The Freyd-Mitchell embedding theorem is a very useful tool for dealing with small abelian categories. However, it does not allow to use "elements" of objects of an abelian category $A$ in those statements that involve "infinite constructions".

So I wonder: for which Grothendieck abelian $A$ (this certainly implies that $A$ is not small if it is non-zero) there exists an exact conservative functor $F$ into abelian groups (this is a certain weak substitute of "having elements")? Does the Gabriel-Popescu theorem help here (so, what can one say if $A$ is described as a "nice" localization of certain category of modules)?

Also, how would you call a functor $F$ possessing this property; does "a stalk functor" sound fine? What is the relation of the existence of $F$ condition to the existence of compact generators for $D(A)$? If $A$ is a category of sheaves for certain Grothendieck topology then can one relate $F$ to the points of this topology?

Any hints, references or examples are very welcome!

  • $\begingroup$ What I know is that there is an analog of Freyd-Mitchell for elementary toposes which does not require any smallness. It is called the Barr cover I think, you embed your topos into double negation sheaves on the (closed) complement of the element "true" in the subobject classifier. $\endgroup$ Dec 6 '16 at 11:23
  • $\begingroup$ If the idea is to reason about "infinite constructions" as though in a category of modules, then why don't you require your embedding to preserve certain (co)limits? $\endgroup$
    – Tim Campion
    Feb 26 '18 at 9:18
  • $\begingroup$ Yes, this is what I want to be true.:) So, which results in this direction are known? $\endgroup$ Feb 27 '18 at 15:09

I am not an expert of the abelian world, but I think I can answer.

Since my background is not precisely abelian, I will start with an example in category theory.

Thm. Let $\mathcal{K}$ be a locally $\kappa$-presentable category, then there is a faithful and conservative functor to Set that preserves $\kappa$-directed colimits.

Proof. Since $\mathcal{K}$ is a locally $\kappa$-presentable category, it has a strong generator $\mathcal{G}$ made by $\kappa$-presentable objects. Then, the functor $\coprod_{G \in \mathcal{G}} \text{hom}_{\mathcal{K}}(G, \_)$ is faithful and conservative. Moreover it preserves $\kappa$-directed colimits and connected limits.

It looks to me that with the same argument one can prove the following statement.

Thm. Let $\mathcal{A}$ be an abelian category with a strong projective generator, then it has an exact, faithful and conservative functor into $\mathbb{Ab}$. Moreover, if the generator is made by $\kappa$-presentable objects, the functor preserves $\kappa$-directed colimits.

Observe that the functor is left exact because $\mathbb{Ab}$ is AB4.

Moreover, the following holds.

Thm. Let $\mathcal{A}$ be a Grothendieck category with a faithful and conservative functor in $\mathbb{Ab}$ which is accessible, exact and preserve connected limits. Then $\mathcal{A}$ has a projective strong generator.

Proof. Call $F$ such a functor, then it must have a multiadjoint. This is equivalent to the fact that $F \cong \coprod_{G \in \mathcal{G}} \text{hom}_{\mathcal{A}}(G, \_)$ for a small family $\mathcal{G}$. Since $F$ is faithful and conservative, $\mathcal{G}$ must be a strong generator.

  • 1
    $\begingroup$ The main shortcoming, I think, is that this functor will typically not be right exact. $\endgroup$
    – Tim Campion
    Sep 27 '18 at 22:04
  • $\begingroup$ I edited, does it help? $\endgroup$ Sep 27 '18 at 22:10
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    $\begingroup$ Sure. I suppose that's the best one can hope for. $\endgroup$
    – Tim Campion
    Sep 27 '18 at 22:12
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    $\begingroup$ $F \cong \text{hom}_{\mathbb{Ab}}(\mathbb{Z}, F) \cong \coprod_{G \in \mathcal{G}(\mathbb{Z})} \text{hom}_{\mathcal{A}}(G, \_)$ $\endgroup$ Sep 28 '18 at 7:10
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    $\begingroup$ If there are enough points (and this is certainly the case for any sort of "ordinary" sheaves) then the direct sum of stalks is an exact conservative functor into abelian groups; yet this functor is only ind-corepresentable. $\endgroup$ Sep 28 '18 at 17:45

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