Timeline for Does the right adjoint of a Quillen equivalence preserve homotopy colimits?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 10, 2011 at 18:46 | vote | accept | Dan Dreiberg | ||
| Dec 9, 2011 at 19:50 | comment | added | Dan Dreiberg | @Justin, yes, you are right, this should be the correct unwinding of Fernando's answer. Anyway, if the diagram $E''$ (obtained from $E'=G\circ R\circ QD:I\to C$ by replacing all objects cofibrantly in $C$) is a homotopy colimit diagram (i.e. $\operatorname{colim}Q'E''\to\operatorname{colim}E''$ is a weak equivalence) then also $E'$ should be a homotopy colimit diagram because of the homotopy invariance of the homotopy colimit. | |
| Dec 9, 2011 at 13:52 | comment | added | Justin Noel | 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. | |
| Dec 8, 2011 at 19:38 | comment | added | Fernando Muro | @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! | |
| Dec 8, 2011 at 19:23 | comment | added | Dan Dreiberg | 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 $C$? | |
| Dec 8, 2011 at 17:38 | comment | added | Fernando Muro | I don't get what you mean. What do you have in mind when you talk about homotopy colimits? | |
| Dec 8, 2011 at 7:10 | comment | added | Dan Dreiberg | 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 $C$? | |
| Dec 6, 2011 at 21:29 | history | answered | Fernando Muro | CC BY-SA 3.0 |