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If $\mathcal C$ is a $\kappa$-accessible 1-category, then the category of morphisms $Mor \mathcal C$ is a $\kappa$-accessible 1-category, with the $\kappa$-presentable objects being those morphisms whose domains and codomains are each $\kappa$-presentable.

In the context of $\infty$-categories, the best result I know of is HTT Proposition 5.4.4.3, which shows that if $\mathcal C$ is a $\kappa$-accessible $\infty$-category and $\kappa \ll \tau$ (meaning that $\lambda < \tau \Rightarrow \kappa^\lambda < \tau$ and $\kappa < \tau$), then $Mor \mathcal C$ is $\tau$-accessible.

Lurie's proof, via HTT Lemma 5.4.4.2 (note that this lemma's proof has been revised since the printed edition), seems to really use the full strength of the assumption $\kappa \ll \tau$. Can this be improved to $\kappa = \tau$? Or at least to the "sharply below" relation $\kappa \triangleleft \tau$ familiar from the theory of accessible 1-categories?

This boils down to asking: if $\mathcal C$ is $\kappa$-accessible, then is every morphism of $\mathcal C$ a levelwise $\kappa$-filtered colimit of morphisms between $\kappa$-presentable objects?

In the case of 1-categories, a follow-your-nose argument works: you just take colimiting diagrams for the domains and codomains and factor the original map through stages of the colimit. I suspect that the same must be true in $\infty$-categories, with the same argument in principle working. But the question seems to be much more subtle $\infty$-categorically.

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    $\begingroup$ By the way -- Lurie systematically uses the relation $\kappa \ll \tau$ in place of the relation $\kappa \triangleleft \tau$ familiar from accessible 1-categories. I don't know whether this is the only place this matters, but I think it does here (in Lemma 5.4.4.2's construction of the $K(\alpha)$'s) $\endgroup$ – Tim Campion Aug 30 at 3:37
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    $\begingroup$ $Mor(\mathcal C)$ is also $\kappa$-accessible by HTT Prop. 5.3.5.15. $\endgroup$ – Marc Hoyois Aug 30 at 8:07
  • $\begingroup$ @MarcHoyois Thanks -- just a few pages from where I was looking, too! $\endgroup$ – Tim Campion Aug 30 at 12:20
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Marc Hoyois answered in the comments: the answer is affirmative, by HTT 5.3.5.15.

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