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?

  • $\begingroup$ (I'm particularly interested in the case where $\mathcal{C}$ is the $(\infty, 1)$-category of $E_\infty$-ring spectra.) $\endgroup$ Aug 17 '13 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... $\endgroup$ Aug 17 '13 at 9:02

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


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.