I really didn’t mean to go silent for so long. But there were problems with getting the internet connection here at PSU to work, and besides I’ve been getting into algebraic topology (which is finally making more sense ever since I found Munkres’s textbook). I may actually do a post about acyclic models or something like that soon. Eventually, I’ll probably post about material related to my project (which I find out about tomorrow). First, however, here is something utterly irrelevant.

Let’s begin with a trivial observation. Suppose is a field and a vector space over , contained in a larger vector space . Given a linear functional , we can extend it to a functional . The proof of this is straightforward: choose a basis for , extend it to a basis for and define as zero on and on . If we want to state this in abstract nonsense, we have just shown that

is exact, the map being the usual restriction map. A look at the proof shows that the key step was that we were able to split a basis of , namely , into two parts ; the first was contained in , while the second was a basis for . The generalization of this is to say that the injection is a split injection.

More generally, if are modules over a ring we can say that is a split submodule if it has a complement such that . Any map for another module extends to by mapping the complement to zero. In this way, we get a map extending .

What I have just described is a special case of the extension problem in mathematics. Suppose are objects in a category, and is a monomorphism (so is a subobject of ). Suppose is a map. Is there a map extending ? The situation may be visualized via a commutative diagram

The goal is to find an arrow from filling in the dotted arrow that makes commutative the diagram. This is the extension problem.

This problem has many guises. For instance, it includes many topological problems, for instance the problem of whether a map is homotopic to a constant. Indeed, let be topological spaces and a map. Let be the cone space of , i.e. the quotient of by ; this contains as a subspace. Then is homotopic to a constant map if and only if extends to .

One of the techniques that makes the problem trivial is if the injection admits a retract such that the composition is the identity.

Then, we just compose to get our extension.

We know that a split injection always allows for this kind of retraction, hence the extension problem is always solvable. In particular, in a semisimple abelian category, it is always solvable. In general, however, it is not. Consider the diagram

Clearly, there is no map filling in the dotted line. And of course, multiplication by 2 is not a split injection.

Motivated by this, we define a module over a ring to be **injective** if whenever is an injection, and one has a map , it can be extended to . In other words, the extension problem is always solvable when is at the bottom. For instance, when is a field, every module is injective because every injection splits. More generally, the same is true for any semisimple ring.

What can we say about injective modules? The notion is dual to projectivity, in some sense, so just as every module admits an epimorphic map for projective, we expect by duality that every module admits a monomorphic map for injective. This is in fact true, but will require some work. We start, first, with a fact about injective abelian groups.

Theorem 1A divisible abelian group (i.e. one where the map for any is surjective) is injective as a -module (i.e. abelian group).

The actual idea of the proof is rather simple, and similar to the proof of the Hahn-Banach theorem. Namely, we extend bit by bit, and then use Zorn’s lemma. (I actually kind of wonder whether this statement might be independent of the axiom of choice, but that’s another story.)

So, the first step is that we have a subgroup of a larger abelian group . We have a map of for some divisible abelian group, and we want to extend it to . Now we can consider the poset of pairs where , and is a map extending . Naturally, we make this into a poset by defining the order as “ if is bigger than and an extension of . It is clear that every chain has an upper bound, so Zorn’s lemma implies that we have a submodule containing , a map extending , such that there is no proper extension of . From this we derive a contradiction unless .

Well, so suppose we have . Pick , and consider the submodule . We are going to show how to extend to this bigger submodule. First, suppose , i.e. the sum is direct. Then we can extend because it is a direct sum: just define it to be zero on .

The slightly harder part is what happens if . In particular, there is an ideal such that if and only if . This ideal, however, is principal; let be a generator. Then . In particular, is defined. We can “divide” this by , i.e. find such that . Now extend to a map from into as follows. Choose . Define . It is easy to see that this is well-defined by the choice of , and gives a proper extension of .

**1. Imbedding in injectives **

In constructing the derived functors of a covariant, left-exact functor, one needs to associate to each -module (where is some fixed ring) an injective resolution (i.e. exact sequence)

where the are injective. It is fairly easy to do this once we have proved that there is an injection for injective, whatever the choice of . (This is sometimes stated as saying that the category of -modules has **enough injectives**. Note that not all abelian categories do.)

Theorem 2Any -module can be imbedded in an injective -module .

Well, first of all, we know that any -module is a quotient of a free -module. (I should add that a free module is necessarily projective, but I don’t want to assume familiarity with projective modules here, since this is a GISLA post.) We are going to show that the **dual** (I’m going to define this shortly) of a free module is injective. And so we’re going to use this basic fact about free modules, together with a dualization argument, to prove this.

