Let $F:\cal C\to D$ be an accessible functor between locally presentable categories. By Theorem 2.19 in Adamek-Rosicky Locally presentable and accessible categories, there exist arbitrarily large regular cardinals $\lambda$ such that $F$ preserves $\lambda$-presentable objects. It is tempting to expect that $F$ should preserve $\lambda$-presentable objects for all sufficiently large $\lambda$, but that is not what the theorem says. However, I do not know a counterexample showing that the stronger claim fails. (For instance, this question asks about this property when $F$ is the pullback functor, and has no answer yet in the general case.)

What is an example of an accessible functor $F$ between locally presentable categories for which there exist arbitrarily large regular cardinals $\mu$ such that $F$ does not preserve $\mu$-presentable objects?

  • $\begingroup$ For fixed $\alpha$, does $\mu^\alpha = \mu$ hold for all sufficiently large regular $\mu$? $\endgroup$ – Reid Barton Mar 13 '19 at 5:17
  • 1
    $\begingroup$ We may assume that $F$ preserves small $\lambda$-filtered colimits. Isn’t it true that, for $\mu$ large enough, an object is $\mu$-presentable if and only if it is a $\mu$-small $\lambda$-filtered colimit of $\lambda$-presentable objects? Another way to put it, is that for $\mu$ large enough (e.g. larger than $\lambda$ and than the set of maps between any two $\lambda$-presentable objects), the property of $\mu$-presentability of an object $X$ is simply the fact that the set of maps from a $\lambda$-presentable object to $X$ is of cardinal $\leq\mu$. $\endgroup$ – Denis-Charles Cisinski Mar 13 '19 at 7:26
  • 3
    $\begingroup$ @Denis-CharlesCisinski As far as I know that is only true if you either remove the $\lambda$-filteredness condition on the colimits (see Remark 1.30 in AR) or add the assumption that $\lambda\lhd\mu$ (which changes it from "for sufficiently large $\mu$" to "for arbitrarily large $\mu$" -- see Remark 2.15 in AR). $\endgroup$ – Mike Shulman Mar 13 '19 at 13:03

An example is given in my paper with Tibor Beke,

Abstract elementary classes and accessible categories, Annals Pure Appl. Logic 163 (2012), 2008-2017, doi:10.1016/j.apal.2012.06.003, arXiv:1005.2910.

see Remark 3.2(4). This is what Reid Barton indicated.

| cite | improve this answer | |
  • 1
    $\begingroup$ Ah, of course. For the non-set-theorist readers, can you give a quick reference or sketch of why $\mu \mapsto \mu^\alpha$ has arbitrarily large non-fixed-points for $\alpha$ infinite? $\endgroup$ – Mike Shulman Mar 13 '19 at 13:11
  • 1
    $\begingroup$ The question requires $\mu$ to be regular, but that example in Remark 3.2(4) doesn't assume $\mu$ is regular, right? It follows from GCH that $\mu^\alpha=\mu$ for all regular $\mu>\alpha.$ @MikeShulman: By a diagonalization argument, if $\mu$ has cofinality $\alpha$ then $\mu^\alpha>\mu.$ A reference is Jech's Set Theory, 3rd Ed, Theorem 3.11. $\endgroup$ – Dap Mar 13 '19 at 15:26
  • 2
    $\begingroup$ So I was slightly confused when I posted my comment and maybe caused more confusion. Let's switch notation and ask whether $\kappa^\alpha > \kappa$ for arbitrarily large $\kappa$. It doesn't matter whether $\kappa$ is regular, because if $\kappa$ is such that $\kappa^\alpha > \kappa$, then the accessible functor $FX = X^\alpha$ fails to preserve $\mu$-presentable objects for the regular cardinal $\mu = \kappa^+$. $\endgroup$ – Reid Barton Mar 13 '19 at 16:21

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.