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The Bergner model structure on $sCat$ (simplicially enriched categories) has a "projective" flavor: fibrations are "levelwise" while cofibrations satisfy stringent relative "freeness" conditions.

This seems to be something inherent to modeling $\infty$-categories using strictly-associative composition. Similarly, Barwick and Kan's model structure on relative categories and Horel's model structure on simplicially-internal categories are induced projectively, and not very many objects are cofibrant.

Is this really a trade-off we're stuck with?

Question 1: Is there a model structure $M$ on $sCat$ other than the Bergner model structure such that the identity functor is a left Quillen equivalence $sCat_M \to sCat_{Bergner}$? More generally, is there a model $M$ of the homotopy theory of $\infty$-categories whose underlying category is $sCat$ which doesn't have fewer cofibrant objects than $sCat_{Bergner}$?

Question 2: Is there such an $M$ with a reasonable supply of cofibrant objects? Say, such that every ordinary category is cofibrant? Or perhaps at least such that every ordinary Reedy category is cofibrant?

Question 3: As above, but for relative categories or simplicially-internal categories?

Two data points:

  • If $C$ is a category and $M$ is a model category, then $M^C$ with levelwise weak equivalences has the correct homotopy theory. But from thinking about the Bergner model structure, you might think you'd have to consider the category of simplicial functors $C' \to M'$ where $C'$ is the standard resolution of $C$ and $M'$ is the simplicial localization of $M$. The fact that you don't have to do this hints at some alternate model category -type structure where $C$ is already cofibrant and $M$ is fibrant -- maybe this would be a model structure on relative categories.

  • A similar phenomenon happens with operads, if I recall correctly (which are also strict monoids for the Kelly tensor product on symmetric sequences) -- the usual model structure (Berger-Moerdijk, I believe) is much more stringent in its cofibrancy requirements than what is needed in practice.

  • As hinted at in the comments, it's a good idea to ask whether the existing "projective" model structures $M$ are left proper. For

    • Left properness of a model structure depends only on the weak equivalences (saying that the co-base change Quillen adjunctions between co-slice categories are Quillen equivalences).

    • A model structure with all objects cofibrant is left proper.

    Well, it turns out that the Bergner, Lack, Barwick-Kan, and Horel model structures are all left proper! (I'm not sure about the Berger-Moerdijk model structure.) So this leaves open the possibility of model structures with the same weak equivalences and all objects cofibrant.

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  • $\begingroup$ In your last remark (about operads), Are you just refering to the fact that the mere $\Sigma$-cofibrant operads already have a lot of the good properties of the actual cofibrant operads have, or do you know another model structure with more cofibrant objects ? $\endgroup$ – Simon Henry Jun 26 '18 at 20:50
  • $\begingroup$ I'm referring to the fact that $\Sigma$-cofibrant operads have good properties. I don't know another model structure where, say, the cofibrant objects are the $\Sigma$-cofibrant operads -- I'd love to learn of one, though! $\endgroup$ – Tim Campion Jun 26 '18 at 20:51
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    $\begingroup$ A simpler version of the same question would be whether the Lack model structure for 2-categories has an injective version. My instinct is no: your data points correspond to the fact that any pseudofunctor $C\to M$ is equivalent to a strict functor when $C$ is arbitrary while $M$ is sufficiently complete or cocomplete (as every model category is), but for a general 2-category $M$ I think there is no way to replace it by an equivalent 2-category with this property, in contrast to how $C$ can be replaced by a cofibrant one. I don't have a proof that it's impossible though; good question! $\endgroup$ – Mike Shulman Jun 27 '18 at 3:47
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    $\begingroup$ @MikeShulman : Good point. Though what you describe would corresponds to a model structure where every object is cofibrant, which I also don't believe is possible (for example I believe it is unknown whether Bergner model structure is left proper or not). But that does not exclude the existence of a model structure with more cofibrant objects, like the case of $\Sigma$-cofibrant operads. $\endgroup$ – Simon Henry Jun 27 '18 at 7:37
  • $\begingroup$ @SimonHenry The Bergner model structure is left proper: see HTT A.3.2.4. (I'm not sure if this is the original reference. Of course, Bergner showed right properness in her original paper.) $\endgroup$ – Tim Campion Jun 27 '18 at 13:25
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EDIT: As Simon Henry points out in the comments, I've been too cavalier with accessibility issues. But since cofibrant generation of the flat maps is the only condition missing in the almost theorem below, one can take any set of flat maps containing generators for the original cofibrations, and its cofibrant closure gives the class of cofibrations for a cofibrantly-generated model structure. In particular, if weak equivalences are stable under finite coproducts, one can ensure that any desired set of objects is cofibrant in the resulting model structure.


It turns out there's a pretty formal affirmative answer using the concept of a flat map (dual to sharp maps).

Almost Theorem: Let $\mathcal K$ be a left proper, combinatorial model category with weak equvialences stable under filtered colimits. Then there is another "flat" model structure on $\mathcal K$ with the same weak equivalences and cofibrations the flat morphisms of the original model structure.

Remark: In a left proper model category, every cofibration is flat. So the flat model structure has more cofibrations than the original one (or else coincides with the original). Moreover, the property of being left proper depends only on being the weak equivalences. So the flat model structure is the one with the given weak equivalences and a maximal number of cofibrations.

Remark: A morphism $\emptyset \to X$ is flat if and only if $X \amalg (-)$ preserves weak equivalences. So under the mild condition that the weak equivalences of $\mathcal K$ are stable under finite coproducts, every object of the flat model structure is cofibrant.

Almost Proof Sketch: By Jeff Smith's theorem, you have to check that (1) the flat maps are stable under cobase change, transfinite composition, and retracts, (2) similarly for acyclic flat maps, and (3) every morphism with the right lifing property with respect to flat maps is a weak equivalence. The first two are pretty straightforward diagram chases (though the transfinite composition part does seem to require the weak equivalences to be stable under filtered colimits). For the last one, by left properness every cofibiration in the original model structure is flat, so every morphism lifting against flat maps is a trivial fibration in the original model structure and hence a weak equivalence.


So this begs the question: what are the flat maps in, say, the Bergner model structure? I think I've identitifed some types, but I'm not sure how to systematically identify them all.

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  • $\begingroup$ Your last para would make a good new question :-) $\endgroup$ – David Roberts Mar 4 at 0:13
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    $\begingroup$ How do you know that flat maps are generated by a set ? Is there a formal accessibility argument ? $\endgroup$ – Simon Henry Mar 4 at 7:08
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    $\begingroup$ @SimonHenry Er-- I hadn't thought of that! I will have to mull it over. At the very least, you can take for the cofibrations the cofibrant closure of any set of flat maps which contains the generators for the original cofibrations and you'll get a model structure. But then it's not clear you can arrange to have every object cofibrant... $\endgroup$ – Tim Campion Mar 4 at 13:37

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