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We know that presheaves of any category have enough projectives and that sheaves do not, why is this, and how does it effect our thinking?

This question was asked(and I found it very helpful) but I was hoping to get a better understanding of why.

I was thinking about the following construction(given during a course);

given an affine cover, we normally study the quasi-coherent sheaves, but in fact we could study the presheaves in the following sense:

Given an affine cover of X,

$Ker_2\left(\pi\right)\rightrightarrows^{p_1}_{p_2} U\rightarrow X$

then we can define $X_1:=Cok\left(p_1,p_2\right)$, a presheaf, to obtain refinements in presheaves where we have enough projectives and the quasi-coherent sheaves coincide. Specifically, if $X_1\xrightarrow{\varphi}X$ for a scheme $X$, s.t. $\mathcal{S}\left(\varphi\right)\in Isom$ for $\mathcal{S}(-)$ is the sheaffication functor, then for all affine covers $U_i\xrightarrow{u_i}X$ there exists a refinement $V_{ij}\xrightarrow{u_{ij}}U_i$ which factors through $\varphi$.

This hinges on the fact that $V_{ij}$ is representable and thus projective, a result of the fact that we are working with presheaves. In sheaves, we would lose these refinements. Additionally, these presheaves do not depend on the specific topology(at the cost of gluing).

In this setting, we lose projectives because we are applying the localization functor which is not exact(only right exact). However, I don't really understand this reason, and would like a more general answer.

A related appearance of this loss is in homological algebra. Sheaves do not have enough projectives, so we cannot always get projective resolutions. They do have injective resolutions, and this is related to the use of cohomology of sheaves rather than homology of sheaves. In paticular, in Rotman's Homological Algebra pg 314, he gives a footnote;

In The Theory of Sheaves, Swan writes "...if the base space X is not discrete, I know of no examples of projective sheaves except the zero sheaf." In Bredon, Sheaf Theory: on locally connected Hausdorff spaces without isolated points, the only projective sheaf is 0

addressing this situation.

In essence, my question is for a heuristic or geometric explanation of why we lose projectives when we pass from presheaves to sheaves.

Thanks in advance!

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You made two mistakes in your statements. In fact, in category of sheaves, we still have this refinement. Because sheafification functor as localization will not kill the morphism from V[ij]-->Ui. This is the reason that we have this refinement in category of schemes. The correct statement is that we might not prove this statement in category sheaves since lost of projectives but can do it within presheaves –  Shizhuo Zhang Jan 2 '10 at 5:10
    
You said sheafification as localization functor is not exact. This is wrong. Sheafification functor is exact functor because the localization is at Serre subcategory of presheaves category –  Shizhuo Zhang Jan 2 '10 at 6:11
    
In fact, there are some other wrong words in your statement such as "because of lost of projective resolution, we use cohomology not homology...." this is not the reason. –  Shizhuo Zhang Jan 2 '10 at 6:14
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4 Answers 4

up vote 17 down vote accepted

This is pretty much Dinakar's answer from a different view point: He says that it is too easy for a sheaf morphism to be an epi, so, since there are so many epis, it is now a stronger requirement that for every epi we find a lift - so strong that is not satisfied most of the times. I just want to call attention to the fact that this problem has nothing to do with module sheaves but is about sheaves of sets - and as such has the following nice interpretation:

The condition of being a projective module sheaf can be split in two conditions: That of existence of the lifting map as a morphism of sheaves of sets and that of it being a morphism of module sheaves.

In the category of sets the first condition is always satisfied; we have the axiom of choice which says that every epimorphism has a section and composing the morphism from our would-be projective with this section produces a lift - set-theoretically. Then one has to establish that one such lift is a module homomorphism.

But in a sheaf category step one can fail. Sheaves (of sets) are objects in the category of sheaves. This category is a topos and can be seen as an intuitionistic set-theoretic universe (in a precise sense: there is a sound and complete topos semantics for intuitionistic logic, see e.g. this book). Now in an intuitionistic universe of sets, the axiom of choice is not valid in general; there might not be a "set-theoretic" section of the epimorphism!

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Thank you for your answer. I REALLY liked your added information. I am surprised that the issue arises from the existence of the morphism rather than the satisfaction of the module condition. I have two more questions(if you feel your answers need another question rather than just comments, let me know). First, we have two places that in sheaf categories this existence of the lift can fail, as you said steps one and two. How often is it the first or second resp? Second, I am a novice in topos theory, could you explain more about the failure of the axiom of choice in these situations? Thanks! –  B. Bischof Dec 3 '09 at 20:47
    
I have no answer to "how often" one or two fail. A module sheaf can fail to be projective because there is a map into it which is stalkwise surjective, but not open-set-wise, as Dinakar said. I don't have a criterion for this but the feeling that this is VERY common. The failure of step two you can recognise as in sets e.g. because there is torsion in the module etc. –  Peter Arndt Dec 4 '09 at 1:38
    
