Suppose that $\mathcal{M}$ is a model category which is not combinatorial, does a homotopy limit in $\mathcal{M}$ correspond to a limit in the associated $\left(\infty,1\right)$-category?

How about when $\mathcal{M}$ doesn't have enough limits and colimits, but is otherwise a model category?

How about when $\mathcal{M}$ is a category of fibrant objects?

If there are counter-examples, under what additional assumptions (besides being combinatorial) will this be true?

  • 1
    $\begingroup$ The theorem relating the two concepts in Lurie's book is very general, using only the concept of homotopy (co-)limits in Kan-complex enriched categories. See here ncatlab.org/nlab/show/limit+in+a+quasi-category#TermsOfHomotopy and here: ncatlab.org/nlab/show/homotopy+Kan+extension#InKanCplCat $\endgroup$ – Urs Schreiber Jul 14 '14 at 18:21
  • 1
    $\begingroup$ In other words, if you have a simplicial model category then you will be fine. But in general it would seem that more work is needed. $\endgroup$ – Zhen Lin Jul 14 '14 at 18:33
  • 1
    $\begingroup$ @Urs: I read the above links before asking this question in fact. I would still need to know that the homotopy limit in the model category (lets say for simplicity a homotopy pullback diagram, which I can arrange to be an ordinary pullback by fibrancy conditions) presents the homotopy limit in the Kan-complex enriched simplicial category associated to it (so, if it's not simplicial, I'd need to take some fibrant replacement of the Hammock localization...) $\endgroup$ – David Carchedi Jul 14 '14 at 19:27

Homotopy limits in any model category always coincide with limits in the associated $(\infty,1)$-category. To see this, you need to know the following (classical) facts:

1) given a cofibrant object $A$, the mapping space functor $Map(A,-)$ (constructed as in Hovey's book, say, using a Reedy cofibrant resolution of $A$ in the category of cosimplicial objects) is a right Quillen functor, and thus commutes with homotopy limits (up to a canonical weak equivalence);

2) furthermore, for any fibrant object $X$, the Kan complex $Map(A,X)$ canonically has the homotopy type of the space of maps from $A$ to $X$ in the simplicial localization of your model category (this due to Dwyer and Kan).

3) the localization in the sense of quasi-categories corresponds via the (homotopy coherent) nerve functor to the simplicial localization, so that you can reinterpret 2) in an appropriate way;

4) in the setting of $(\infty,1)$-categorie, the Yoneda embedding preserves limits.

Therefore, by the Yoneda lemma (I mean the very usual one, applied to the homotopy category of the model category you are dealing with), to prove the property you want, it is sufficient to consider the case of the usual model category structure on simplicial sets, which obviously falls in the setting considered by Lurie (but can also be easily deduced from earlier results of Dwyer and Kan).

[Edit] I realize I did not write anything about the case where $\mathcal M$ is not a bicomplete model category.

In the case of a model category with only finite limits, the same is true for finite homotopy limits (same proof). More generally, finite homotopy limits in categories of fibrants objects coincide with the corresponding finite limits in the associated $(\infty,1)$-category (which is always finitely complete). For a proof, we proceed as follows. It is sufficient to consider the case of a small category of fibrant objects (this is an easy exercise, and you don't need Grothendieck universes to do so). Then Remark 3.13 in my paper [Invariance de la $K$-théorie par équivalences dérivées, J. K-theory 6 (2010), 505-546] (available here) reduces the problem to the case of a combinatorial proper simplicial model category, in which case you can apply Lurie's result. For a general discussion about homotopy limits in categories of fibrant objects, there is my paper [Catégories dérivables, Bull. SMF 138 (2010), 317-393] (available here), or Radulescu-Banu's paper.

| cite | improve this answer | |

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.