This post is part of a series on the Sullivan conjecture in algebraic topology. The Sullivan conjecture is a topological result, which remarkably reduces — as H. Miller showed —  to a purely algebraic computation in the category of unstable modules (and eventually algebras) over the Steenrod algebra, and in particular an injectivity assertion. This is a rather formidable category, but work of Kuhn enables one to identify a quotient category of it with the category of “generic representations” of the general linear group, which can be studied using different (and often easier) means. Kuhn’s work provides an approach to proving much of the algebraic background that goes into the Sullivan conjecture. In this post, I’ll describe one of the important ingredients.

The Gabriel-Popsecu theorem is a structure theorem for Grothendieck abelian categories, a version of which will be useful in understanding the structure of the category of unstable modules over the Steenrod algebra. The purpose of this post is to discuss this result and its many-object version due to Kuhn, from the paper “Generic Representations of the Finite General Linear Groups and the Steenrod Algebra: I.” Although the proof consists mostly of a series of diagram chases, there are some subtleties that I found rather difficult to grasp, and I thought it would be worthwhile to go through it in detail here.

1. Grothendieck abelian categories

Let ${\mathcal{A}}$ be an abelian category. Then ${\mathcal{A}}$ is Grothendieck abelian if

• ${\mathcal{A}}$ has a generator: that is, there is an object ${X \in \mathcal{A} }$ such that every object in ${\mathcal{A}}$ can be built up from colimits starting with ${X}$. (More precisely, the smallest subcategory of ${\mathcal{A}}$, closed under colimits, that contains ${X}$ is ${\mathcal{A}}$ itself)
• Filtered colimits in ${\mathcal{A}}$ exist and are exact.

Many categories occurring in “nature” (e.g., categories of modules over a ring of sheaves on a site) are Grothendieck, and it is thus useful to have general results about them. The goal of this post is to describe a useful structure theorem for Grothendieck abelian categories, which will show that they are the quotients of categories of modules by Serre subcategories.

One can give a simple categorical condition on ${\mathcal{A}}$ in order for it to be equivalent to a category of modules over a ring. This is essentially Morita theory, and is as follows:

Theorem 17 Suppose ${\mathcal{A}}$ has a compact, projective generator ${X}$, and let ${R = \mathrm{End}(X)}$. Then the functor

$\displaystyle \mathcal{A} \rightarrow \mathrm{Mod}(R^{op}), \quad C \mapsto \hom_{\mathcal{A}}(X, C)$

is an equivalence of categories.

In particular, Morita theory states that what distinguishes categories of modules over a ring is the existence of a compact, projective generator.

If ${X}$ is no longer compact and projective, one can still say something. It turns out that ${\mathcal{A}}$ will be expressed as the localization of right ${R}$-modules by a Serre subcategory. In particular, any Grothendieck abelian category can be expressed in this way.

Theorem 18 (Gabriel-Popescu) Let ${X}$ be a generator for ${\mathcal{A}}$ and let ${R = \mathrm{End}(X)}$ as before. There is an adjunction:

$\displaystyle \otimes_R X: \mathrm{Mod}(R^{op}) \rightleftarrows \mathcal{A}: \hom(X, \cdot)$

such that:

• The functor ${\otimes_R X}$ is exact.
• The right adjoint ${\hom(X, \cdot)}$ is fully faithful (that is, ${\mathcal{A}}$ is a localization of ${\mathrm{Mod}(R^{op})}$).

If ${\mathcal{N} \subset \mathrm{Mod}(R^{op})}$ is the (Serre) subcategory killed by ${\otimes_R X}$, then there is induced an equivalence

$\displaystyle \mathrm{Mod}(R^{op})/\mathcal{N} \simeq \mathcal{A}.$

2. The first step in the proof

The existence of the functor ${\otimes_R X}$ follows from the adjoint functor theorem. More explicitly, given a right ${R}$-module ${M}$, the functor

$\displaystyle M \mapsto M \otimes_R X$

is a right-exact, additive functor with the property that ${R \otimes_R X = X}$. So to compute ${M \otimes_R X}$, choose a free presentation

$\displaystyle R^I \rightarrow R^J \rightarrow M \rightarrow 0$

and define ${M \otimes_R X}$ to be the cokernel of ${X^I \rightarrow X^J}$.

Our first goal is now to check that ${\hom_{\mathcal{A}}(X, \cdot)}$ is fully faithful, which is equivalent to the condition that for any ${Y \in \mathcal{A}}$,

$\displaystyle X \otimes_R \hom_{\mathcal{A}} (X, Y) \rightarrow Y$

be an isomorphism. In fact, most of the argument is quite general. One of the key steps in this process is that ${\mathcal{A}}$ is an abelian category, which means that to check that a map is an isomorphism, then it suffices to show that it is a monomorphism and an epimorphism.

