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Let $\mathcal{C}$ be a compactly generated presentable $(\infty, 1)$-category. Consider the functor $$ \Phi: \mathcal{C} \to \mathrm{Cat}_\infty, \quad x \mapsto (\mathcal{C}_{x/})^\omega,$$ that sends an object $x \in \mathcal{C}$ to the $(\infty, 1)$-category of compact objects in the undercategory $(\mathcal{C}_{x/})$. I am interested in when this functor commutes with filtered colimits.

I believe that if $\{x_i\}_{i \in I}$ is a filtered diagram of objects in $\mathcal{C}$, then the induced functor $$\varinjlim \Phi(x_i) \to \Phi( \varinjlim x_i),$$ is (for formal reasons) always fully faithful, and that the image contains those objects of $\mathcal{C}_{\varinjlim x_i/}$ which are both compact and $n$-cotruncated. However, I do not know if this functor is essentially surjective in general: the problem is that the filtered colimit of idempotent complete categories need not be idempotent complete (I think?); the "image" of an idempotent is not a finite colimit.

For ordinary categories, the above displayed functor is an equivalence of categories (for instance, if $\mathcal{C}$ consists of commutative rings, and compact objects correspond to finitely presented algebras); I'm wondering if something goes wrong with $(\infty, 1)$-categories?

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  • $\begingroup$ (I'm particularly interested in the case where $\mathcal{C}$ is the $(\infty, 1)$-category of $E_\infty$-ring spectra.) $\endgroup$
    – Akhil Mathew
    Commented Aug 17, 2013 at 2:55
  • $\begingroup$ I think this is true for $\kappa$-filtered colimits where $\kappa>\omega$... but it's definitely not obvious (to me!) what to do when $\kappa = \omega$. As you said, it seems like Idem is not $\omega$-compact in $\text{Cat}_{\infty}$ which is causing issues... $\endgroup$
    – Dylan Wilson
    Commented Aug 17, 2013 at 9:02

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In fact, a filtered colimit of idempotent complete $\infty$-categories is idempotent complete. See Lemma 7.3.5.16 of Higher Algebra (in the 2014 edition). This gives an affirmative answer to the above question.

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  • $\begingroup$ I looked in the latest version of Higher Algebra and it doesn't seem to have such a lemma and the 2014 edition isn't available. Do you know another reference for this result? $\endgroup$
    – user39598
    Commented Sep 23 at 12:06

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