism $f_\ast: E \to F$ is essential if and only if $f^\ast$ preserves arbitrary products. (Of course $f^\ast$ is already left exact and so preserves equalizers, if it preserves small products, then it preserves small limits. Using the fact that Grothendieck toposes are cototal, this is enough to ensure that $f^\ast$ has a left adjoint.)

  • Then in particular for a small site $(C, J)$, the inclusion functor $i: Sh(C, J) \to Set^{C^{op}}$ is an essential geometric morphism if and only if sheafification $a: Set^{C^{op}} \to Sh(C, J)$ preserves small products. Thus, putting the last two facts together, the composite geometric morphism $$Sh(C, J) \stackrel{i}{\to} Set^{C^{op}} \stackrel{\Gamma}{\to} Set$$ is essential, i.e., $Sh(C, J)$ is locally connected, if $a$ preserves products.

  • In the case where the site is $(\text{Open}(X), J)$ where $J$ is the canonical Grothendieck topology given by covering families, $Sh(X) = Sh(C, J)$ is locally connected if and only if $X$ is a locally connected space. Thus $a: Set^{\text{Open}(X)^{op}} \to Sh(X)$ preserves small products only if $X$ is locally connected.

share|improve this answer
add comment

Let $X$ be a space. An abelian group (or a set, if you prefer) $G$ determines a constant presheaf, call it $C_G$, and the associated sheaf is the sheaf of locally constant maps into $G$. Given a family $G_i$ with product $G$, the product of the presheaves $C_{G_i}$ is the presheaf $C_G$, but the product of the associated sheaves is not in general the sheaf of locally constant maps to $G$. (If $X$ is not locally connected then one can easily have a map to $G$ which projects to a locally constant map to $G_i$ for each $i$ but which is not locally constant itself. )

This is presumably a down-to-earth special case of what Todd Trimble is saying.

share|improve this answer
    
Not only down-to-earth, but also easy to understand: 1+ –  Martin Brandenburg Nov 2 '11 at 19:18
    
It could be more down-to-earther if $G$, $X$ and $G_i$'s were specified... –  Andrej Bauer Oct 8 '12 at 15:46
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.