The story of how I ended up writing this post is a bit roundabout. I was trying to figure out whether a technical lemma in Hartshorne on the cohomology of inductive limits of sheaves could be proved using spectral sequences, and I thought I had it, but I then realized that I had no justification for asserting that the category in question had enough injectives (which is necessary to apply the Grothendieck spectral sequence). So I tracked down the original reference to the result, which was–conveniently enough–in Grothendieck’s famous Tohoku paper (available openly). It turns out that there is a way to see that the category I was interested in had enough injectives, but it is a fairly interesting and involved result. I will explain what I learned from reading the section of the paper today. Next time, I will explain my thoughts that led me here. Also, before I proceed, here’s a PDF of the post.

It is known that the category of modules over a ring has enough injectives, i.e. any object can be imbedded as a subobject of an injective object. This is one of the first things one learns about injective modules, though it is a nontrivial fact and requires some work. Similarly, when introducing sheaf cohomology, one has to show that the category of sheaves on a given topological space has enough injectives, which is a not-too-difficult corollary of the first fact.

However, it is more difficult to see that, for instance, the category of inductive diagrams of sheaves (within a fixed inductive system) has enough injectives. For this a general result that states that wide classes of abelian categories (satisfying minor categorical conditions) have enough injectives is handy and convenient. This result is in Tohoku, but Grothendieck claims it is not his.

I will assume familiarity with basic diagram-chasing in abelian categories (e.g. the notion of the inverse image of a subobject).

**1. Preliminaries **

Let be an abelian category. An object is called **injective** if the functor is exact; it is always left-exact. This is the same as saying that if is a subobject of , then any can be extended to a morphism . We’d like to be able to embed every object in an injective one; this is necessary to define right-derived functors. If so, we say that the category **has enough injectives**.

Let us consider the following conditions on an abelian category . First, assume admits all filtered colimits. So, for instance, arbitrary direct sums exist.

Moreover, if the are subobjects of , then the **sum** makes sense; it is the image of , and is a sub-object of .

We start with the condition

for any family of subobjects of such that form a filtered increasing family. Note that there is always a canonical monomorphism because each injects into . We want this to be an isomorphism. This is really a condition on the lattice of subobjects of .

Note that (A) is always satisfied for modules over a ring. In addition, it is satisfied for sheaves on a topological space. Moreover, suppose it is satisfied for the abelian category . Then for any category , it is satisfied for the functor category ; this is because intersections, direct sums, etc. are calculated “pointwise” in a functor category.

Let us analyze some of the consequences of (A). In the lattice of subobjects of , the sum now corresponds to the taking the sup. So the lattice admits sups. Indeed, suppose (in this subobject lattice) that for each in some index set . Then . In particular, dominates . Conversely, if dominates , then obviously dominates each .

Next, suppose given a morphism . There is an order-preserving map from the subobject lattice of to the subobject lattice of . I claim that:

Proposition 1commutes with the operator (i.e., sups). More precisely, I claim that

for any family of subobjects of .

This follows directly from the characterization of as supremums in the lattice of subobjects. Indeed, let be a subobject of . Then dominates if and only if dominates , or if and only if dominates each . Similarly, dominates if and only if dominates each , i.e. if and only dominates each .

So the same objects dominate and . This means that they must be equal. (Indeed, the first dominates itself, hence dominates the second, so we have the relation ; and vice versa.)

Indeed, this follows because commutes with join and meet in the lattices of subobjects, so that we have

in any abelian category. (Proof: a morphism factors through if and only if factors through , i.e. if it factors through both and . This is equivalent to factoring through and .)

The general result then follows from the interpretation of arbitrary sums as a sup.

**2. Generators **

In the category of -modules, any module admits a surjection for some index set . There is a generalization of this to abelian categories.

Definition 2Let be an abelian category with inductive limits. Then is called ageneratorif to every , there is an index set and an epimorphism

Any nonzero free module is a generator in the category of -modules. In the category of sheaves on a topological space, the direct sum of constant sheaves for an open set is a generator. Indeed, by homming out of such sheaves into some sheaf , you can have each section over each open set covered in the image—so you get an epimorphism of sheaves.

**3. The main result **

The main result is as follows:

Theorem 3Suppose is an abelian category admitting inductive limits which satisfies (A) and admits a generator. Then has enough injectives.

The proof of this is a bit messy, but not terrible. The first idea is to show that being injective can be reduced to a statement that involves only the generator (as opposed to every object in the category). The next is to use some set-theoretic messiness that is made possible by only having to consider the generator.

**4. The main lemma **

Lemma 4Suppose . Suppose whenever is a subobject of , then any morphism can be extended to some . Then is injective.

