6
$\begingroup$

There is a well-known theorem for transporting a model category structure along a left adjoint $F:\mathcal{M}\to \mathcal{N}$ which is explained here and which is due to Sjoerd Crans.

The difficult part is to check the third condition. By Axiomatic homotopy theory for operads (2.6), it suffices to check that $\mathcal{N}$ has a functorial fibrant replacement and a functorial path-object for fibrant objects. This paper cites as references:

  1. D. G. Quillen, Homotopical algebra, Lect. Notes Math. 43 (1967) Theorem II.4
  2. Every homotopy theory of simplicial algebras admits a proper model, Theorem 7.6
  3. Algebras and modules in monoidal model categories, Theorem A.3

The first reference is about the construction of a model category structure on the category of simplicial objects of a category satisfying some conditions. The second reference is a "useful lemma" Lemma 7.6. As for the third reference, I just can't find where is Theorem A.3.

Could someone give a reference for the proof of this fact (or the proof) ?

Improve this question
$\endgroup$
5
  • $\begingroup$ The transport theorem is also in the book "Model Categories and Their Localizations" by P. Hischhorn (Theorem 11.3.2). $\endgroup$
    – Philippe Gaucher
    Commented May 16, 2018 at 9:24
  • $\begingroup$ The first reference is incorrect, it should refer to the last paragraph in the proof of Theorem 4 of Section II.4. $\endgroup$
    – Dmitri Pavlov
    Commented May 17, 2018 at 14:28
  • $\begingroup$ Apart from Quillen's original argument contained in the above reference, there is also an exposition in my draft, see Theorem 3.9 dmitripavlov.org/cooperads.pdf (no claims of originality). Quillen's argument does not require all objects to be fibrant. $\endgroup$
    – Dmitri Pavlov
    Commented May 17, 2018 at 17:52
  • 2
    $\begingroup$ @DmitriPavlov Yes but did you read my answer ? The argument of Theorem 3.9 of your paper has exactly the same structure as the argument of Theorem 2.2.1 in HKRS's paper, which is indeed the dual of -something I did not know- Quillen's path object argument. That answers my question in full generality. I thank the other people for the very interesting references, I was not aware at all that so many works were done about transport of model category structures along left or right adjoints. $\endgroup$
    – Philippe Gaucher
    Commented May 17, 2018 at 20:07
  • $\begingroup$ Yes, I essentially copied 3.9 from Quillen's book (and reversed the direction of all arrows). I think you should just cite Quillen for this argument, because everything is already there. $\endgroup$
    – Dmitri Pavlov
    Commented May 17, 2018 at 23:32

3 Answers 3

Reset to default
6
$\begingroup$

Instead of Theorem A.3 in the Schwede-Shipley paper, it should be Lemma 2.3. They prove that it suffices if all objects in $M$ are fibrant, and if every object in $N$ has a path object. I think the other references are not necessary. Probably the numbering convention had to do with an older version, where section 2 was an appendix. By the way, this result you mention by Berger-Moerdijk has been strongly generalized by Johnson and Yau in their paper on model structures for PROPs, and also in their 2nd book. A summary of Yau's work in this direction can be found in Theorem 5.7 of this paper of mine with Yau. As far as I know, that's about as general as you can make it.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Okay but can we remove the hypothesis that all objects of $\mathcal{M}$ are fibrant ? $\endgroup$
    – Philippe Gaucher
    Commented May 16, 2018 at 11:51
  • $\begingroup$ And also by reading the papers, it seems that the adjunction must be monadic. $\endgroup$
    – Philippe Gaucher
    Commented May 16, 2018 at 12:39
  • $\begingroup$ @PhilippeGaucher: You do not need all objects in M to be fibrant, see the two references that I posted under the main post. $\endgroup$
    – Dmitri Pavlov
    Commented May 17, 2018 at 17:52
6
$\begingroup$

My go-to reference for inducing model structures along an adjunction is Hess, Kedziorek, Riehl, and Shipley's A necessary and sufficient condition for induced model structures, which works in great generality.

