15
$\begingroup$

For a model category $C$, I'm denoting $h_\infty(C)$ the associated $\infty$-category (for example its Dwyer-Kan localization at weak equivalences, or if $C$ is simplicial the simplicial nerve of the category of bifibrant objects, or any other equivalent construction...).

If $C$ is a combinatorial model structure and $I$ is any small category, then the category $C^I$ carries projectives and injectives model structures ( that are clearly equivalent).

It then happens that (in this case) we have an equivalence $h_\infty(C^I) \simeq h_\infty(C)^I$ using one of these model structures on $C^I$.

So informally this model structure on $C^I$ does model all homotopy coherent I-diagram in $C$. I feel this is used in many places, but I never got a good understanding of that result. It seems like a very non-trivial "rectification" theorem (where you turn homotopy coherent diagram $ I \to C$ into actual functors). For example if I replace $C$ by a general relative category this is obviously not the case. I know proofs of this fact, but they are all fairly indirect, with some technical steps, and I feel they don't really explain why such a result is true, or at least I don't get the explanation.

So, my question : can someone give some sort of intuition of why this is true or maybe a more "high level" proof that this is true ? Maybe someone just knows a simpler/more direct proof than the ones I may have seen ?

Basically, I'd be interested in hearing any good answer to the question why is it the case that $h_\infty(C^I) \simeq h_\infty(C)^I$ ?

Maybe a more precise way to ask this : What I'd be especially interested in is an explanation that would give some intuition of when more general statement of this kind are true. For e.g, I have little intuition on the following questions :

  • Is it true if $C$ is a model category, that is not combinatorial, but for which I do have a satisfying model structure on $C^I$ ? Is there some condition I can add to make it true ?

  • what if $C$ is only a fibration category or a cofibration category and $I$ is nice enough ? ( Okay here the answer is clearly No, so, I'm refining it in the next point)

  • What if $C$ is a cofibration or fibration category with some additional conditions on infinite limits / colimits that among other things guarantee that $h_\infty(C)$ has infinite limits/colimits ?

  • What kind of conditions on a general relative category $C$ might make this true ?

$\endgroup$
3
  • $\begingroup$ Is it reasonable to ask in general for what Dwyer–Kan localization $\mathcal C\to\mathcal C[W^{-1}]$ of $\infty$-categories does this hold? $\endgroup$
    – Z. M
    Commented Oct 28, 2021 at 6:20
  • $\begingroup$ @Z.M Why would we not? $\endgroup$ Commented Oct 28, 2021 at 6:30
  • $\begingroup$ @Z.M A fully general of course not. I'm only wondering if there are nice sufficient condition that make it true. ( Something like "being a complete fibration category", but maybe a little more general) $\endgroup$ Commented Oct 28, 2021 at 13:09

1 Answer 1

11
$\begingroup$

If we look at the proof that I know of in the case this is true - see below - we need in fact that, for any small $1$-category $I$, $h_\infty(C^I)$ has small (co)limits and that they can be computed termwise, and we need to prove the particular case where $I$ is discrete as a first step, which usually comes from a variant of calculus of fractions ($C$ could be an $\infty$-category to begin with). The proof can be formulated in a model free way once we have a good theory of pointwise Kan extensions.

Anyway, this is true for any model category (in Quillen's original sense) $C$ which has either small colimits or small limits. For a category with fibrations and weak equivalences $C$ (such as a category of fibrant objects à la Brown), and if small products of (trivial) fibrations between fibrant objects exist and are (trivial) fibrations (this already implies that $h_\infty(C)$ has small limits), this is true if one of the following further conditions is satisfied:

  • limits of countable towers of (trivial) fibrations between fibrant objects are (trivial) fibrations;
  • the factorization of any map with fibrant codomain into a weak equivalence followed by a fibration can be made functorially.

In fact the last two conditions are only there to ensure that $C^I$ is a category with weak equivalences and fibrations defined level-wise (the tricky part being the existence of factorizations if ever these were not functorial in $C$ to begin with, but this is documented in the work of Radulescu-Banu). The principle of all the proofs I know of is that one proves that both $\infty$-categories $h_\infty(C^I)$ and $h_\infty(C)^I$ have small limits (it sufficices to prove this is true for $h_\infty(C)$ for a generic $C$), and to prove that the comparison functor $h_\infty(C^I)\to h_\infty(C)^I$ commutes with small limits (=with small products and with pullbacks). One can then check fully faithfulness on objects which are homotopy right Kan extensions of objects of $C$ along an object of $I$ (seen as a functor from the terminal category), and prove through cofinality arguments that any $I$-indexed functor is an homotopy limit of such thing to finish the proof. This kind of approach survives very well if we allow $C$ to be an $\infty$-category to begin with (but $I$ must be a $1$-category), so that any reasonnable notion of model $\infty$-category (e.g. Mazel-Gee's) is eligible for such rectification theorem. The case where $I$ is a finite direct category holds with a much greater generality (no need to have small homotopy (co)limits, finite ones are good enough), but the idea of the proof is similar (with simpler technical arguments, though). References can be found in my book on higher categories: Theorem 7.9.8 (with Example 7.9.6 and Remark 7.9.7. to get easy sufficient conditions in practice) if $I$ is small and Theorem 7.6.17 if $I$ is a direct finite category. It is striking to me that this kind of rectification is the best explanation of why homotopy limits in $C$ coincide with limits in $h_\infty(C)$; see Remark 7.9.10. In the same direction, there is Balzin's work which is more general because it deals with sections of general fibred model structures but also less general because it restricts to the case where the indexing small category is Reedy. The case of a cofibration categories is also documented in this paper of Tobias Lenz with a different proof from the one refered to above (he does this in the language of derivators, but it is easy to extract an $\infty$-categorical statement from his proof).

$\endgroup$
4
  • $\begingroup$ That's very interesting. The proofs I knew looked fairly different. In genral it is based on the idea to start with a class of categories $I$ such that the theorem can be proved by "induction" on $I$ for e.g. when $I$ is a direct or Reedy category or a Bergner-cofibrant simplicial category (in this case we can even avoid the induction and rely on the Bergner model structure, assuming C is simplicial) and then extend to a general I by showing that if $I \simeq J$ in an appropriate sense (e.g. in the Bergner model structure, or as relative categories) $\endgroup$ Commented Oct 28, 2021 at 13:16
  • $\begingroup$ ... Then $C^I$ and $C^J$ are Quillen equivalent. $\endgroup$ Commented Oct 28, 2021 at 13:26
  • $\begingroup$ Regarding Balzin's work there is also a more recent work by Harpaz (arxiv.org/abs/1902.04867) That proves the same result for a general base category, but at the cost of stronger assumption on the fibration ( each fiber needs to a simplicial combinatorial model structure and each transition a simplicial Quillen adjunction) $\endgroup$ Commented Oct 28, 2021 at 13:27
  • $\begingroup$ @SimonHenry There is still an induction hidden in the proof I sketch above: I use that commuting with small limits is the same as commuting with small sums and with pullbacks, while proving this equivalence consists in proving that any small $\infty$-category is a localization of a direct category $J$ (and that any localization is both final and cofinal) and then to do an induction on $J$. $\endgroup$ Commented Oct 30, 2021 at 12:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .