It's truly bizarre that there exists no full modern exposition of this theorem, as noted elsewhere. Anyway, I thought I'd poke through and see if I could get the gist of how it works as somebody who has familiarity with categorical techniques, if not abelian techniques.

Here's how it goes, following Swan and wikipedia. We have a small abelian category $\mathcal{A}$, and we want a full exact embedding into the category of modules for some ring. The first step is to take the Yoneda embedding $\mathcal{A} \to \mathcal{L}^\mathrm{op}$, where $\mathcal{L} = \mathrm{Lex}(\mathcal{A},\mathsf{Ab})$. There are other ways to denote $\mathcal{L}$ -- it's $\mathrm{Ind}(\mathcal{A}^\mathrm{op})$, or $\mathrm{Pro}(\mathcal{A})^\mathrm{op}$. So it's a general fact that this embedding is exact, and I totally believe that $\mathcal{L}^\mathrm{op}$ is abelian.

But unless I'm reading something wrong, the point is that $\mathcal{L}^\mathrm{op}$ actually is a category of modules over a ring -- one constructs a projective generator in it. This can't be right. Because $\mathcal{L}^\mathrm{op} = \mathrm{Ind}(\mathcal{A}^\mathrm{op})^\mathrm{op}$ is the opposite of a locally presentable category! So $\mathcal{L}^\mathrm{op}$ can't be locally presentable (the opposite of a locally presentable category is never locally presentable unless the category is a preorder -- cf the nlab Counterexample 7, or Thm 1.64 in Adámek and Rosický), and hence it can't be the category of modules over a ring.

What am I misunderstanding? The obvious thing to do is to dualize and embed $\mathcal{A}$ into $\mathrm{Lex}(\mathcal{A}^\mathrm{op},\mathsf{Ab}) = \mathrm{Ind}(\mathcal{A})$, which is locally presentable. But if you do this it seems it would take some kind of miracle for the generator to be projective.


You don't show that $\mathcal{L}^{op}$ actually is a category of modules, just that any small subcategory of it fully and exactly embeds in one. Indeed, an abelian category with a projective generator is not usually equivalent to a category of modules: it only is if the projective generator is compact (and the category is cocomplete). In the case of $\mathcal{L}^{op}$, there is no reason to expect the projective generator to be compact.

Note that in general, a projective generator doesn't even necessarily give a full embedding into a category of modules, let alone one that is essentially surjective on objects. However, if you have a projective generator $P$ that can actually surject onto every object, then $\operatorname{Hom}(P,-)$ does give a fully faithful exact embedding into $R\text{-Mod}$ where $R=\operatorname{End}(P)$ (but then it won't be essentially surjective on objects, since if such a $P$ exists your category won't be cocomplete). Such a $P$ will exist for any small subcategory of $\mathcal{L}^{op}$ (just take a direct sum of enough copies of your projective generator), and in particular for the image of $\mathcal{A}$.

Here's a simple example where a projective generator does not give a full embedding (a special case of Jeremy Rickard's suggestion in the comments). Let $k$ be a field and consider $k$ as an object of $k\text{-Mod}^{op}$. Then $k$ is a projective generator, and the embedding it gives is just the duality functor $k\text{-Mod}^{op}\to k\text{-Mod}$. This is not full: if $V$ and $W$ are infinite-dimensional vector spaces, then not every linear map $W^*\to V^*$ is the dual of a linear map $V\to W$.

Here is a different sort of example that I think is also instructive. Take the category of countable abelian groups, and by Lowenheim-Skolem take a countable subcategory $\mathcal{C}$ that is an elementary substructure (over the language of categories) and contains the objects $\mathbb{Z}$, $\bigoplus_\mathbb{N}\mathbb{Z}$, and all homomorphisms $\mathbb{Z}\to\bigoplus_\mathbb{N}\mathbb{Z}$. Then $\mathbb{Z}$ will still be a projective generator for $\mathcal{C}$, and so it gives a faithful exact embedding $\mathcal{C}\to\mathcal{Ab}$ which sends $\bigoplus_\mathbb{N}\mathbb{Z}$ to $\operatorname{Hom}_{\mathcal{C}}(\mathbb{Z},\bigoplus_\mathbb{N}\mathbb{Z})=\bigoplus_\mathbb{N}\mathbb{Z}$. But the inclusion is not full, since $\mathcal{C}$ is countable and hence $\operatorname{Hom}_{\mathcal{C}}(\bigoplus_\mathbb{N}\mathbb{Z},\bigoplus_\mathbb{N}\mathbb{Z})$ is countable, but $\operatorname{Hom}_{\mathcal{Ab}}(\bigoplus_\mathbb{N}\mathbb{Z},\bigoplus_\mathbb{N}\mathbb{Z})$ is uncountable.

