4
$\begingroup$

We know that, in a combinatorial simplicial model category $\mathbf{M}$, we can find a regular cardinal $\lambda$ large enough that all $\lambda$-filtered homotopy colimits can be computed as strict colimits. I'm wondering if it is possible to transfer this result to a given (full) subcategory $\mathbf{A} \subseteq \mathbf{M}$, i.e.

is it true that $\lambda$-filtered homotopy colimits (in the enriched sense) can be computed as strict colimits in $\mathbf{A}$, whenever they exist? Possibly enlarging $\lambda$, closing $\mathbf{A}$ under weak equivalences, or assuming that all objects of $\mathbf{A}$ are (co)fibrant.

Improve this question
$\endgroup$
4
  • $\begingroup$ Do you want to assume Vopenka? I believe that under VP, "the same proof" as Adamek and Rosicky, Thm 6.9 (performed in a presheaf $\infty$-category rather than the category of graphs), shows that every full subcategory inclusion into a presentable $\infty$-category preserves $\lambda$-filtered colimits for some $\lambda$. Then you can translate this into a model-categorical statement with enough co/fibrancy hypotheses. $\endgroup$
    – Tim Campion
    Commented Aug 10, 2022 at 15:38
  • $\begingroup$ I'm not so sure that that proof generalizes to $\infty$-categories, not in the straightforward way at least. However, I have an alternative proof that works up to my question above, so in the end you may use Vopenka if you want, but also I'd like to know if that holds in general. $\endgroup$
    – Giulio Lo Monaco
    Commented Aug 11, 2022 at 8:21
  • $\begingroup$ What would you like to assume about 𝐀? Could it be reflective? $\endgroup$
    – Boris Chorny
    Commented Aug 12, 2022 at 18:29
  • 1
    $\begingroup$ That would be too strong for me, unfortunately. I can assume that $\mathbf{A}$ is precisely the subcategory of $\mathbf{M}$ inducing a given full subcategory $\mathcal{C} \subseteq N_{\Delta}(\mathbf{M}^{\circ})$, but this in turn should be a general subcategory $\endgroup$
    – Giulio Lo Monaco
    Commented Aug 15, 2022 at 9:12

0

Reset to default

You must log in to answer this question.