For convenience, denote:

$\displaystyle F \stackrel{\mathrm{def}}{=} \cdot\otimes_R X, \quad G \stackrel{\mathrm{def}}{=} \hom_{\mathcal{A}}(X, \cdot)$

for the two adjoint functors that enter.

We will start by proving:

Proposition 19 Let ${f: M \rightarrow G Y}$ be a morphism in ${\mathrm{Mod}(R^{op})}$ for some ${Y \in \mathcal{A}}$. If ${f}$ is injective, then its adjoint ${\widetilde{f}: FM \rightarrow Y}$ is also a monomorphism.

Proof: The condition of the proposition (that ${f}$, adjoint to ${\widetilde{f}}$, be a monomorphism of modules) implies that if ${FM' \rightarrow FM}$ is any morphism in the image of ${F}$, then the composite

$\displaystyle FM' \rightarrow FM \rightarrow Y$

is zero only if

$\displaystyle FM' \rightarrow FM$

is zero. The condition “in the image of ${F}$” is the snag, and the point of the proof is to get around it.

To prove that ${\widetilde{f}}$ is a monomorphism, we may choose an epimorphism ${u: Q \twoheadrightarrow FM}$ in ${\mathcal{A} }$, and show that the map

$\displaystyle \mathrm{ker}(u) \rightarrow \mathrm{ker}(\widetilde{f} \circ u)$

is an isomorphism.

For ${Q}$, we can choose a free ${R}$-module ${R^I}$ and a surjection ${u': R^I \twoheadrightarrow M}$. We can take ${u}$ to be ${F(u')}$, so that we need to prove that

$\displaystyle \mathrm{ker}( F(u'): X^I \rightarrow FM ) \rightarrow \mathrm{ker}( \widetilde{f} \circ F(u'): X^I \rightarrow Y).$

Since filtered colimits are exact, it suffices to prove the result with ${I}$ replaced by a finite subset thereof (although we may lose the epimorphism property).

So we may reduce to proving:

Lemma 20 Let ${f, \widetilde{f}}$ be as above. Let ${R^J}$ be a finite free ${R}$-module and let ${v: R^J \rightarrow M}$ be a map of ${R}$-modules. Then

$\displaystyle \mathrm{ker}( F(v)) = \mathrm{ker}( \widetilde{f} \circ F(v)).$

To prove this lemma in turn, it suffices to prove that if

$\displaystyle X \rightarrow X^J$

is any morphism in ${\mathcal{A}}$, then ${X \rightarrow X^J \rightarrow FM}$ is zero if and only if ${X \rightarrow X^J \rightarrow FM \rightarrow Y}$ is zero. (We can do this because ${X}$ generates ${\mathcal{A}}$.) However, we observe that any map ${X \rightarrow X^J}$ is in the image of ${F}$, and ${X^J \rightarrow FM}$ is in the image of ${F}$. We have already seen that, by adjointness, if ${X \rightarrow FM}$ is any map in the image of ${F}$, then ${X \rightarrow FM}$ is zero if and only if ${X \rightarrow FM \stackrel{\widetilde{f}}{\rightarrow} Y}$ is zero. This completes the proof.

$\Box$

Using this, we will show that ${G}$ is fully faithful.

Proposition 21 The functor ${G}$ is fully faithful.

Proof: To see that ${G}$ is fully faithful, it suffices by a diagram chase to show that

$\displaystyle FG Y \rightarrow Y$

is an isomorphism for any ${Y \in \mathcal{A}}$. It is “formal” that ${FG Y \rightarrow Y}$ is an epimorphism. In fact, the claim that ${FG Y \rightarrow Y}$ is an epimorphism for every ${Y}$ is equivalent to the faithfulness of ${G}$. But if ${ Y \rightarrow Y'}$ is any morphism in ${\mathcal{A}}$ and

$\displaystyle \hom_{\mathcal{A}}(X, Y) \rightarrow \hom_{\mathcal{A}}(X, Y')$

is zero, then ${Y \rightarrow Y'}$ is zero as ${X}$ generates ${\mathcal{A}}$.

To see that ${FG Y \rightarrow Y}$ is a monomorphism, we appeal to the preceding lemma and note that the adjoint map is the identity ${GY \rightarrow GY}$, which is evidently a monomorphism. $\Box$

3. Completion of the proof

The final step is to show that ${F = \otimes_R X}$ is an exact functor — it is already right exact as a left adjoint. Since ${\mathrm{Mod}(R^{op})}$ has enough projectives, we can talk about the the left derived functors of ${F}$, and it suffices to show that they vanish. For this, in turn, it suffices to show that if

$\displaystyle M' \subset M$

is an inclusion of modules, with ${M}$ free, then

$\displaystyle FM' \rightarrow FM$

is a monomorphism in ${\mathcal{A}}$. Using a filtered colimit argument, we can even reduce to the case of ${M}$ finitely generated.