The axiom of choice (AC) fails for sheaves over every non-discrete T_1-space: AC implies Booleanness (Moerdijk/MacLane, chap.VI, exercise 16), i.e. that the lattice of open sets is a Boolean algebra; for a T_1-space this means that it is discrete (MM,ch.VI ex.3) –  Peter Arndt Dec 4 '09 at 1:41
    
An example for Dinakar's situation: Let V be a fixed open set of some space X and consider sheaves on X. Let F,G be the sheaves with F(U):={continuous real-valued functions on U} and G(U):={cont. real-valued functions on (U intersected with V)}. The morphism F->G given by restricting functions to V is not surjective on every open set U, since there could exist a function on (U \cap V) which tends to infinity towards the border and thus is not extendable to U, i.e. is not a restriction of something defined on all of U. But surjectivity at stalks is clear by looking at small enough open sets... –  Peter Arndt Dec 4 '09 at 2:11
    
I don't think the map is surjective on stalks on the boundary of V. Do you have another example in mind? –  Steven Gubkin Jan 8 '10 at 13:15
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One reason is that surjectivity of a map of sheaves is a weaker condition than surjectivity of a map of presheaves. For a map of sheaves to be surjective, it need only be surjective on stalks.

Recall the definition of a projective sheaf $\mathcal{P}$: Suppose $\mathcal{N} \rightarrow \mathcal{M}$ is a surjective map of sheaves and $\mathcal{P} \rightarrow \mathcal{M}$ is a sheaf map. Then we require that there exists a lifting $\mathcal{P} \rightarrow \mathcal{N}$ making the obvious diagram commute. Because of the definition of surjectivity for sheaves, there's probably an open set $U$ for which the map $\mathcal{N}(U) \mapsto \mathcal{M}(U)$ isn't surjective. So if $\mathcal{P}(U)$ doesn't map into the image, then there is no hope for a lifting. In all but the trivial cases (like discrete spaces), it will be easy to cook up a map $\mathcal{N} \rightarrow \mathcal{M}$ to do this.

For presheaves, surjectivity means surjectivity on each open set, so this problem doesn't happen. But presheaves as an abelian category aren't very interesting. For example, the strictness of surjectivity means there is no cohomology.

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This was exactly what I was looking for, Thank you! –  B. Bischof Dec 1 '09 at 19:06
    
Your answer was excellent, but I have switched the accepted answer to Peter's due to it being a bit more specific and providing a bigger and more complete picture. Thank you very much for this explanation as it has helped very much. –  B. Bischof Dec 3 '09 at 20:49
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Here is an answer to a slightly different question, namely, what can one do once it is established that projective sheaves very often do not exist. For a nonaffine scheme, there is no known analogue of projective quasi-coherent sheaves on an affine scheme, but there is an analogue of the unbounded homotopy category of complexes of projectives.

Namely, Amnon Neeman has proven that the homotopy category of complexes of projective modules over an (arbitrary, noncommutative) ring is equivalent to the quotient category of the homotopy category of complexes of flat modules by the triangulated subcategory of acyclic complexes of flat modules with flat modules of cocycles. Building upon this result, Daniel Murfet in his Ph.D. thesis studies the mock homotopy category of projectives on a separated Noetherian scheme, defined as the quotient category of the homotopy category of unbounded complexes of flat quasi-coherent sheaves by the triangulated subcategory of pure acyclic complexes.

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Thank you for this, as you said, not what I was looking for, but this helped in another way. –  B. Bischof Dec 1 '09 at 19:07
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Sorry if this is silly; but might it have something to do with needing in the sheaf category to consider the sheafified presheaf cokernel in order to talk about projections? that is, I (think I) can imagine a nontrivial preasheaf with only trivial stalks, so that its sheafification is trivial; on the other hand, the sheaf morphisms are just the same as presheaf morphisms between sheaves. Hence there are probably too many sheaf morphisms with trivial cokernel.

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Consider the sheaves $C^1:U\mapsto\mathcal{F}(U)$ and $C^2:U\mapsto\mathcal{F}(U^2)$, the free abelian groups on the points and pairs of points in $U$ respectively. There is a sheaf morphism $C^1\rightarrow C^2$ induced by the diagonal. The presheaf cokernel at $U$ is isomorphic to the free abelian group generated by pairs $(u,v)$ with $u\neq v$; this clearly has trivial stalks, so the sheaf cokernel is trivial. But I'm afraid I don't know how this picture generalizes when points aren't the right thing to talk about. –  some guy on the street Dec 1 '09 at 17:29
    
Wait: is $C^2$ a sheaf? dear me... I'm afraid maybe not. In fact the stalks of $C^2$ look isomorphic to those of $C^1$ ... –  some guy on the street Dec 2 '09 at 0:00
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