  • 2
    $\begingroup$ Another example of a projective generator giving a non-full embedding: If $I$ is an injective cogenerator in a module category $R\text{-Mod}$, with endomorphism ring $E$, then the functor $\operatorname{Hom}(-,I)$ from $R\text{-Mod}$ to $E\text{-Mod}$ is not usually full. So just take the opposite category of $R\text{-Mod}$. $\endgroup$ – Jeremy Rickard Jul 11 '16 at 10:17
  • $\begingroup$ Thanks so much! It's funny, in a nonabelian context I've been thinking lately about regular generators which are not dense, but the classic example in the non-additive context is $\mathbb{Z} \in \mathsf{Ab}$, and this problem is fixed by only taking additive functors. An explicit example along the lines of Jeremy Rickard's suggestion would be $\mathbb{Z}$ in compact Hausdorff abelian groups, I think. $\endgroup$ – tcamps Jul 11 '16 at 11:33
  • $\begingroup$ Did I just claim that $\mathbb{Z}$ is compact? Ouch! $\endgroup$ – tcamps Jul 20 '16 at 18:14

I just wanted to outline a proof of the Freyd-Mitchell embedding theorem that even I can understand.

Proposition 1. If $\mathcal{A}$ is an abelian category, then $\mathrm{Ind}(\mathcal{A})$ is abelian, and the inclusion $\mathcal{A} \to \mathrm{Ind}(\mathcal{A})$ is fully faithful, exact, takes values in compact objects, and preserves generators and projective objects.

Proof. The fully faithfulness, exactness, and compactness are standard (true for any category $\mathcal{A}$ whatsoever). For abelianness, see here. The preservation of generators is not hard to see using the "quotient of a coproduct" characterization of generators in a category with coproducts. The preservation of projective objects follows from the fact that every epimorphism in $\mathrm{Ind}(\mathcal{A})$ is the cokernel of a levelwise map of filtered colimits of objects of $\mathcal{A}$.

Proposition 2. Any locally finitely presentable category $\mathcal{C}$ contains an injective cogenerator $I$.

Proof. By the small object argument, it's easy to construct an object $I$ which is injective with respect to monomorphisms between small objects. Then an induction on the rank of objects, similar to the one here, shows that $I$ is actually injective with respect to all monomorphisms. In the abelian case, this is also an old theorem of Grothendieck.

Proposition 3. A cocomplete abelian category $\mathcal{A}$ with a compact projective generator $P$ is equivalent to the category of $\mathrm{End}(P)$-modules.

Proof. The functor $\mathcal{A}(P,-): \mathcal{A} \to \mathrm{End}(P)\mathrm{-Mod}$ preserves all colimits and is fully faithful on the object $P$. One shows that the subcategory of $\mathcal{A}$ on which the functor is fully faithful is closed under colimits, so is all of $\mathcal{A}$. So this functor embeds $\mathcal{A}$ as a full subcategory of $\mathrm{End}(P)\mathrm{-Mod}$ closed under colimits and containing $\mathrm{End}(P)$, which thus is all of $\mathrm{End}(P)\mathrm{-Mod}$.

Theorem. If $\mathcal{A}$ is a small abelian category, then there is a fully faithful, exact embedding of $\mathcal{A}$ into a category of modules over a ring.

Proof. First embed $\mathcal{A}$ into $\mathrm{Pro}(\mathcal{A}) = \mathrm{Ind}(\mathcal{A}^\mathrm{op})^\mathrm{op}$; by Proposition 1 this category is abelian and the embedding is fully faithful and exact. It is also the opposite of the locally (finitely) presentable category $\mathrm{Ind}(\mathcal{A}^\mathrm{op})$, so it contains an projective generator $P$ by Proposition 2. Let $\mathcal{B} \subset \mathrm{Pro}(\mathcal{A})$ be closure of $\mathcal{A} \cup \{P\}$ under finite limits and colimits; this is a full abelian subcategory, and the inclusion $\mathcal{A} \to \mathcal{B}$ is fully faithful and exact. By Proposition 1, so is the inclusion $\mathcal{B} \to \mathrm{Ind}(\mathcal{B})$, and moreover, $P$ is a compact projective generator in $\mathrm{Ind}(\mathcal{B})$. So by Proposition 3, the emedding $\mathcal{A} \to \mathrm{Ind}(\mathcal{B})$ is the desired embedding.

  • 1
    $\begingroup$ Conclusion: you don't need to learn what a "weakly effaceable functor" is to prove the embedding theorem! $\endgroup$ – tcamps Dec 19 '16 at 20:01
  • $\begingroup$ Proposition 2 should probably include the hypothesis that for sufficiently large $\kappa$, the $\kappa$-small objects are closed under strong epimorphic images. But the full generality of this statement is not needed for the abelian case. $\endgroup$ – tcamps Dec 27 '17 at 5:35
  • 2
    $\begingroup$ If you want to do the embedding theorem without worrying about actual abelian category concepts, you can just do it for Barr exact categories, which are the strongest non-additive generalization. You get the same theorem. $\endgroup$ – Kevin Carlson Feb 4 '18 at 21:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.