# Elementary questions on the Freyd-Mitchell embedding theorem

I have a few elementary questions about the Freyd-Mitchell embedding theorem which I can't see answered elsewhere here and which all arise from some confusions I ran into.

1. Can someone point out why (and where in the proof) exactly you need that the abelian category $$\mathcal{A}$$ is small? Is it to show that the functor category $$\mathsf{Fun}(\mathcal{A},\mathsf{Ab})$$ is Grothendieck?

2. Does the FH embedding theorem also hold if $$\mathcal{A}$$ is only essentially small? I would say so but have seen this nowhere written out. If $$\mathcal{A}$$ is essentially small, there is an equivalence $$F \colon \mathcal{A} \to \mathcal{A}'$$ into a small category $$\mathcal{A}'$$. The category $$\mathcal{A}'$$ is automatically abelian and $$F$$ is an exact equivalence. Composed with the FH embedding of $$\mathcal{A}'$$ we get an exact embedding of $$\mathcal{A}$$ into a module category. Where's the mistake?

3. This one really confuses me. I thought that an exact embedding $$F \colon \mathcal{A} \to R\text{-}\mathsf{Mod}$$ induces an isomorphism $$\mathrm{Sub}(X) \to \mathrm{Sub}(F(X))$$ between the lattices of subobjects of $$X$$ and $$F(X)$$, therefore maps simple objects to simple modules etc., and I can deduce the Jordan–Hölder theorem for a finite length object in $$\mathcal{A}$$ simply from the one for modules. But the comments here seem to indicate that this is wrong. Why is it so? I must be stupid here.

1. You need this to know that $$Fun(A,Ab)$$ is locally small; and also indeed to check that $$Fun(A,Ab)$$ has nice properties such as Grothendieck-ness, or the existence of injective cogenerators etc. In other words it's needed for most of the steps that lead to the reduction to the case of a small subcategory of a nice Grothendieck abelian category.

2. Yes of course it works, and your argument shows that. I guess it's never written out because category theorists don't really care about the difference between small and essentially small. I would be surprised though if it were actually never mentioned.

3. It doesn't induce an isomorphism, only an embedding: given a subobject $$Y\to F(X)$$, how do you cook up a $$Z$$ such that $$F(Z) \cong Y$$ ? For instance look at the exact embedding $$\mathbb Q-Mod\to \mathbb Z-Mod$$, it certainly does not map simples to simples.

• Re: 1, what happens in the proof of Freyd-Mitchell if you replace $[A, \text{Ab}]$ with the small presheaves on $A^{op}$? – Qiaochu Yuan Dec 6 '20 at 21:33
• @QiaochuYuan : I guess it depends on what you mean by small presheaves. $A$ might be big but not accessible, so maybe those don't behave appropriately. But it's a good question ! Certainly there'll still be issues with Grothendieck-ness since Grothendieck categories must be presentable, but small presheaves need not be as far as I can tell – Maxime Ramzi Dec 6 '20 at 21:35
• @QiaochuYuan : one big thing for which smallness is used is the creation of an injective cogenerator/projective generator, and this uses a direct sum indexed by $A$ or something of a similar size, so one would have to find a way to bypass that – Maxime Ramzi Dec 6 '20 at 21:35
• By small I just mean a small colimit of representables. And yes, that makes sense, I guess I wouldn't be surprised if there wasn't an injective cogenerator. – Qiaochu Yuan Dec 6 '20 at 21:36
• @QiaochuYuan : yeah I guess if you really really want to make everything work without the smallness assumption, the point where you will really be stuck is the point where you need a co/generator, the rest can probably be worked around by taking small presheaves and similar techniques – Maxime Ramzi Dec 6 '20 at 21:38

Here is an explicit counterexample when $$A$$ is not assumed to be small. Any abelian category admitting an exact (fully faithful) embedding into $$\text{Mod}(R)$$ must be well-powered, meaning every object must have a set of subobjects (since the same is true in $$\text{Mod}(R)$$ and an exact embedding induces an embedding on posets of subobjects, but not, as Maxime points out, an isomorphism). There are abelian categories that are not well-powered; you can see some examples in this MO question. I particularly like Jeremy Rickard's example of the (still locally small!) category of eventually constant functors $$\text{Ord} \to \text{Ab}$$.

• Yes, exactly that came to my mind as well and I liked Rickard's example because it's down to earth in a sense. – user170048 Dec 7 '20 at 7:08
• Is there an example of a locally small well-powered abelian category that does not admit an exact embedding into a module category? – user78294 Dec 13 '20 at 13:31