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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeJul 21st 2010
   • (edited Jul 21st 2010)
   • PermaLink
   Author: Urs
   Format: MarkdownItexI am a bit stuck/puzzled with the following. Maybe you have an idea: I have a group object $G$ and a morphism $G \to Q$. I have a model for the universal $G$-bundle $\mathbf{E}G$ (an object weakly equivalent to the point with a fibration $\mathbf{E}G \to \mathbf{B}G$). I have another object $\mathbf{E}Q$ weakly equivalent to the point such that I get a commuting diagram $$ \array{ G &\to& Q \\ \downarrow && \downarrow \\ \mathbf{E}G &\to& \mathbf{E}Q } $$ Here $Q$ is not groupal and i write $\mathbf{E}Q$ only for the heck of it and to indicate that this is contractible and the vertical morphisms above are monic (cofibrations if due care is taken). So I have $G$ acting on $\mathbf{E}G$ and the coequalizer of that action exists and is $\mathbf{B}G$ $$ G \times \mathbf{E}G \stackrel{\to}{\to} \mathbf{E}G \to \mathbf{B}G $$ I can also consider the colimit $K$ of the diagram $$ G \times \mathbf{E}G \stackrel{\to}{\to} \mathbf{E}G \to \mathbf{E}Q \,. $$ That gives me a canonical morphism $\mathbf{B}G \to K$ fitting in total into a diagram $$ \array{ G &\to& Q \\ \downarrow && \downarrow \\ \mathbf{E}G &\to& \mathbf{E}Q \\ \downarrow && \downarrow \\ \mathbf{B}G &\to& K } \,. $$ Now here comes finally the question: I know that the coequalizer of $G \times \mathbf{E}G \stackrel{\to}{\to} \mathbf{E}G$ is a model for the _homotopy colimit_ over the diagram $$ \cdots G \times G \stackrel{\to}{\stackrel{\to}{\to}} G \stackrel{\to}{\to} * $$ as you can imagine. But I am stuck: what intrinsic $(\infty,1)$-categorical operation is $K$ a model of? I must be being dense....

   I am a bit stuck/puzzled with the following. Maybe you have an idea:

   I have a group object $G$ and a morphism $G \to Q$. I have a model for the universal $G$-bundle $\mathbf{E}G$ (an object weakly equivalent to the point with a fibration $\mathbf{E}G \to \mathbf{B}G$).

   I have another object $\mathbf{E}Q$ weakly equivalent to the point such that I get a commuting diagram

   $$\array{ G &\to& Q \\ \downarrow && \downarrow \\ \mathbf{E}G &\to& \mathbf{E}Q }$$

   Here $Q$ is not groupal and i write $\mathbf{E}Q$ only for the heck of it and to indicate that this is contractible and the vertical morphisms above are monic (cofibrations if due care is taken).

   So I have $G$ acting on $\mathbf{E}G$ and the coequalizer of that action exists and is $\mathbf{B}G$

   $$G \times \mathbf{E}G \stackrel{\to}{\to} \mathbf{E}G \to \mathbf{B}G$$

   I can also consider the colimit $K$ of the diagram

   $$G \times \mathbf{E}G \stackrel{\to}{\to} \mathbf{E}G \to \mathbf{E}Q \,.$$

   That gives me a canonical morphism $\mathbf{B}G \to K$ fitting in total into a diagram

   $$\array{ G &\to& Q \\ \downarrow && \downarrow \\ \mathbf{E}G &\to& \mathbf{E}Q \\ \downarrow && \downarrow \\ \mathbf{B}G &\to& K } \,.$$

   Now here comes finally the question: I know that the coequalizer of $G \times \mathbf{E}G \stackrel{\to}{\to} \mathbf{E}G$ is a model for the homotopy colimit over the diagram

   $$\cdots G \times G \stackrel{\to}{\stackrel{\to}{\to}} G \stackrel{\to}{\to} *$$

   as you can imagine. But I am stuck: what intrinsic $(\infty,1)$-categorical operation is $K$ a model of?

   I must be being dense….

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeJul 21st 2010
   • (edited Jul 21st 2010)
   • PermaLink
   Author: Urs
   Format: MarkdownItexNot sure if it helps to see the pattern or distract from its general structure: but my detailed setup is described in a bit more detail (though still in a rough fashion) <a href="http://ncatlab.org/schreiber/show/differential+cohomology+in+an+(%E2%88%9E%2C1)-topos#UniversalConn">here</a>.

   Not sure if it helps to see the pattern or distract from its general structure: but my detailed setup is described in a bit more detail (though still in a rough fashion) here.

3. • CommentRowNumber3.
   • CommentAuthorMike Shulman
   • CommentTimeJul 21st 2010
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexI don't understand the definition of K; is there a typo in the displayed equation after "I can also consider the colimit K of the diagram"?

   I don’t understand the definition of K; is there a typo in the displayed equation after “I can also consider the colimit K of the diagram”?

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeJul 21st 2010
   • (edited Jul 21st 2010)
   • PermaLink
   Author: Urs
   Format: MarkdownItex$K$ was supposed to be the coequalizer of the two composite maps $$ G \times \mathbf{E}G \stackrel{\to}{\to} \mathbf{E}Q $$ But wait, I guess I am being stupid....

   $K$ was supposed to be the coequalizer of the two composite maps

   $$G \times \mathbf{E}G \stackrel{\to}{\to} \mathbf{E}Q$$

   But wait, I guess I am being stupid….

5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeJul 21st 2010
   • PermaLink
   Author: Urs
   Format: MarkdownItexRight, so I guess i simply mean that the square $$ \array{ \mathbf{E}G &\to& \mathbf{E}Q \\ \downarrow && \downarrow \\ \mathbf{B}G &\to& K } $$ is a pushout.

