Let $Ab$ be the category of abelian groups, and let $F: Ab \to Ab$ be a covariant functor which is leftexact and limitpreserving. Is $F$ necessarily naturally equivalent to a functor of the form $\mathrm{Hom}(A,)$ for some $A\in Ab$?

2$\begingroup$ If $F$ also preserves $\kappa$filtered colimits for some cardinal $\kappa$ the answer is `yes'. In this case, $F$ admits a left adjoint $G$ and the object in question is $G(\mathbb{Z})$. $\endgroup$ – Dylan Wilson Mar 23 '16 at 18:51

2$\begingroup$ and since every corepresentable functor has this property, your question can be translated into: "Do there exist limit preserving functors from Ab to Ab which don't preserve filtered colimits?" seems like the answer should be yes but I'll have to think of a counterexample... $\endgroup$ – Dylan Wilson Mar 23 '16 at 18:53

$\begingroup$ I guess it's not so obvious since the result is maybe true for Set... $\endgroup$ – Dylan Wilson Mar 23 '16 at 18:57

2$\begingroup$ Limitpreserving implies left exact. $\endgroup$ – Qiaochu Yuan Mar 23 '16 at 18:57

1$\begingroup$ @QiaochuYuan As can be seen from the answer below, this is closely related to the (non)existence of a cogenerator. Arbitrarily large simple groups are used in the proof that nonabelian groups do not have a cogenerator (just looked up this proof (1.64 on p.104) in Manes' "Algebraic Theories"). And the same are used to construct a nonrepresentable limitpreserving functor in the nlab link of yours. $\endgroup$ – მამუკა ჯიბლაძე Mar 23 '16 at 19:43
The category of abelian groups is smallcomplete, wellpowered, and has a cogenerator (e.g., $\mathbb{Q}/\mathbb{Z}$). It follows from the Special Adjoint Functor Theorem that any limitpreserving functor $G: Ab \to Ab$ has a left adjoint $F$. (A proof of the SAFT may be found on this nLab page.) And as Dylan Wilson pointed out in a comment, we then have $G \cong \text{Hom}(\mathbb{Z}, G) \cong \text{Hom}(F(\mathbb{Z}), )$, so $G$ is representable.
Referring to another comment by Dylan: all such functors are necessarily accessible, since any abelian group is $\kappa$presentable for some $\kappa$.

$\begingroup$ It's good to know there isn't a counterexample after all. Is there a nice equivalent condition on a, say presentable category, for every continuous functor out being accessible? I guess I want the target to be nice as well. It feels like this is equivalent to being cototal or something... $\endgroup$ – Dylan Wilson Mar 24 '16 at 0:00

$\begingroup$ @DylanWilson Cototality was actually the first word that popped into my mind when I read the question; it was just a matter of recalling why $Ab$ was cototal. As you may know, a functor from a cototal category to a locally small category has a left adjoint iff it preserves all small limits. Locally presentable categories need not be cototal (as witnessed by $Grp$, as Qiaochu was saying), but any accessible limitpreserving functor between locally presentable categories has a left adjoint. Is this responsive to your question (not sure I parsed it right)? $\endgroup$ – Todd Trimble♦ Mar 24 '16 at 0:30

1$\begingroup$ Sorry I wasn't very clear. My question is: suppose C has the property that it is presentable AND every limitpreserving functor out of C is accessible. Is it the case that C is cototal? $\endgroup$ – Dylan Wilson Mar 24 '16 at 1:46


1$\begingroup$ Note that it is a particular case of a theorem from homological algebra known as Watts theorem: every limitpreserving functor from the category of left $R$modules to abelian groups is representable. $\endgroup$ – Philippe Gaucher Mar 24 '16 at 9:14