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 ${F}$ is a field and ${W}$ a vector space over ${F}$, contained in a larger vector space ${V}$. Given a linear functional ${f: V \rightarrow K}$, we can extend it to a functional ${\tilde{f}:W \rightarrow K}$. The proof of this is straightforward: choose a basis ${B_1}$ for ${W}$, extend it to a basis ${B_2}$ for ${V}$ and define ${\tilde{f}}$ as zero on ${B_2}$ and ${f}$ on ${B_1}$. If we want to state this in abstract nonsense, we have just shown that

$\displaystyle \hom_k(V,K) \rightarrow \hom_k(W, K) \rightarrow 0$

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 ${V}$, namely ${B_2}$, into two parts ${B_1, B_2 =B_1}$; the first was contained in ${W}$, while the second was a basis for ${V/W}$. The generalization of this is to say that the injection ${W \rightarrow V}$ is a split injection.

More generally, if ${M,N}$ are modules over a ring ${A}$ we can say that ${M \subset N}$ is a split submodule if it has a complement ${R \subset N}$ such that ${N = M \oplus R}$. Any map ${M \rightarrow Q}$ for another module ${Q}$ extends to ${M \oplus R}$ by mapping the complement ${R}$ to zero. In this way, we get a map ${N \rightarrow Q}$ extending ${M}$.

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

The goal is to find an arrow from ${y \rightarrow z}$ 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 ${X,Z}$ be topological spaces and ${f: X \rightarrow Z}$ a map. Let ${CX}$ be the cone space of ${X}$, i.e. the quotient of ${X \times I}$ by ${X \times \{1\}}$; this contains ${X}$ as a subspace. Then ${f}$ is homotopic to a constant map if and only if ${f }$ extends to ${CX}$.

One of the techniques that makes the problem trivial is if the injection ${x \rightarrow y}$ admits a retract ${y \rightarrow x}$ such that the composition ${x \rightarrow x}$ is the identity.

Then, we just compose ${y \rightarrow x \rightarrow z}$ 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 ${Q}$ over a ring ${R}$ to be injective if whenever ${M \rightarrow N}$ is an injection, and one has a map ${M \rightarrow Q}$, it can be extended to ${N \rightarrow Q}$. In other words, the extension problem is always solvable when ${Q}$ is at the bottom. For instance, when ${R}$ 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 ${M}$ admits an epimorphic map ${P \rightarrow M}$ for ${P}$ projective, we expect by duality that every module admits a monomorphic map ${M \rightarrow Q}$ for ${Q}$ injective. This is in fact true, but will require some work. We start, first, with a fact about injective abelian groups.

Theorem 1 A divisible abelian group (i.e. one where the map ${x \rightarrow nx}$ for any ${n \in \mathbb{N}}$ is surjective) is injective as a ${\mathbb{Z}}$-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 ${M }$ of a larger abelian group ${N}$. We have a map of ${f:M \rightarrow Q}$ for ${Q}$ some divisible abelian group, and we want to extend it to ${N}$. Now we can consider the poset of pairs ${(\tilde{f}, M')}$ where ${M' \supset M}$, and ${\tilde{f}: M' \rightarrow N}$ is a map extending ${f}$. Naturally, we make this into a poset by defining the order as “${(\tilde{f}, M') \leq (\tilde{f}', M'')}$ if ${M'' }$ is bigger than ${M'}$ and ${\tilde{f}'}$ an extension of ${\tilde{f}}$. It is clear that every chain has an upper bound, so Zorn’s lemma implies that we have a submodule ${M' \subset N}$ containing ${M}$, a map ${\tilde{f}: M' \rightarrow N}$ extending ${f}$, such that there is no proper extension of ${\tilde{f}}$. From this we derive a contradiction unless ${M' = N}$.

Well, so suppose we have ${M' \neq N}$. Pick ${m \in N - M'}$, and consider the submodule ${M' + m \subset N}$. We are going to show how to extend ${\tilde{f}}$ to this bigger submodule. First, suppose ${\mathbb{Z}m \cap M' = \{0\}}$, i.e. the sum is direct. Then we can extend ${\tilde{f}}$ because it is a direct sum: just define it to be zero on ${\mathbb{Z}m}$.

The slightly harder part is what happens if ${\mathbb{Z} m \cap M' \neq \{ 0\}}$. In particular, there is an ideal ${I \subset \mathbb{Z}}$ such that ${n \in I}$ if and only if ${nm \in M'}$. This ideal, however, is principal; let ${g}$ be a generator. Then ${gm = p \in M'}$. In particular, ${\tilde{f}(gm)}$ is defined. We can “divide” this by ${g}$, i.e. find ${u \in Q}$ such that ${gu = \tilde{f}(gm)}$. Now extend to a map ${\tilde{f}'}$ from ${\mathbb{Z} m + M'}$ into ${Q}$ as follows. Choose ${m' \in M', k \in \mathbb{Z}}$. Define ${\tilde{f}'( m' + km) = \tilde{f}(m') + k u}$. It is easy to see that this is well-defined by the choice of ${u}$, and gives a proper extension of ${\tilde{f}}$.

