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

While reading Brylinski I am trying to understand the descent of morphisms of sheaves.

In trying to form a new definition of a presheaf $A$ over a space $X$, we associate to each surjective local homeomorphism $f:Y \to X$ a set, denoted $A(Y\xrightarrow{f}X)$. The "restriction" condition of a presheaf amounts to: given a surjective local homeomorphism $g:Z \to Y$ we have a pullback map $g^{-1}:A(Y\xrightarrow{f}X) \to A(Z \xrightarrow{fg}X)$. The transitivity property for these "restriction" (pullback) maps is that given any diagram $$W \xrightarrow{h} Z \xrightarrow{g} Y \xrightarrow{f} X$$ having $(gh)^{-1} = h^{-1} \circ g^{-1}$ as pullbacks $A(Y\xrightarrow{f}X) \to A(W \xrightarrow{fgh} Z)$. $\\ \\$

If $A$ is already a presheaf, in the good 'ol fashioned sense, then we can define our assignment $A(Y\xrightarrow{f}X)$ to be the global sections of $Y$ given by the inverse image of $A$ on $X$, i.e. $\Gamma(Y, f^{-1}A)$

I have 2 questions:

  1. Is it true that if $A$ is already a sheaf in the good 'ol fashioned sense, then the above property (transitivity of the "restriction") is satisfied? My proof feels trivial, hence my worry. Also, I am uneasy since Brylisnki doesn't state this fact but instead says it "should" be true.

  2. He later comments that as functors from the category of sheaves on $Y$ to the category of sheaves on $W$ , $h^{-1}\circ g^{-1}$ and $(gh)^{-1}$ are NOT equal; but there is a natural transformation. Why are these two functors not equal? It seems like they send the same sheaves to the same places, unless of course I am making identifications of categories that I don't realize?

share|improve this question
    
First; are you aware of the notions of pseudo-functors and stacks? –  Martin Brandenburg May 9 '12 at 21:21
    
No, to be honest, I am really trying to do this all without talking about stacks (or schemes, etc). The goal is to define these descent properties in a "hands-on" way so that I can understand the definition of a "gerbe" in a "hands-on" way. –  cheyne May 9 '12 at 22:38
2  
If you want anyone to attempt to answer this question, I suggest you say what this "new" definition of a sheaf is. –  David Carchedi May 9 '12 at 23:42
    
@David Above is the "new" definition for a presheaf, which is all I am concerned about at the moment. I will be more clear now. @Martin: Turns out the book is using all of this exposition to define a stack in this context, hence why I am unfamiliar with it up to this point. –  cheyne May 10 '12 at 12:53
    
@Cheyne: Your notation is bad. The functor from $\textbf{Sh}(W)$ to $\textbf{Sh}(Z)$ should be denoted $h_*$, etc. The notation $h^{-1}$ (or $h^*$) is reserved for the one going in the opposite direction. And it is true that this "inverse image" functor does not compose strictly: $h^{-1} g^{-1} \ne (g \circ h)^{-1}$. There is, however, a natural isomorphism. –  Zhen Lin May 10 '12 at 20:00
show 1 more comment

1 Answer

up vote 3 down vote accepted

Yes, the property is satisfied (note that $\Gamma(Y,f^{-1}A)=A(f(Y))$, since $f$ is open). I didn't read the book, but I imagine that the point is not to give a strange definition of a sheaf on a topological space, but rather to motivate the generalisation to situations where you know what maps you wish to consider to be local homeomorphisms, but they are not actually local homeomorphisms for any (classical) topology. The classical example is the etale (Grothendieck) topology on schemes. The book "Sheaves in geometry and logic" has a good exposition of these ideas.

share|improve this answer
    
OK thanks. Now, I am thinking about my second question. Is it a bad idea to call yours an "answer" before the second part is answered? Then people won't offer answers to the second question? I'm new to this forum. And agree with you that the idea is to motivate more general concepts. –  cheyne May 10 '12 at 16:28
    
I'm not sure about the etiquette, I'm also not a very active participant... As for the second part, this is the same as saying that $(A\times B)\times C$ is not equal to $A\times(B\times C)$. This is formally true, but mostly irrelevant, since there is a canonical isomorphism. The same happens with pullbacks of sheaves, but in the abstract framework you have to be given this isomorphism, and I guess the book is trying to motivate this. In my personal opinion, this point is often stressed way beyond its importance... –  Moshe May 10 '12 at 16:41
    
thank you for confirming all of my beliefs/suspicions!! –  cheyne May 10 '12 at 20:07
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.