So in other words, we just have to check the meaning of injectivity on the subobjects of . To prove this, suppose satisfies the condition of the lemma, and is a subobject of . We have a morphism that we want to extend to .

Consider the set of pairs of subobjects of containing such that extends to . We can make this into a poset in the obvious manner, and every chain has an upper bound since the category admits inductive limits. To show that extends to , we have to show (by Zorn’s lemma) that any element can be extended to a bigger subobject. This will imply that the maximal element in this poset is a morphism out of .

OK. So, we have ; we will extend it to a bigger subobject. Choose whose image is not contained in ; this must be possible, since is a generator. Let be the pre-image of .

Then we have a map , so a map . By assumption, there is an extension . We now want to piece together the extension with .

There is a map coming from . The image of this is precisely the sum , which is a proper superobject of , and which we will extend to. First, however, we need to understand the kernel. Then, to hom out of will be the same thing as homming out of while annihilating the kernel.

I claim that the kernel is precisely the image of , via the inclusion and the map (restricting ). In other words, the sequence

is exact, where we are given that and are monomorphisms. This will follow now from the next lemma.

Lemma 5Let be an abelian category. Suppose given objects , . Suppose there is a morphism such that is the pre-image of in . The sequence

is exact (where is the combination of the inclusion and the restriction of ).

For this, we note that one can reduce to the category of abelian groups. Indeed, for each object , we have to show that is exact. Moreover, is a subgroup of , and is the preimage of under . So we have the exact same hypotheses of the lemma but in the category of abelian groups and with hom-sets as the new objects. (This is a special case of a standard “Yoneda trick.”) In particular, we just have to consider the category of abelian groups.

So let be abelian groups. Suppose is annihilated. Then, first of all, the image of in must be , so lies in . Moreover, must be the image of under , so it follows that is in the image of . (This is basically immediate.) Thus, the lemma is true for abelian groups, hence for all abelian categories.

So, return to the proof of the main lemma. We know that is the quotient of by . Now consider the map piecing together (remember that this is extended) and . These patch on by assumption, so it extends to But this is just ; thus we get an extension of to . And, by previous remarks, the lemma is proved.

**5. Proof of the theorem **

Fix an object . We have to embed in an injective object . First, we construct . Note that to prove that is in fact injective, we just need to show (by the previous lemma) that any map into from a subobject of the generator extends to all of .

We shall define a sequence as follows—in fact, we will define for any ordinal number . First, we define . We will do so in such a way that any map of a subobject into extends to a map .

Consider the set of all mappings of subobjects into , for all subobjects of . There is a map

that sends to the combination of its image in and the inclusion in (at the -th factor). Consider the quotient; we call this . There is obviously a map from the inclusion on the first factor (and then the quotient). Then, if is a map for , we can extend it to by using the appropriate factor in the second part of the sum. (This is why the quotienting was necessary—to make sure that this procedure indeed gives an extension.)

Moreover, is a monomorphism, because fibered coproducts preserve monicness in any abelian category. Note that is the push-out of , and is monic. (I’m skipping a few nontrivial details here—this can be seen by diagram-chasing, or by invoking the Freyd-Mitchell embedding theorem, though this is deprecated.) So we have embedded in a suitable object , but it is not injective in general.

Define inductively , and so on. We define for an ordinal by transfinite induction. Assume that for is defined and there are monomorphisms for . We will now construct .

If has an immediate predecessor , then set . If is a limit ordinal, then note that the set of objects is an inductive system and we can take the inductive limit of these to be .

Now, choose to be the smallest infinite ordinal whose cardinality is strictly greater than the cardinality of the set of subobjects of . Then is a limit ordinal; it has no immediate predecessor. There is an injection .

I claim now that is injective. Indeed, let , and consider . There is the sequence of subobjects for . We have that is the union (inductive limit) of the subobjects . The axiom (A) implies that is the union of the , since inverse images commute with sups.

But the sequence must stabilize before since there are less than subobjects of . In particular, the image of is contained in a subset for . But we know that can be extended to a map from into by construction. So it can be extended from a map of into . This proves injectivity.

Moreover, can be embedded in each . This completes the proof.

August 23, 2010 at 10:05 am

[…] condition (A) is preserved under functor categories, as remarked there). So by the theorem of the previous post, admits enough […]

August 15, 2012 at 1:47 pm

Dear Akhil,

You wrote in section 1 that condition (A) is satisfied for $R$ modules. If we take $R$ to be a field and consider the category of vector spaces, then we can find a vector space V and sub vector spaces $A_1,A_2,B$ such that $A_1+A_2=V$ and $A_i\cap B=0$, which would show that (A) is not satisfied.

August 15, 2012 at 10:21 pm

Thanks for the correction. It should have been “for filtered unions”; I’ve just fixed it.