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We know that whenever we have a Bousfield localization between two simplicial model categories, this gives rise to a reflective subcategory in $\infty$-categories (or coreflective, depending on the direction of the Bousfield localization).

  1. I'm interested in what happens if the model categories at issues are not simplicial, or even in the intermediate case when the categories themselves are simplicial, but we don't know that the functors that comprise the Bousfield localization are simplicial. Does this still give rise to a reflection of $\infty$-categories? At least, when the model categories are combinatorial?

  2. Related question: we know by a result by Dugger that every combinatorial model category is Quillen equivalent to a combinatorial simplicial model category. Is this assignation functorial? Do functors and adjunctions between combinatorial model categories get promoted to simplicial functors and adjunctions?

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  • $\begingroup$ Another option is to avoid model categories and $\infty$-categories and do Bousfield localisation in the framework of derivators instead. There is some work of that type at arxiv.org/abs/1907.07801. $\endgroup$
    – Neil Strickland
    Commented Jun 15, 2021 at 9:01
  • $\begingroup$ What do you mean by Bousfield localisation? In my understanding, a left Bousfield localisation of a model category is a model structure on the same category with the same cofibrations and perhaps more weak equivalences. In particular the "underlying" adjunction is the identity adjunction, which is automatically enriched! $\endgroup$
    – Zhen Lin
    Commented Jun 15, 2021 at 9:38
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    $\begingroup$ Yeah, sorry, that was unclear from my phrasing, I actually meant something that is Quillen equivalent to a Bousfield localization, but where the Quillen equivalence is not necessarily simplicial. Anyway, the question remains still valid when the categories are not simplicial, or at least I fail to see a trivial reason why they would still give rise to a reflection of $\infty$-categories. $\endgroup$
    – Giulio Lo Monaco
    Commented Jun 15, 2021 at 10:24

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This is not a full answer, but too long for a comment.

Here's a relevant paper. Mazel-Gee proves the folklore claim that a Quillen adjunction induces an adjunction on underlying $\infty$-categories with very little assumption on the model categories involved.

I think it's proved (or mentioned) in the appendix (at least) that upon passing to homotopy categories, the co/unit is the same as the one in the derived adjunction introduced by Quillen, in particular, in the case of a Bousfield localization, the (co)unit is an equivalence, and so it is an equivalence also at the level of $\infty$-categories ($C\to ho(C)$ is conservative), and so it also induces a reflective subcategory.

In other words, I think you don't need properness or anything like that.

For your question 2, I'm not sure about functoriality per se, but Quillen adjunctions definitely induce Quillen adjunctions: that's because Dugger's universal homotopy theory satisfies some almost universal property.

You can have a look e.g. at propositions 2.3, 5.10 and theorem 6.3 in Dugger's paper : suppose $M\rightleftarrows N$ is a Quillen adjunction, then by 6.3 you can make it into a Quillen adjunction $UC/S \rightleftarrows N$ for some $C$ and $S$ and then with 5.10 you can lift it to $UC/S \rightleftarrows UD/T$

To ask whether it can be made simplicial is to ask whether 2.3 can be made simplicial, i.e. suppose given $\gamma : C\to M$ where $M$ is a simplicial model category, can the functor $Re: UC\to M$ from 2.3 be made simplicial ?

Now I'm not entirely sure the answer is yes, but I would guess that it is if $M$ is nice enough. At least, Lurie seems to indicate something similar in the last sentence of the proof of A.3.7.6. from Higher topos theory : "The proof given in [19] [ Dugger's paper ] shows that when $\mathbf A$ is a simplicial model category , then $F$ and $G$ can be chosen to be simplicial functors" - the context is not quite the same, so that is not claimed in HTT, but something similar is.

Hopefully someone can comment on whether this is the case (that's why this answer is not a complete answer)

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    $\begingroup$ The functor $U \mathcal{C} \to \mathcal{M}$ is defined by taking a cosimplicial resolution (= componentwise Reedy-cofibrant replacement) of $\gamma : \mathcal{C} \to \mathcal{M}$ and then applying unenriched left Kan extension. I remember reading somewhere – one of Dugger's papers, I'm sure – that this is not necessarily simplicially enriched. But if it happens that $\gamma (-) \otimes \Delta^\bullet$ is a cosimplicial resolution of $\gamma$ then I think you do get a simplicially enriched functor. $\endgroup$
    – Zhen Lin
    Commented Jun 15, 2021 at 11:24
  • $\begingroup$ @ZhenLin : ah, so my guess would be wrong ! I think it would be lovely if you (or someone else) could find that "somewhere" to correct my answer - but to complete it one would need a proof that there is no alternative definition that can make it simplicial (I must confess I haven't read Dugger's paper in detail so I don't know if there's a uniqueness statement that would forbid this) $\endgroup$
    – Maxime Ramzi
    Commented Jun 15, 2021 at 11:57

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