Let $\mathsf{M}$ be a simplicial model category presenting an $\infty$-category $\mathcal{M}$. I'm interested in a general statement relating compact objects in $\mathcal{M}$ (in the $\infty$-categorical sense) with the compact objects in $\mathsf{M}$. Here's roughly what I expect to be true but if i'm missing some assumptions or if some are redundant feel free to phrase the correct statement as an answer.

Suppose further that $\mathsf{M}$ satisfies that weak equivalences between fibrant objects are stable under filtered colimits. Then is the following true

  1. Let $X$ be a compact cofibrant object in $\mathsf{M}$. Is $X$ compact as an object in $\mathcal{M}$?

  2. Is every compact object in $\mathcal{M}$ a retract of (the image in $\mathcal{M}$) of some compact object in $\mathsf{M}$?

  • 1
    $\begingroup$ Here's a related question in the stable setting: mathoverflow.net/questions/289950/… $\endgroup$ – David White May 31 '19 at 12:27
  • 1
    $\begingroup$ I suppose your boldface condition rules out $Top$ as an example, but it's worth recalling that in $Top$, the only compact objects are the finite discrete spaces. $\endgroup$ – Tim Campion Jun 3 '19 at 3:06
  • $\begingroup$ @TimCampion Oh right! Seems like in $Top$ what makes the statement work is that finite CW complexes look compact when mapping into a filtered diagram of cofibrations. That might certainly complicate the general statement I was hoping for :( $\endgroup$ – Saal Hardali Jun 3 '19 at 10:47

If $X$ is such that $X \times \Delta^n$ is compact for every $n$ then yes. This happens, for example, if the cotensor functor $(-)^{\Delta^n}$ preserves filtered colimits, a condition which is quite common. Sufficient conditions of a similar nature are described in Proposition 5.3.1 of this paper. That said, you should expect the answer to your question to be negative in general (though I don't have a counter-example of the top of my head, at least not with $X$ cofibrant).

| cite | improve this answer | |
  • $\begingroup$ That's cool! Just to make sure I understood correctly the linked proposition. In particular the Joyal model structure on simplicial sets satisfies the conditions of 5.3.1. as well, is this correct? $\endgroup$ – Saal Hardali Jun 18 '19 at 8:05
  • $\begingroup$ Yes, though verifying the assumption there is not completely obvious. You need to check that any map $f: X \to Y$ of finite simplicial sets can be factored as a cofibration $f: X \to Z$ with $Z$ finite, followed by a categorical equivalence $Z \to Y$. One way to do this is to choose a finite simplicial set $I$ which is categorically equivalent to $\Delta^0$ and which contains a non-degenerate edge $\rho:\Delta^1 \hookrightarrow I$ (e.g., take $I=\Delta^0\coprod_{\Delta^{\{1,3\}}}\Delta^3\coprod_{\Delta^{\{0,2\}}} \Delta^0 $) and set $Z = X \times I \coprod_{X \times \{\rho(1)\}} Y$. $\endgroup$ – Yonatan Harpaz Jun 18 '19 at 11:53

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.