2024-09-17
Abstract nonsense
Definition (Additive category)
An additive category
- For all
, is an abelian group and composition is bilinear, i.e., and . (Technically this is extra data.) - There exists a zero object
such that (the abelian group). - For all
, there exists
which is a coproduct:
and also a product:
Example
The category of projective
Kernels and cokernels
Let
Motivating example:
.
Cokernels
If a cokernel exists, it is the pair
Therefore,
Kernels
If a kernel exists, it is the pair
Therefore,
These definitions only require the existence of the
morphism.
Image and coimage
Suppose every morphism has a kernel and cokernel.
There is a canonical map
Example: In
There is an isomorphism
Exercise 1
We have
In particular, there is a canonical element
The canonical element
Example
Let
We can check that
is not an isomorphism in
Definition (Abelian category)
An abelian category is an additive category such that every morphism has a kernel and cokernel, and for all
Sheaves on
Let
Recalling the definition
Recall that a sheaf
- an assignment of an abelian group
for all open - with restriction maps
for all satisfying the compatibility condition
and if
- (locality) if
such that for all then , and - (gluing) given
for all that glue: then there exists such that for all .
Kernels
Let
is a sheaf. Exercise: check this.
On each open set
where the existence and uniqueness of the map
Cokernels
The definition
is not always a sheaf. We have to sheafify:
where
Therefore,
giving us
where the map
1st Isomorphism Theorem
Typically done with stalks. We can do this with an explicit model of sheafification.
The picture:
In this diagram
We want to know if
On the other hand,
Theorem ( is abelian)
The category
Exercise 2
Prove that are not many projective objects in
Moving sheaves
Let
Pushforward
We define
This definition makes sense because if
Pullback
This is harder to define: we can’t just put in
Q: Is
a sheaf without sheafifying?
A: Not in general.
Claim
Proof
An element
plus compatibility with restriction.
An element
But to map out of a direct limit, we need to specify a map
) Given a , we can construct the map in the position of bysince . ) Given a , we define . Note that this makes sense because