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Let $I$ be a small category and $\mathcal{K}$ be a combinatorial model category.

Is it known a model category structure on the functor category $\mathcal{K}^I$ such that a map of diagrams $D\to E$ is a weak equivalence if and only if $\mathrm{colim} D \to \mathrm{colim} E$ is a weak equivalence of $\mathcal{K}$ ? (so not the objectwise weak equivalences, and by $\mathrm{colim}$, I mean the colimit)

I have no trace of a thing like that in the nLab or in the MathReview. I don't know what keyword to use in fact. Since there is no reason for a colimit of and objectwise weak equivalence to be a weak equivalence, it is not possible to see it as a localization of the projective or the injective model category structure.

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    $\begingroup$ One thing to try would be to "left induce" a model structure along the colimit functor since it is a left adjoint: define the cofibrations as well as the weak equivalences to be created by colimits. There are some links to papers about left induced model structures at the bottom of ncatlab.org/nlab/show/transferred+model+structure. $\endgroup$ – Mike Shulman Jul 20 '18 at 8:21
  • $\begingroup$ @MikeShulman Naively I believed that the transport of model structures would give the projective model structure since the right adjoint (the constant diagram functors) takes (trivial) fibrations to objectwise (trivial) fibrations. And a model structure is determined by these two classes of maps. $\endgroup$ – Philippe Gaucher Jul 20 '18 at 8:43
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    $\begingroup$ It takes fibrations and trivial fibrations to very specific objectwise fibrations and trivial fibrations that are from generating the whole projective model structure. So It is not sufficient to deduce that it is the projective model structure. If the structure that Mike mention exists then it answer your questions, but I guess it is not going to exists in general. $\endgroup$ – Simon Henry Jul 20 '18 at 14:46
  • $\begingroup$ Right - what "left induced" means is that, as I said, we define a map of diagrams to be a cofibration or a weak equivalence if its colimit is such. This means such a model structure would not be the projective one, and would be an answer to your question, if it exists. But you'd have to look at the conditions for such a model structure to exist and check whether they hold. $\endgroup$ – Mike Shulman Jul 20 '18 at 14:58
  • $\begingroup$ @SimonHenry okay I understand the mistake in my intuition now. $\endgroup$ – Philippe Gaucher Jul 23 '18 at 7:58
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Following the suggestion of Mike Shulman to check the condition for existence of the left-induced model structure: Under the additional assumptions that $I$ be connected and that $I$-colimits commute with pull-backs, the so-called left-acyclicity condition is indeed fulfilled. I just did the exercise figuring out some assumptions we have to impose so that it becomes really easy. Presumably, they can be weakened quite considerably, but I will leave this to others. (E.g., a very particular kind of pull-back is enough, which is more likely to commute with colimits, I guess.)

Let us call a morphism $\varphi$ in $K^I$ a cofibration (resp. a weak equivalence) if $\mathrm{colim}(\varphi)$ is a cofibration (resp. a weak equivalence) in $K$. From Garner, Kedziorek, Riehl, Lifting accessible model structures, arXiv:1802.09889, Corollary 2.7, and the assumption that the model category $K$ is combinatorial (or more generally, accessible) it follows that we can left-induce the model structure along $\mathrm{colim}\colon K^I\to K$ if and only if the left-acyclicity condition holds:

If a morphism $\varphi$ in $K^I$ satisfies the right-lifting property against all cofibrations in $K^I$, then it is a weak equivalence.

Let $\varphi\colon A\to B$ be a morphism in $K^I$ satisfying the right-lifting property against all cofibrations in $K^I$. We have to show that it is a weak equivalence. That is, we have to show that $\mathrm{colim}(\varphi)\colon \mathrm{colim}(A)\to\mathrm{colim}(B)$ is a weak equivalence in $K$. But for this, it is enough that $\mathrm{colim}(\varphi)$ satisfies the right-lifting property agains all cofibrations in $K$. So, let $\gamma\colon c\to d$ be a cofibration in $K$ and suppose we are given any pair of morphisms $c\to \mathrm{colim}(A)$ and $d\to \mathrm{colim}(B)$ such that the following diagram commutes.

$$\require{AMScd}\begin{CD} c @>>> \mathrm{colim}(A)\\ @V\gamma VV @VV\mathrm{colim}(\varphi)V\\ d @>>> \mathrm{colim}(B) \end{CD}\quad\quad(\ast)$$

I will denote the constant diagram-functor $K\to K^I$ by underlining. Consider $C := \underline c\times_{\underline{\mathrm{colim}(A)}}A$ and $D := \underline d\times_{\underline{\mathrm{colim}(B)}}B$, where $A\to \underline{\mathrm{colim}(A)}$ and $B\to \underline{\mathrm{colim}(B)}$ are the universal morphisms. Then we get a commutative square $$\begin{CD} C @>>> A\\ @V\underline{\gamma}\times\varphi VV @VV\varphi V\\ D @>>> B \end{CD}\quad\quad(\ast\ast)$$ in $K^I$, where the horizontal morphisms are the natural projections. Now, assuming that $\mathrm{colim}(\underline c\times_{\underline{\mathrm{colim}(A)}}A) = c$, naturally in $c$ and $A$, (e.g., if $I$ is connected and $I$-colimits commute with pull-backs,) this commutative square gets mapped to $(\ast)$ under $\mathrm{colim}$. In particular, $\underline{\gamma}\times\varphi$ is a cofibration. Thus, $\mathrm{colim}(\varphi)$ satisfies the right-lifting property agains all cofibrations as soon as $\varphi$ does, but then $\varphi$ is a weak equivalence, as claimed.

I think through a refinement of the argument, one can get rid of the connectedness-assumption. But I don't know whether this really depends on some sort of continuity.

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  • $\begingroup$ I am just discovering and reading your answer. In my case, $I$ is connected. I have no idea concerning the other condition. $\endgroup$ – Philippe Gaucher Aug 20 '18 at 8:33
  • $\begingroup$ And I don't know how to weaken the continuity condition. Maybe it would be enough that (in $K^I$) pull-backs of weak equivalences are weak equivalences. $\endgroup$ – Ben Aug 20 '18 at 9:56

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