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I would like to know for what kind of model category finite homotopy limits commute with sequential homotopy colimits. Would cofibrantly generated and finitely locally presentable be enough? It seems to be true in simplicial sets.

I found a proof of something close in Stephan Schwede's paper "Spectra in model categories and applications to the algebraic cotangent complex", but I wasn't able to deduce the statement above from his. Jacob Lurie has a proof of the analogous statement for colimits and limits of quasi-categories, in "Higher topos theory".

Anything related would be much appreaciated!

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  • $\begingroup$ I bet you need a combinatorial finely presentable model category. $\endgroup$
    – Fernando Muro
    Commented Sep 27, 2013 at 18:26
  • $\begingroup$ From what I could see by googling around, I bet you're right. I'm afraid this could mean that the proof is very indirect... By the way what does finely mean? $\endgroup$
    – Emanuele Dotto
    Commented Sep 27, 2013 at 18:52
  • $\begingroup$ Probably "finitely presentable". The corresponding statement is true for ordinary categories, i.e. in a locally finitely presentable category, sequential (indeed, filtered) colimits preserve finite limits. $\endgroup$
    – Zhen Lin
    Commented Sep 27, 2013 at 20:09
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    $\begingroup$ What do you mean by "finite homotopy limit"? If you mean a homotopy limit indexed by an ordinary category with finitely many morphisms, it is not true even in spaces. Take $S^1 \subset S^2 \subset \cdots$ each with the antipodal action of $\mathbb{Z}/2$. Taking homotopy fixed points of the action gives the empty set in each case (the homotopy fixed points are sections of $S^n \to \mathbb{R}{P}^n \to \mathbb{R}{P}^{\infty} = B \mathbb{Z}/2$), but for the sequential homotopy colimit $S^\infty$, which is contractible, taking homotopy fixed points gives a one point space. $\endgroup$
    – Omar Antolín-Camarena
    Commented Sep 27, 2013 at 22:34
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    $\begingroup$ On the other hand, homotopy limits indexed by simplicial sets with finitely many nondegenerate simplices do commute with filtered homotopy colimits in spaces. (Notice that while $\mathbb{Z}/2$ thought of as a category with a single element has only finitely many morphisms, what matters here is that its nerve has infinitely many nondegenerate simplices.) $\endgroup$
    – Omar Antolín-Camarena
    Commented Sep 27, 2013 at 22:37

1 Answer 1

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In combinatorial model categories finite limits commute with (sufficiently large) filtered homotopy colimits. Suppose, for simplicity, that the combinatorial model category is simplicial and generating cofibrations have $\lambda$-presentable domain and codomain. In this case $\lambda$-filtered colimits are homotopy colimits. Suppose, in addition, that the underlying locally presentable model category is $\lambda$-locally presentable. Then $\lambda$-filtered colimits commute with $\lambda$-small limits. Finite limits are $\lambda$-small for all $\lambda$.

Lets say that the category $J$ indexing the homotopy limit is finite if it has finitely many objects and morphisms, and the diagram $EJ$ of simplicial sets serving as a cofibrant replacement of the constant diagrams of points indexed by $J$ in the projective model structure on ${\cal S}^J$ has a finite simplicial set in each entry. For example, a finite group is not a finite category by this definition. A finite homotopy limit is a homotopy limit over a finite diagram.

Suppose that $\cal M$ is a simplicial $\lambda$-combinatorial model category and $F\colon J\to \cal M$ a finite diagram. Then $\mathrm{holim}_J F$ may be computed as a weighted limit with the weight $EJ$, in other words this is an end construction: $$ \mathrm{holim}_J F = \mathrm{hom}(EJ, F), $$ which is a finite weighted limit commuting with $\lambda$-filtered colimits, hence, commuting with $\lambda$-filtered homotopy colimits. In particular, if $\lambda=\aleph_0$, then filtered homotopy colimits commute with finite homotopy limits.

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  • $\begingroup$ Boris, is really necessary that the category be cofibrant? $\endgroup$
    – Fernando Muro
    Commented Sep 29, 2013 at 10:42
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    $\begingroup$ Of cause not, Fernando, cofibrant generation is a luxury. We just need it to compare the homotopy filtered colimits with the strict filtered colimits. For this purpose it suffices to assume that trivial fibrations are closed under $\lambda$-filtered colimits. But I'd rather keep this answer less technical. $\endgroup$
    – Boris Chorny
    Commented Sep 29, 2013 at 11:50
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    $\begingroup$ No, this assumption can also be removed using a framing and applying the respective formulas from Hirschhorn's book for the computation of homotopy limits. Alternatively we can replace our combinatorial model category by a simplicial one in a homotopy meaningful way and conclude that finite homotopy limits commute with filtered homotopy colimits since they commute in the simplicial replacement. $\endgroup$
    – Boris Chorny
    Commented Sep 29, 2013 at 19:05
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    $\begingroup$ Boris, thanks for your answer. If $\lambda$ is bigger than $\aleph_0$, is it then not true in general that sequential homotopy colimits commute with finite homotopy limits? (thanks for clarifying what finite means in this context) $\endgroup$
    – Emanuele Dotto
    Commented Sep 30, 2013 at 15:25
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    $\begingroup$ I think that you are right. In a more general situation it is probably possible to find a counterexample, but every combinatorial model category is Quillen equivalent, by a theorem of Dugger, to a left Bousfield localization of simplicial presheaves over some small category equipped with the projective model structure. But in such categories generating cofibrations are the same as the projective generating cofibrations. In particular, they have $\aleph_o$-presentable domains and codomains. $\endgroup$
    – Boris Chorny
    Commented Sep 30, 2013 at 18:46

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