1. Imbedding in injectives

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

$\displaystyle 0\rightarrow M \rightarrow Q_1 \rightarrow Q_2 \rightarrow \dots ,$

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

Theorem 2 Any ${R}$-module ${M}$ can be imbedded in an injective ${R}$-module ${Q}$.

Well, first of all, we know that any ${R}$-module ${M}$ is a quotient of a free ${R}$-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 ${G}$, define the dual group as ${G^\vee = \hom_{\mathbb{Z}}(G, \mathbb{Q}/\mathbb{Z})}$. By basic algebra, an exact sequence of groups

$\displaystyle 0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$

induces an exact sequence

$\displaystyle 0 \rightarrow C^\vee \rightarrow B^\vee \rightarrow A^\vee .$

However, I claim that the map ${B^\vee \rightarrow A^\vee}$ is surjective. (This is not the case when ${\mathbb{Q}/\mathbb{Z}}$ is replaced by something else.) The reason is that any element of ${A^\vee}$ is just a homomorphism from ${A}$ into the divisible, and consequently injective, abelian group ${\mathbb{Q}/\mathbb{Z}}$. Whence, any such homomorphism can be extended to one of ${B}$ into ${\mathbb{Q}/\mathbb{Z}}$. So that means that the element in question fo ${A^\vee}$ comes from one of ${B^\vee}$. We have shown:

Proposition 3 Dualization 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 ${R}$-modules. The dual fo a left ${R}$-module is acted upon by ${R}$, but it is not quite a left ${R}$-module. The action, which is natural enough, is as follows. Let ${f: M \rightarrow \mathbb{Q}/\mathbb{Z}}$ be a homomorphism of abelian groups (since, obviously, ${\mathbb{Q}/\mathbb{Z}}$ has in general no ${R}$-module structure), and ${r \in R}$; then ${rf}$ is the map ${M \rightarrow \mathbb{Q}/\mathbb{Z}}$ defined via

$\displaystyle (rf)(m) = f(rm).$

The reason is the contravariance of the hom in the first variable. Basically, if ${f: M \rightarrow \mathbb{Q}/\mathbb{Z}}$ as above, we’d have ${r_1 (r_2 f) = (r_2 r_1)f}$ instead of the normal way. In short, ${M^\vee}$ is a right ${R}$-module. In case you haven’t heard about these, a right ${R}$-module is basically a left-module over the opposite ring ${R^{op}}$, 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) ${R}$-module can be imbedded in a left injective ${R}$-module. (By the same observation, all this goes through for right modules too…)

Here’s a naive approach, which doesn’t work: So let ${M}$ be a left ${R}$-module, and take a surjection ${F \rightarrow M}$ for ${F}$ a free module, i.e. an exact sequene ${F \rightarrow M \rightarrow 0}$. Dualization gives an exact sequence of right ${R}$-modules

$\displaystyle 0 \rightarrow M^{\vee} \rightarrow F^{\vee}.$

Now this means that the right ${R}$-module ${M^{\vee}}$ was imbedded in a divisible abelian group ${F^{\vee}}$. (Why is ${F^{\vee}}$ divisible? Because given ${f: F \rightarrow \mathbb{Q}/\mathbb{Z}}$, you just divide ${f}$ on a basis, using the fact that ${\mathbb{Q}/\mathbb{Z}}$ is divisible.) Nah, this isn’t right though. We wanted ${M}$ to be imbedded in something; we got ${M^{\vee}}$ imbedded in something.

So start over. We’ll delay introducing ${F}$ until the time is right. Let ${M}$ be as before, and now consider the right ${R}$-module ${M^{\vee}}$. We can take a free right module ${F}$ and a surjection

$\displaystyle F \rightarrow M^{\vee} \rightarrow 0.$

Now dualizing gives an exact sequence

$\displaystyle 0 \rightarrow M^{\vee \vee} \rightarrow F^{\vee}.$

But, by Pontryagin duality, we have ${M \simeq M^{\vee \vee}}$ (indeed, as ${R}$-modules—note that the dual of a right-module is a left-module!). I claim now that ${F^{\vee}}$ is injective. This will prove the theorem.

Lemma 4 The dual of a right free ${R}$-module ${F}$ is an injective ${R}$-module.

Let ${0 \rightarrow A \rightarrow B }$ be exact; we have to show that

$\displaystyle \hom_R( B, F^\vee) \rightarrow \hom_R(A, F^\vee) \rightarrow 0 .$

is exact. Now we can reduce to the case where ${F}$ is the right ${R}$-module ${R}$ itself (right action!). Indeed, ${F}$ is a direct sum of ${R}$‘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 ${R^{\vee}}$ is injective. Now I claim that

$\displaystyle \hom_R(B, R^{\vee}) = \hom_{\mathbb{Z}}(B, \mathbb{Q}/\mathbb{Z})$

In particular, ${\hom_R( --, R^\vee)}$ is an exact functor because ${\mathbb{Q}/\mathbb{Z}}$ is an injective abelian group. The proof of this result is actually “trivial.” For instance, a ${R}$-homomorphism ${f: B \rightarrow R^\vee}$ induces ${\tilde{f}: B \rightarrow \mathbb{Q}/\mathbb{Z}}$ by sending ${b \rightarrow (f(b))(1)}$. One checks that this is bijective.