Improve this answer
$\endgroup$
7
  • $\begingroup$ It is important to mention that this reference merely combines the Smith recognition theorem for combinatorial model categories with a recent result by Makkai and Rosický that shows that cofibrations transferred along a left adjunction are generated by a set. $\endgroup$
    – Dmitri Pavlov
    Commented May 16, 2018 at 17:40
  • 3
    $\begingroup$ @DmitriPavlov The main theorem in the paper Tim mentions is strictly more general than you suggest, as it applies to the more general class of accessible model categories. It may also be important to mention that the proof given in that paper is incorrect; this proof was corrected and replaced by a more elementary proof (avoiding the use of algebraic weak factorisation systems) in arXiv:1802.09889. $\endgroup$
    – Alexander Campbell
    Commented May 17, 2018 at 6:39
  • $\begingroup$ @AlexanderCampbell: Well, a theorem with a wrong proof is no theorem at all, isn't it? The correct part does amount to what I suggested. And the newer paper that you cited explicitly attributes the breakthrough to Makkai and Rosický [22] on page 3: "Very recently, [22] showed that in this same setting, left-lifted factorizations also exist". $\endgroup$
    – Dmitri Pavlov
    Commented May 17, 2018 at 14:13
  • 2
    $\begingroup$ @DmitriPavlov And yet the theorem is true, as is everything I said in my previous comment. In the sentence you quote, "this same setting" refers to the setting of combinatorial model categories discussed in that paragraph; the main result of the paper applies in the more general setting of accessible model categories. Furthermore, they refer to "this breakthrough result" of Makkai--Rosicky in that setting, which is not to say that it is "the" breakthrough of the more general result we are discussing. $\endgroup$
    – Alexander Campbell
    Commented May 17, 2018 at 14:44
  • 2
    $\begingroup$ @DmitriPavlov True, maybe this answer could have been a comment, and certainly it was not as well-informed as David White's very topical answer. It was maybe a lucky guess based on Philippe's comments and the question title that perhaps a different approach might be helpful. And indeed, as I read Philippe's answer, the HKRS paper did in fact turn out to at least inspire a solution. Consider also the value of an answer not just to the OP, but to a general MO user -- particularly because of the more general title, a user is likely to find this question looking for more general information. $\endgroup$
    – Tim Campion
    Commented May 17, 2018 at 15:02
2
$\begingroup$

It suffices to dualize the proof of Theorem 2.2.1 in Necessary and sufficient conditions for induced model structures. It uses indeed an argument coming from Quillen's book "Homotopical Algebra", II page 4.9 (the diagram in the bottom part of the page). And it is also necessary to use the fact that the class of weak equivalences satisfies the 2-to-6 property.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ I should mark this post as the answer but it is not very ethical :-). So I won't do it. $\endgroup$
    – Philippe Gaucher
    Commented May 17, 2018 at 20:10
  • $\begingroup$ Sometimes if I feel embarrassed about accepting my own answer, I make it community wiki before accepting. It's a bit irrational, but it makes me feel better about it :-). Or sometimes I accept an an answer that was helpful and explain the final resolution of the question in a comment. $\endgroup$
    – Tim Campion
    Commented May 18, 2018 at 14:06
  • 1
    $\begingroup$ @TimCampion I don't think it is irrational. For the benefit of future readers, you should accept what you believe to be the best answer. If that answer happens to be yours (and is not merely a trivial modification of someone else's answer), you should still accept it. Making your answer CW before accepting it removes the potential conflict of interest in the decision of which answer to accept, ensuring that the decision really is based on which answer is best. $\endgroup$
    – Mike Shulman
    Commented May 18, 2018 at 17:46
  • 1
    $\begingroup$ @TimCampion done. $\endgroup$
    – Philippe Gaucher
    Commented May 19, 2018 at 5:47

You must log in to answer this question.

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