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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?

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  • $\begingroup$ For fixed $\alpha$, does $\mu^\alpha = \mu$ hold for all sufficiently large regular $\mu$? $\endgroup$ – Reid Barton Mar 13 '19 at 5:17
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    $\begingroup$ 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
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    $\begingroup$ @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
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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.

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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