An Abelian category is an additive category with kernels and cokernels, and .

( a ring) and ( a ringed space) are abelian categories.

Kernels and cokernels are defined categorically: they are pullbacks and pushouts respectively.

Let be a ringed space. A sequence of -module homomorphisms

is exact in if and only if the sequence of -module homomorphisms

is exact for all .

An -module is:

By the following Lemma 1Lemma 1, is a full subcategory of quasi-coherent sheaves.

Let be a map of ringed spaces. Then the restriction of to factors through , i.e. there exists a pullback .

Let . It suffices to prove that . This is a local property on , so it suffices to check this on an open covering.

Consider

We note that we have

We now choose an open covering such that locally there exists a presentation:

We have the following diagram:

Then we apply (which is a left adjoint to the pushforward, so it preserves all colimits, and specifically cokernels, and therefore right exact sequences) to this exact sequence to find that

where the vertical isomorphisms are by commutativity of (pullback of the structure sheaf is the structure sheaf). So is quasi-coherent.

Let be a ring. Then there is an adjoint pair

Moreover, we have the following properties for:

Let be an -module and be an -module. There is a map

Let’s describe an inverse: for , we have

But now is an -module, i.e., acts invertibly on . By the universal property of localisation, there exists a unique map in the diagram above for every . So

is a -module homomorphism. It is clear that this association is inverse to the previous map. This shows adjunction.

Because is left adjoint, it preserves all colimits. For exactness, we get right exactness for free since is a left adjoint. We just need left exactness: we need to show that if

is an exact sequence of -modules, then

is an exact sequence of -modules. To show this, we can check on stalks and use the fact that for all . So it suffices to prove that

is exact for all . But this is clear from the properties of localisation.

For full faithfulness, it suffices to show that , but this is trivial.

Finally, for any -module , there exists a presentation

But now, this just turns into

since . Therefore .

Let be a ring and be a quasi-coherent -module. Then there exists an -module and a -module homomorphism

Assume that there exists a global presentation, i.e., some such that

is exact. Then .

Let (since ) by full faithfulness of . That is, and the following sequence is exact

Applying , we get

Since , there exists a unique between the cokernels.

Now we have to patch these together. admits an open cover on which admits presentations.

We can find with (by quasi-compactness) such that

is a presentation for each . We apply Step 1Step 1 to and . This implies that there exists

for some -module . Now note that we have on the intersection:

and in fact for an isomorphism by full faithfulness of . Also note that

as we have an -module isomorphism . We now define via the exact sequence

That is, is the kernel of . Let and be the inclusions. Then (on the LHS is regarded as an -module via , and on the RHS is regarded as an -module), and similarly for the other inclusions. We thus have by applying to the sequence above defining :

The bottom sequence is exact because is a sheaf. The square on the right commutes because the map is by construction the map from the gluing condition for sheaves. Since we have isomorphisms in the right square, there exists a unique isomorphism between the kernels.