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.


2 Answers 2

  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.

  • $\begingroup$ Re: 1, what happens in the proof of Freyd-Mitchell if you replace $[A, \text{Ab}]$ with the small presheaves on $A^{op}$? $\endgroup$ Dec 6, 2020 at 21:33
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    $\begingroup$ @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 $\endgroup$ Dec 6, 2020 at 21:35
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    $\begingroup$ @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 $\endgroup$ Dec 6, 2020 at 21:35
  • $\begingroup$ 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. $\endgroup$ Dec 6, 2020 at 21:36
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    $\begingroup$ @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 $\endgroup$ Dec 6, 2020 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 $\text{Ord} \to \text{Ab}$.

  • $\begingroup$ Yes, exactly that came to my mind as well and I liked Rickard's example because it's down to earth in a sense. $\endgroup$
    – user170048
    Dec 7, 2020 at 7:08
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    $\begingroup$ Is there an example of a locally small well-powered abelian category that does not admit an exact embedding into a module category? $\endgroup$
    – user78294
    Dec 13, 2020 at 13:31

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