First, for any abelian group , define the **dual group** as . By basic algebra, an exact sequence of groups

induces an exact sequence

However, I claim that the map is surjective. (This is not the case when is replaced by something else.) The reason is that any element of is just a homomorphism from into the divisible, **and consequently injective**, abelian group . Whence, any such homomorphism can be extended to one of into . So that means that the element in question fo comes from one of . We have shown:

Proposition 3Dualization preserves exact sequences (but reverses the order).

If you prefer, dualization is a contravariant exact functor on the category of abelian groups.

Now, we’re going to apply this to -modules. The dual fo a left -module is acted upon by , but it is not quite a left -module. The action, which is natural enough, is as follows. Let be a homomorphism of abelian groups (since, obviously, has in general no -module structure), and ; then is the map defined via

The reason is the contravariance of the hom in the first variable. Basically, if as above, we’d have instead of the normal way. In short, is a **right** -module. In case you haven’t heard about these, a right -module is basically a left-module over the opposite ring , so the same formalism (kernels, cokernels, injections, etc.) goes through the same as before.

Now, let’s remember this, and return to the original goal: prove that any (left) -module can be imbedded in a left injective -module. (By the same observation, all this goes through for right modules too…)

Here’s a naive approach, which doesn’t work: So let be a left -module, and take a surjection for a free module, i.e. an exact sequene . Dualization gives an exact sequence of right -modules

Now this means that the right -module was imbedded in a divisible abelian group . (Why is divisible? Because given , you just divide on a basis, using the fact that is divisible.) Nah, this isn’t right though. We wanted to be imbedded in something; we got imbedded in something.

So start over. We’ll delay introducing until the time is right. Let be as before, and now consider the right -module . We can take a free **right** module and a surjection

Now dualizing gives an exact sequence

But, by Pontryagin duality, we have (indeed, as -modules—note that the dual of a right-module is a left-module!). I claim now that is injective. This will prove the theorem.

Lemma 4The dual of a right free -module is an injective -module.

Let be exact; we have to show that

is exact. Now we can reduce to the case where is the right -module itself (right action!). Indeed, is a direct sum of ‘s by assumption, and taking hom’s turns them into direct products by basic algebra; moreover the direct product of exact sequences is exact.

So we are reduced to showing that is injective. Now I claim that

In particular, is an exact functor because is an injective abelian group. The proof of this result is actually “trivial.” For instance, a -homomorphism induces by sending . One checks that this is bijective.

August 7, 2010 at 5:09 am

Munkres’ book is really great for algebraic topology, but a really nice supplement is John Baez’s notes:

http://math.ucr.edu/home/baez/algebraic_topology/

I find reading them is like having an iron veil lifted from upon mine eyes.

August 7, 2010 at 8:10 am

They look interesting, but probably more as a supplement to Munkres’s “Elements of Algebraic Topology” than a complement (I mean, EAT covers everything (basic) but homotopy theory–these cover only homotopy theory). It looks like Munkres’s other topology book was the textbook.

August 9, 2010 at 11:32 pm

Have you tried Whitehead’s tome, “Elements of Homotopy Theory”? That’s a great read though you’ll need to invest at least a year for that and only that (literally). The other way

to go is Allen Hatchers “Algebraic Topology” which is pretty sweet and geometrically

motivated. If you like to think in terms of geometry you’ll find that book nice. Also, the

book avoids the generally monstrous amount of homological algebra one needs to get

into algebraic topology. (Like the monstrous amount of comm. algebra one needs to get

into algebraic geometry.)

August 11, 2010 at 7:26 am

I saw it, but it looked pretty intimidating. Prerequisites in terms of homological algebra (or commutative algebra) are generally acceptable (I still have to properly learn spectral sequences, which I should do sometime).

At any rate, Munkres looks eminently understandable, so after that perhaps I will be able to follow these more advanced texts.

August 16, 2012 at 1:10 pm

Not sure if anybody’s gonna read this but it’s worth a try.

Your overview is quite interesting, I came here while trying to solve an exercice on Lang’s Algebra and he follows more or less your same approach to the topic. Still, he says that usually M -> M^^ is a monomorphism, while it seems to me you’re claiming they are isomorphic by Pontryagin duality. However, in the proof of the claim, everything you need is that M injects canonically into M^^ and this is trivial, so I am not sure about the meaning of the “isomorphic” symbol.