I came to the following question when thinking about the (infinitely generated) tilting-cotilting correspondence, where it appears to be relevant.

Does there exist a locally presentable abelian category with enough injective objects that is not a Grothendieck category?

Background: an abelian category is called Grothendieck if it has a generator and satisfies the axiom Ab5. The latter means that (the category is cocomplete and) the filtered colimit functors are exact. Any locally presentable category has a generator, so the question is really about existence of locally presentable abelian categories with enough injective objects, but nonexact filtered colimits.

An abelian category is said to satisfy Ab3 if it is cocomplete. Any locally presentable category is cocomplete by definition. An abelian category is said to satisfy Ab4 if it is cocomplete and the coproduct functors are exact. Any cocomplete abelian category with enough injective objects satisfies Ab4.

The category opposite to the category of vector spaces over some fixed field is cocomplete and has enough injective objects, but it does not satisfy Ab5 (because the category of vector spaces does not satisfy Ab5*, i.e., does not have exact filtered limits). The category opposite to the category of vector spaces is not locally presentable, though.

Any locally finitely presentable abelian category is Grothendieck, and any Grothendieck abelian category is locally presentable, but the converse implications do not hold. Any Grothendieck abelian category has enough injective objects. Does a locally presentable abelian category with enough injective objects need to be Grothendieck?

EDIT: In response to one of the comments, let me add that the thematic example of a locally presentable abelian category that is not locally finitely presentable and, moreover, not Grothendieck is the category of Ext-$p$-complete (weakly $p$-complete) abelian groups, as mentioned in Martin Frankland's answer to this question -- What's an example of a locally presentable category "in nature" that's not $\aleph_0$-locally presentable? . This category is not Grothendieck because the colimit of the sequence of monomorphisms $\mathbb Z_p\overset{p}\longrightarrow \mathbb Z_p\overset{p}\longrightarrow\mathbb Z_p\overset{p} \longrightarrow\dotsb$ vanishes in it (hence filtered colimits do not preserve monomorphisms and do not commute with the kernels).

The Ext-$p$-complete abelian groups are otherwise known as "$p$-contramodule $\mathbb Z$-modules" or "contramodules over the topological ring of $p$-adic integers $\mathbb Z_p$". Generally, various contramodule categories provide examples of locally presentable (often, but not always, locally $\aleph_1$-presentable) abelian categories which are not locally finitely presentable, not Grothendieck, and, typically, have enough projective objects but no nonzero injectives.

Let me cite our paper with Jiří Rosický "Covers, envelopes, and cotorsion theories in locally presentable abelian categories and contramodule categories", Journ. of Algebra 483 (2017), arXiv version at https://arxiv.org/abs/1512.08119 , and my preprint "Abelian right perpendicular subcategories in module categories", https://arxiv.org/abs/1705.04960 , as references on locally presentable abelian categories. In particular, according to Example 4.1(4) in the former paper, there are no nonzero injective objects in the category of Ext-$p$-complete abelian groups.

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    $\begingroup$ @TheoJohnson-Freyd The category of Banach spaces is not abelian. $\endgroup$ – Jeremy Rickard Dec 28 '17 at 9:05
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    $\begingroup$ @IvanDiLiberti : I guess yes? 1.9 here ncatlab.org/nlab/show/injective+object#in_abelian_categories $\endgroup$ – Fosco Dec 28 '17 at 11:20
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    $\begingroup$ I don’t know if it’s helpful but Grothendieck categories are exactly the localizations of the category of additive presheaves over any full generating subcategory. On the other side locally presentable categories are reflective localizations of a category of presheaves. So a suggestion would be looking at this latter category of presheaves. $\endgroup$ – user40276 Dec 29 '17 at 10:53
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    $\begingroup$ Neeman's counterexample to Roos' theorem (or rather its dual) is locally presentable and AB4 but not AB5. I don't know if it has enough injectives, though. Also, Grothendieck shows (Tohoku p.137, Rmk 1) that if a category is AB4 and AB3* and has a cogenerator, then it has "injective effacements", a bit weaker than enough injectives. Roos shows ("Derived functors of inverse limits revisited") that if a category is AB4 and AB3* and has a cogenerator then filtered colimits of monos are exact. $\endgroup$ – Tim Campion Jan 20 '18 at 6:59
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    $\begingroup$ Also, one easily observes that if a category has a set of generators and enough injectives, then it has a set of injective cogenerators (embed every quotient object of every object in the set of generators into an injective object, and such injective objects will form a set of injective cogenerators). If the category is also Ab3*, it has a single injective cogenerator. $\endgroup$ – Leonid Positselski Jan 20 '18 at 11:54

Here is a an idea of example (I haven't checked all the details).

Let $\lambda < \kappa$ be two inaccessible cardinals and consider $\mathbf{Vect}_\lambda$ the category of $\kappa$-small vector spaces (over a given field). Then consider $$\mathcal C = \mathrm{Ind}_\lambda^\kappa\left(\mathbf{Vect}_\lambda^\mathrm{op}\right)$$ obtained by addding freely all $\kappa$-small and $\lambda$-filtered colimits.

Since $\mathbf{Vect}_\lambda$ has all $\lambda$-small limits, $\mathcal C$ has all $\kappa$-small colimits and is then $\lambda$-presentable. Moreover $\mathcal C$ is abelian and all exact sequences in $\mathcal C$ split, so all objects of $\mathcal C$ are injective.

The embedding $$\mathbf{Vect}_\lambda^\mathrm{op} \hookrightarrow \mathrm{Ind}_\lambda^\kappa\left(\mathbf{Vect}_\lambda^\mathrm{op}\right)$$ is exact and commutes with all $\lambda$-small colimits. Since $\mathbf{Vect}_\lambda^\mathrm{op}$ is not AB5, then $\mathcal C$ is not AB5.

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  • $\begingroup$ Why do you think that all exact sequences in $\mathcal C$ split? Generally speaking, if you have an abelian category $\mathcal A$ in which all exact sequences split, the category $\operatorname{Ind}(\mathcal A)$ may have nonsplit exact sequences. $\endgroup$ – Leonid Positselski Jan 14 '19 at 12:41
  • $\begingroup$ I spent a long time trying to show that this example worked about a year ago, with no luck. It's not hard to show that this category is abelian, locally presentable, with monomorphisms stable under transfinite composition, AB4, and not AB5. But I was unable to show that it had enough injectives (even under large cardinal hypotheses). If you like, I can undelete my answer so you can see what I tried (I say this with some trepidation because I have already deleted and undeleted my answer many times). $\endgroup$ – Tim Campion Jan 14 '19 at 21:55

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