3
$\begingroup$

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 $A\in Ab$?

$\endgroup$
  • 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
    $\begingroup$ Limit-preserving 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 non-representable limit-preserving functor in the nlab link of yours. $\endgroup$ – მამუკა ჯიბლაძე Mar 23 '16 at 19:43
9
$\begingroup$

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 $\kappa$.

$\endgroup$
  • $\begingroup$ 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 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 '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 limit-preserving functor out of C is accessible. Is it the case that C is cototal? $\endgroup$ – Dylan Wilson Mar 24 '16 at 1:46
  • $\begingroup$ (But thank you for your comments too!) $\endgroup$ – 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 limit-preserving functor from the category of left $R$-modules to abelian groups is representable. $\endgroup$ – Philippe Gaucher Mar 24 '16 at 9:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.