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

Let $T_1$ and $T_2$ be two Grothendieck topologies on the same small category $C$, and let $T_3 = T_1 \cup T_2$ (by which I mean the smallest Grothendieck topology on $C$ containing $T_1$ and $T_2$). If $X$ is a $T_1$-sheaf, is its $T_2$-sheafification still a $T_1$-sheaf (and therefore a $T_3$-sheaf)?

If the answer is "not always," then are there conditions one can impose on $T_1$ and $T_2$ to make it true? Does it matter if $X$ is already $T_2$-separated?

(I'm really interested in the analogous question for stacks, but I'm guessing the answers will be pretty much the same.)

share|improve this question
    
Just to make things more concrete, have you tried taking T_1 = the canonical topology and T_2 some topology where Hom(-,u) is not a sheaf and seeing if this gives you a counterexample? I am still trying to develop a nice collection of examples for this stuff, so I can't pick out anything specific. I am thinking about it. –  Steven Gubkin Dec 7 '09 at 18:29
    
mathoverflow.net/questions/80074/… –  name Nov 6 '11 at 19:55
add comment

2 Answers

up vote 9 down vote accepted

I think my answer to this question provides a counterexample: Let C be the category a → b, and consider the topologies T1 generated by the single covering family {a → b} and T2 generated by declaring the empty family to be a covering of a. (Caution: In my other answer I used covariant functors for some reason, so I hope I didn't err in translating the example.)

Note that if I switch T1 and T2, though, the condition you ask for is satisfied, even though neither class of sheaves contains the other. So there may be some interesting conditions under which it is true.

share|improve this answer
    
If I have parsed correctly what you say, a T_1-presheaf F is a sheaf if and only if the map F(b) -> F(a) induced by the only nonidentity covering a->b is a bijection, and a T_2-presheaf is always a sheaf, so that T_2-sheafification is just the identity functor. I feel uneasy about the empty family being a covering: it seems to me that the sheaf condition is trivial for such a thing... –  Alberto García-Raboso Dec 7 '09 at 19:17
    
That's what I intended for T_1. For T_2, the sheaf condition is supposed to say that F(a) is a point. –  Reid Barton Dec 7 '09 at 19:23
2  
If you prefer, we could take C = z -> a -> b, T_1 generated by the covering a -> b, T_2 generated by the covering z -> a. –  Reid Barton Dec 7 '09 at 19:24
    
Thanks! Maybe we need some kind of "commutativity" condition for the two topologies. –  Mike Shulman Dec 7 '09 at 19:51
1  
At ncatlab.org/nlab/show/superextensive+site I attempted to write out a proof that this does hold in one particular case (which happens to be the case I care about). –  Mike Shulman Dec 7 '09 at 23:26
show 1 more comment

I am not sure whether I will answer your question properly. I just tell some facts which might be useful

I know there are following categories equivalence Qcoh(X),Qcoh(X,t),Qcoh(aX,t)

X is a presheaf Qcoh(X) is quasi coherent modules on X,t is a grothendieck topology. aX is the sheafification according to topology t of X.

If t is coarser than topology of effective descent, then those three categories are equivalent. Which means Qcoh(X)=Qcoh(X,t1)=Qcoh(X,t2)=Qcoh(aX,t1)=Qcoh(aX,t2) are the same if t1 t2 are coarser than effective descent topology. This fact shows that it is not necessary to consider sheaf but presheaf. Their descent theory can be described by category of quasi coherent sheaves.

Similar results applied to stack. These facts are indicated in Giraud's book,but not wrote it out. They were proved by Orlov in his paper quasi coherent sheaves in commutative and noncommutative geometry and Kontsevich-Rosenberg preprint in MPIM "noncommutative stack"

share|improve this answer
1  
This doesn't really look relevant to me. –  Mike Shulman Dec 20 '09 at 6:57
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.