Let $Ab$ be the category of abelian groups, and let $F: Ab \to Ab$ be a covariant functor which is left-exact and limit-preserving. Is $F$ necessarily naturally equivalent to a functor of the form $\mathrm{Hom}(A,-)$ for some $A\in Ab$?
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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 WilsonMar 23, 2016 at 18:51
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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 WilsonMar 23, 2016 at 18:53
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$\begingroup$ I guess it's not so obvious since the result is maybe true for Set... $\endgroup$– Dylan WilsonMar 23, 2016 at 18:57
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2$\begingroup$ Limit-preserving implies left exact. $\endgroup$– Qiaochu YuanMar 23, 2016 at 18:57
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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 non-representable limit-preserving functor in the nlab link of yours. $\endgroup$– მამუკა ჯიბლაძეMar 23, 2016 at 19:43
1 Answer
The category of abelian groups is small-complete, well-powered, and has a cogenerator (e.g., $\mathbb{Q}/\mathbb{Z}$). It follows from the Special Adjoint Functor Theorem that any limit-preserving 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 $\kappa$.
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$\begingroup$ 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... $\endgroup$ Mar 24, 2016 at 0:00
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$\begingroup$ @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 limit-preserving 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, 2016 at 0:30
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1$\begingroup$ Sorry I wasn't very clear. My question is: suppose C has the property that it is presentable AND every limit-preserving functor out of C is accessible. Is it the case that C is cototal? $\endgroup$ Mar 24, 2016 at 1:46
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$\begingroup$ (But thank you for your comments too!) $\endgroup$ Mar 24, 2016 at 1:46
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1$\begingroup$ Note that it is a particular case of a theorem from homological algebra known as Watts theorem: every limit-preserving functor from the category of left $R$-modules to abelian groups is representable. $\endgroup$ Mar 24, 2016 at 9:14