Elementary questions on the Freyd-Mitchell embedding theorem

Question

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.

Answer 1

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.

Answer 2

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 $\text{Ord} \to \text{Ab}$.