2024-09-17

Abstract nonsense

Definition (Additive category)

An additive category is a category such that:

  1. For all , is an abelian group and composition is bilinear, i.e., and . (Technically this is extra data.)
  2. There exists a zero object such that (the abelian group).
  3. For all , there exists
X©YXYpXpYiXiY

which is a coproduct:

XX©YZYiXf9!iYg

and also a product:

XX©YWYpXpYf9!g

Example

The category of projective -modules is an additive category.

Kernels and cokernels

Let be an additive category. Let be a morphism in .

Motivating example: .

Cokernels

If a cokernel exists, it is the pair satisfying the universal property:

XYcokerfZf0¼fg9!eg

Therefore,

Kernels

If a kernel exists, it is the pair satisfying the universal property:

WkerfXY9!g0iff

Therefore,

These definitions only require the existence of the morphism.

Image and coimage

Suppose every morphism has a kernel and cokernel.

kerfker¼fimfXYcoimfcokerifcokerfif:=f¼f:=f

There is a canonical map given by described below.

Example: In

kerfimfXYX=kerfcokerf=Y=imfiff¼f»=

There is an isomorphism by the 1st isomorphism theorem.

Exercise 1

We have

In particular, there is a canonical element : it satisfies the listed properties by putting and in the definitions of kernel and cokernel above. (Done in Assignment 3.)

The canonical element is what induces the isomorphism in the st isomorphism theorem. It is not an isomorphism in general.

Example

Let . We have , but there is a map

We can check that and , and we find that , . The map

is not an isomorphism in (the inverse is not continuous).

Definition (Abelian category)

An abelian category is an additive category such that every morphism has a kernel and cokernel, and for all , the canonical map is an isomorphism.

Sheaves on

Let be a topological space. The category of sheaves on is “obviously” additive.

Recalling the definition

Recall that a sheaf is:

  • an assignment of an abelian group for all open
  • with restriction maps for all satisfying the compatibility condition
F(U)F(V)F(W)

and if is an open cover:

  • (locality) if such that for all then , and
  • (gluing) given for all that glue: then there exists such that for all .

Kernels

Let be a morphism of sheaves. Then

is a sheaf. Exercise: check this.

On each open set , we have

W(U)(kerf)(U)F(U)G(U)9!gU0fU

where the existence and uniqueness of the map comes from the universal property of kernels in the category of abelian groups (which are indeed the kernels in the category theoretic definition).

Cokernels

The definition

is not always a sheaf. We have to sheafify:

where denotes sheafification. Recall that we had the adjoint pair

PreSh(X)Sh(X)leftadjoint!shea¯¯cationforgetÃrightadjoint

Therefore,

giving us

F(U)G(U)(cokerPSf)(U)Z(U)(cokerf)(U)f0g9!9!

where the map uniquely determines the map by adjunction.

1st Isomorphism Theorem

Typically done with stalks. We can do this with an explicit model of sheafification.

The picture:

kerfimfFGcoimfcokerff

In this diagram and require sheafification. So when we are describing the kernel of the morphism , we will also need to take sheafification into account.

We want to know if . For , working with an explicit model of sheafification, we have

On the other hand, consists of , where is an open cover, plus gluing. One sees that the condition is the same as the condition by the 1st isomorphism theorem for abelian groups.

Theorem ( is abelian)

The category of sheaves on a topological space is an abelian category.

Exercise 2

Prove that are not many projective objects in .

Moving sheaves

Let be a continuous function.

Pushforward

We define

This definition makes sense because if is an open then is open.

Pullback

This is harder to define: we can’t just put in because it may not be open. Instead, we define

Q: Is a sheaf without sheafifying?
A: Not in general.

Claim

is left adjoint to , i.e.,

Proof

An element is a family of maps

plus compatibility with restriction.

An element is a family of maps

©U:(f¤F)(U)G(U)lim¡!Vf(U)F(V)F(V)=?

But to map out of a direct limit, we need to specify a map for every (as depicted by above).

  • ) Given a , we can construct the map in the position of by
    since .
  • ) Given a , we define . Note that this makes sense because