2024-08-27

Sheaves

Let be a presheaf. We say that is a sheaf if for every open cover where open, the sequence

F(U)YiF(Ui)Yi;jF(Ui\Uj)ij

is an equalising sequence.

Idea: Whenever two sections agree on some overlap, they “glue”.

Examples

  1. Let be a set. Then
    is the constant presheaf. It is not usually a sheaf. This is because the sheaf condition requires the following to be equal:
    Instead, we define the constant sheaf via
    where denotes the set of continuous functions and is given the discrete topology. This is the sheafification of the constant presheaf.
  2. Let . Define the sections of

We define the presheaf

This is a sheaf.
3. It turns out that if is any sheaf on , then there exists which is a local homeomorphism (for every point in , there is an open neighbourhood homeomorphic to an open neighbourhood of ) such that (c.f. espace étalé).

Notation

We will denote:

  • ; and similarly
  • .
  • We may also write or where is or etc.

Remark

We write for a presheaf when we want to vary . We also write

when we want to let vary as well, e.g., for a fixed .

Presheaf on a base of opens

Let be a ring. We defined previously the presheaves

and for an -module,

These functors are only defined on a basis , and so are not sheaves under our definition.

Extending to all opens

Let be a base of opens. Suppose we have a presheaf

We only know how to evaluate on:

  • an open set ; and
  • an open cover of consisting of basis elements;

This is enough to construct the left of the diagram:

We now need to deal with the intersections (in fact, because bases are closed under intersection). But we would like to be able to cover these with basis elements too:

  • an open cover of , , consisting of basis elements.

Now, we want the following sequence to be equalising:

Under this condition, there is an equivalence

Example: is a sheaf

Let be a ring and and -module. Then the presheaf

is a sheaf on (in the sense above).

Proof

To show this, we need to show that given:

  • , with ,
  • with ,

we must show that the sequence

is exact. We use the following facts (due to Corollary 3):

  • if and only if for some . Therefore .
  • implies . Therefore .
  • Because , we have .
  • Because , there exist such that .

The point here is that we want to bring things to “common looking denominators”.

Hence, we can put

  • in the place of , and
  • in the place of ; which in turn puts
  • in the place of , and
  • in the place of .

Since and are quasi-compact opens, we can reduce the indexing sets and to finite sets. Then:

  • ; and
  • .

We need to show that

is exact.

Claim (Locality/separated)

is injective.

Proof

If and in each , then there exists such that for all by finiteness of . But , and therefore . So

Gluing

We apply this to the map

for each pair : since in , is injective. Via the factorisation

it suffices to prove that

is exact.

We already know that is injective. It remains to show that:

  • given such that in for all ,
  • there exists such that for all .

Because is finite, we can assume that for all . So from , there exist with

We can again assume that for all by finiteness of , and further that . So we have the equation

Again, we have in , and so there exist such that

Set . So

So in . Therefore

Alternative approach

It is possible to check exactness of

at every maximal ideal: localising at every maximal ideal, the complex becomes degenerate. But this doesn’t give an explicit way of gluing the functions together.

Remarks

  1. If is a sheaf of sets, then the sheaf conditions imply that (terminal object of the category).
  2. If , then
    is an equivalence of categories. Note that this is not an equivalence for , because for a presheaf is unconstrained.