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Call a diagram $E$ in a model category a homotopy colimit diagram if the morphism $$\mathrm{hocolim}~E\to \mathrm{colim}~ E$$ is a weak equivalence. A homotopy colimit is defined as the categorical colimit of a cofibrant replacement of the diagram in the projective model structure and this is where the morphism comes from.

Let $F:C\rightleftarrows D:G$ be a Quillen equivalence between model categories $C$ and $D$. The (Edit: derived!) left adjoint $F$ preserves homotopy colimits, i.e. if $E$ is a homotopy colimit diagram in $C$, then $F\circ Q\circ E$ is a homotopy colimit diagram in $D$ where $Q$ denotes a cofibrant replacement.

Does the (Edit: derived!) right adjoint $G$ preserve homotopy colimits if the adjunction is a Quillen equivalence?

To be more precise, if $E$ is a homotopy colimit diagram in $D$, is $G\circ R\circ E$ is a homotopy colimit diagram in $C$ where $R$ denotes a fibrant replacement?

I suppose that this is true since the notion of homotopy colimit should depend only on the homotopy category and not on the model, I guess, but I cannot think of an argument.

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Hello Daniel. Got hungry. Helped myself to some beans. Hope you don't mind. –  Yemon Choi Dec 6 '11 at 19:01
    
Didn't this user ask a question last week which vanished? Something about Quillen equivalences and units? I seem to recall thinking about the problem and leaving a comment, then Tom Goodwillie left a comment, and now the question is gone. Is this question going to vanish too? –  David White Dec 6 '11 at 19:22
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Dan, I didn't see your previous question, but generally people prefer it if you don't delete questions (even if the answer turns out to be trivial). David's comment gives one reason why. And even a question with a trivial answer can be useful for bystanders to learn from. –  Tom Leinster Dec 6 '11 at 19:55
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Left Quillen functors don't generally preserve homotopy colimits in the sense you describe. The correct statement is that the derived left Quillen functor preserves hocolims. (Think of $B\otimes_A: Ch(A)\to Ch(B)$, where $A\to B$ isn't flat.) –  Charles Rezk Dec 6 '11 at 20:47
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No, please feel free to undelete it. –  Dan Dreiberg Dec 11 '11 at 13:14
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1 Answer

up vote 1 down vote accepted

The homotopy colimit functor $Ho(D^I)\rightarrow Ho(D)$ is the left adjoint of the constant diagram functor $Ho(D)\rightarrow Ho(D^I)$. Quillen equivalences induce Quillen equivalences between diagram categories, you you can replace $D$ with $C$, hence you're done by uniqueness of adjoints.

PS Don't worry about the fact that $D^I$ may not be a model category if $D$ is not cofibrantly generated. You can work with weaker axioms and convenient replacement of the notion of Quillen equivalence.

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That you for answering. Does this imply that the right derived functor of $G$ preserves homotopy colimits in the abovementioned sense, i.e. if $E$ is a homotopy colimit diagram in $D$, you replace the objects $E(x)$ all fibrantly, then the map $\mathrm{hocolim}~G\circ E\to \mathrm{colim}~G\circ E$ is a weak equivalence in $C$? –  Dan Dreiberg Dec 8 '11 at 7:10
    
I don't get what you mean. What do you have in mind when you talk about homotopy colimits? –  Fernando Muro Dec 8 '11 at 17:38
    
I am not criticizing your answer, I think it implies exactly what I am asking for but I just want to be sure that I get precisely what you mean: So let $E:I\to D$ be a diagram in $D$ and $QE:I\to D$ its cofibrant replacement in the projective structure on $\mathrm{Fun}(I,D)$. Let $R$ be a fibrant replacement of $D$ and consider the diagram $E′=G\circ R\circ QD:I\to C$ in $C$. Let $Q′E′:I\to C$ its cofibrant replacement in the projective structure on $\mathrm{Fun}(I,C)$. My question is: Is the map $\mathrm{colim} Q′E′\to\mathrm{colim} E′$ a weak equivalence in $C$? –  Dan Dreiberg Dec 8 '11 at 19:23
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@Dan, Yes, I believe so. Now I realize that you phrase your question in terms of a specific construction of hocolim, while I answered in terms of its universal characterization. I think they're just two point of views on the same topic. I like to think of the universal characterization because it's available for different kinds of homotopic categories, even for things such as derivators! –  Fernando Muro Dec 8 '11 at 19:38
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I think unwinding your answer in terms of model categories we find that these equivalences on the homotopy categories are given by total derived functors. So we don't just take a bifibrant replacement before applying our functor, but we also apply an appropriate (co)fibrant replacement after (so we are mapping between the subcategory of bifibrant objects). The functor $ho(D^I)\rightarrow ho(C^I)$ that is part of your equivalence is $RQE\mapsto Q^\prime E^prime$. The total derived functor back is $RF$ and will induce an equivalence in homotopy categories of diagrams. –  Justin Noel Dec 9 '11 at 13:52
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