Heuristic explanation of why we lose projectives in sheaves.

Question

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!

Answer 1

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!

Answer 2

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.

Answer 3

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.

Answer 4

We can turn the question around to ask: why do we have projectives in module categories? One answer is that we know we have a plentiful supply of projectives because free modules are projective. Abstractly, we have a forgetful functor $U : \text{Mod}(R) \to \text{Set}$ with left adjoint $F : \text{Set} \to \text{Mod}(R)$. Then we have the following two results:

1. By the axiom of choice, every set is projective (in the sense that homs out of it preserve epimorphisms).
2. If a functor $U$ with a left adjoint preserves epimorphisms, then its left adjoint $F$ preserves projectives.

Finally, it is straightforward to verify that $U$ in fact preserves epimorphisms (that is, epimorphisms of $R$-modules are surjective on underlying sets).

Now what is the analogous situation for sheaves? We still have a forgetful functor $U : \text{Sh}(X) \to \text{Psh}(X)$, and it still has a left adjoint, namely sheafification. However, $U$ no longer preserves epimorphisms (this is exactly Dinakar's observation that an epimorphism of sheaves need not be an epimorphism of presheaves), so the above argument doesn't go through.

For sheaves what we can instead do is the following. There is a different forgetful functor sending a sheaf on $X$ to its stalks; it can be thought of as pullback $p^{\ast}$ along the map $p : X_d \to X$ where $X_d$ denotes $X$ with the discrete topology. As a pullback, this functor has a right adjoint, namely pushforward $p_{\ast}$. The composite $p_{\ast} p^{\ast}$ is the Godement construction. In any case, a dual argument to the above shows that because pullback $p^{\ast}$ preserves monomorphisms, its right adjoint $p_{\ast}$ preserves injectives. So now instead of a plentiful supply of projectives we have a plentiful supply of injectives.

Answer 5

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.