But in this case, the map ${M' \hookrightarrow M}$ can be identified with a map ${M' \hookrightarrow G(X^I)}$ for some finite index set ${I}$. The above lemma shows that the adjoint map ${F(M') \rightarrow X^I \simeq FG(X^I)}$ is a monomorphism, which completes the proof. This in particular proves the Gabriel-Popescu theorem.

The Gabriel-Popescu theorem has another interpretation. Let ${\mathcal{A}}$ be as above and let ${\mathcal{D} \subset \mathcal{A}}$ be the full subcategory spanned by ${X}$. Then ${\mathcal{D}}$ and ${\mathcal{A}}$ are categories enriched over abelian groups, and the second statement is equivalent to saying that

$\displaystyle \mathcal{D} \rightarrow \mathcal{A} \stackrel{\mathrm{id}}{\rightarrow} \mathcal{A}$

exhibits the identity functor ${\mathcal{A} \rightarrow \mathcal{A}}$ as a left Kan extension. This is equivalent to the condition that any element of ${\mathcal{A}}$ can be built up as a “canonical colimit” of elements of ${\mathcal{D}}$, or the fact that

$\displaystyle X \otimes_R \hom_{\mathcal{A}}(X, Y) \rightarrow Y$

is an isomorphism for all ${Y \in \mathcal{A}}$. We knew at the beginning that every object of ${\mathcal{A}}$ could be obtained form ${X}$ using colimits; this condition states that it can be done so in a canonical way.

Corollary 22 If ${I \in \mathcal{A}}$ is injective, then so is ${GI}$ in ${R}$-modules.

This is a formal argument: a right adjoint whose left adjoint is exact preserves injectives.

4. Kuhn’s many-object variant

The purpose of this section is to explain a many-object version of the Gabriel-Popescu theorem due to Kuhn. Suppose ${\mathcal{A}}$ is Grothendieck abelian and ${\mathcal{S} \subset \mathcal{A}}$ is a collection of objects which generates ${\mathcal{A}}$ under colimits. It is possible but technically inconvenient to give ${\mathcal{A}}$ a single generator by taking the direct sum of everything in ${\mathcal{S}}$.

Kuhn’s idea is to generalize the Gabriel-Popescu theorem in the following manner. A ring is the same thing as a category enriched in abelian groups, containing only one object. A module over that ring is an enriched functor from that category to abelian groups. Hence:

Definition 23 We define the category ${\mathrm{Rep}(\mathcal{S}^{op})}$ to be the category of (additively) enriched functors

$\displaystyle \mathcal{S}^{op} \rightarrow \mathbf{Ab}.$

Given a single object ${X \in \mathcal{A}}$, we got an adjunction between right ${R}$-modules (for ${R = \mathrm{End}(X,X)}$) and ${\mathcal{A}}$ itself. The many-object analog is the right adjoint

$\displaystyle \mathcal{A} \rightarrow \mathrm{Rep}(\mathcal{S}^{op})$

which is really the Yoneda functor: it sends an object ${Y}$ to the representation ${S \mapsto \hom_{\mathcal{A}}(S, Y)}$. The adjoint functor theorem gives a left adjoint

$\displaystyle \mathrm{Rep}(S^{op}) \rightarrow \mathcal{A}$

which sends a representation ${F: \mathcal{S}^{op} \rightarrow \mathbf{Ab}}$ to the functor tensor product (see for instance this MO post)

$\displaystyle F \otimes_{\mathcal{S}^{op}} \mathrm{Id}$

where ${\mathrm{Id}: \mathcal{S} \rightarrow \mathcal{S}}$ is the identity functor.

Now, one has:

$\displaystyle F: \mathrm{Rep}(\mathcal{S}^{op})\rightleftarrows \mathcal{A}: G$

has the following properties:

• ${F}$ is exact.
• ${G}$ is fully faithful.

In particular, ${\mathcal{A}}$ is a localization of ${\mathrm{Rep}(\mathcal{S}^{op})}$ by a Serre subcategory.

The proof of Kuhn’s theorem is a generalization of the proof of the Gabriel-Popescu theorem. The key idea is to replace “free modules” with representable functors. For any ${S \in \mathcal{S}}$, there is a representation ${h_S \in \mathrm{Rep}(\mathcal{S}^{op})}$ given by

$\displaystyle S' \mapsto \hom_{\mathcal{A}}(S', S).$

The Yoneda lemma shows that these are projective objects in ${\mathrm{Rep}(\mathcal{S}^{op})}$, and ${h_S}$ maps under the functor ${F}$ to ${S}$ itself. (In fact, this is one way to describe the functor tensor product.) One can now run more or less the above proof, noting that ${F}$ is fully faithful on these representable objects ${h_S}$.