   Right, so I guess i simply mean that the square

   $$\array{ \mathbf{E}G &\to& \mathbf{E}Q \\ \downarrow && \downarrow \\ \mathbf{B}G &\to& K }$$

   is a pushout.

6. • CommentRowNumber6.
   • CommentAuthorMike Shulman
   • CommentTimeJul 22nd 2010
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexIs $EG \to EQ$ a cofibration? If not, then it seems that that pushout has no homotopical meaning. If so, then it's an acyclic cofibration, since both EG and EQ are contractible, hence $BG \to K$ is also an acyclic cofibration and thus a weak equivalence, so K is the same as BG. Regardless, it doesn't seem that K can contain any information about Q, since that pushout doesn't contain Q anywhere.

   Is $EG \to EQ$ a cofibration? If not, then it seems that that pushout has no homotopical meaning. If so, then it’s an acyclic cofibration, since both EG and EQ are contractible, hence $BG \to K$ is also an acyclic cofibration and thus a weak equivalence, so K is the same as BG. Regardless, it doesn’t seem that K can contain any information about Q, since that pushout doesn’t contain Q anywhere.

7. • CommentRowNumber7.
   • CommentAuthorUrs
   • CommentTimeJul 22nd 2010
   • PermaLink
   Author: Urs
   Format: MarkdownItexYeah, it looks puzzling. I should turn the question around: given a morphism in an $(\infty,1)$-category $G \to Q$, where $G$ happens to be a group object. Does this induce any canonical morphism out of the delooping $\mathbf{B}G$? (Feel free to assume some extra properties if that helps make you think of something.)

   Yeah, it looks puzzling.

   I should turn the question around:

   given a morphism in an $(\infty,1)$-category $G \to Q$, where $G$ happens to be a group object. Does this induce any canonical morphism out of the delooping $\mathbf{B}G$?

   (Feel free to assume some extra properties if that helps make you think of something.)

8. • CommentRowNumber8.
   • CommentAuthorUrs
   • CommentTimeJul 22nd 2010
   • (edited Jul 22nd 2010)
   • PermaLink
   Author: Urs
   Format: MarkdownItexWell, I should maybe add the following: that morphsm $G \to Q$ is part of a fiber sequence $$ G \to Q \to L \to \mathbf{B}G \,. $$ In the case that $G$ happens to be twice deloopable, this continues as $$ G \to Q \to L \to \mathbf{B}G \to \mathbf{B}Q $$ and that $\mathbf{B}G \to \mathbf{B}Q$ is what I need. So one way to ask what I am asking is: in the case that $G$ is not twice deloopable, what's a reaonable universal approximation to the non-existent $\mathbf{B}G \to \mathbf{B}Q$ here? (Actually, I have a guess for that, too, using some extra structure I have availabke, but I can't show that my guess reproduces the above ordinary pushout construciton.)

   Well, I should maybe add the following: that morphsm $G \to Q$ is part of a fiber sequence

   $$G \to Q \to L \to \mathbf{B}G \,.$$

   In the case that $G$ happens to be twice deloopable, this continues as

   $$G \to Q \to L \to \mathbf{B}G \to \mathbf{B}Q$$

   and that $\mathbf{B}G \to \mathbf{B}Q$ is what I need.

   So one way to ask what I am asking is: in the case that $G$ is not twice deloopable, what’s a reaonable universal approximation to the non-existent $\mathbf{B}G \to \mathbf{B}Q$ here?

   (Actually, I have a guess for that, too, using some extra structure I have availabke, but I can’t show that my guess reproduces the above ordinary pushout construciton.)

9. • CommentRowNumber9.
   • CommentAuthorMike Shulman
   • CommentTimeJul 23rd 2010
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexShould it be obvious that such a fiber sequence can be continued if G is twice deloopable? Or is this a special characteristic of your situation?

   Should it be obvious that such a fiber sequence can be continued if G is twice deloopable? Or is this a special characteristic of your situation?

10. • CommentRowNumber10.
    • CommentAuthorUrs
    • CommentTimeJul 23rd 2010
    • (edited Jul 23rd 2010)
    • PermaLink
    Author: Urs
    Format: MarkdownItex> Should it be obvious that such a fiber sequence can be continued if G is twice deloopable? Or is this a special characteristic of your situation? This is special for my situation. The fiber sequence that i am considering is that induced by the counit $$ LConst \Gamma \to Id $$ of the terminal geometric morphism. I am working in an "locally $\infty$-connected" situation, so that $\mathbf{\flat} := LConst \Gamma$ is a right adjoint. As such it preserves looping, and hence I get a fiber sequence $$ G \to \mathbf{\flat}_{dR} \mathbf{B}G \to \mathbf{\flat}\mathbf{B}G \to \mathbf{B}G \to \mathbf{\flat}_{dR} \mathbf{B}^2 G \to \mathbf{\flat}\mathbf{B}^2 G \to \mathbf{B}^2 G $$

    Should it be obvious that such a fiber sequence can be continued if G is twice deloopable? Or is this a special characteristic of your situation?

    This is special for my situation.

    The fiber sequence that i am considering is that induced by the counit

    $$LConst \Gamma \to Id$$

    of the terminal geometric morphism. I am working in an “locally $\infty$-connected” situation, so that $\mathbf{\flat} := LConst \Gamma$ is a right adjoint. As such it preserves looping, and hence I get a fiber sequence

    $$G \to \mathbf{\flat}_{dR} \mathbf{B}G \to \mathbf{\flat}\mathbf{B}G \to \mathbf{B}G \to \mathbf{\flat}_{dR} \mathbf{B}^2 G \to \mathbf{\flat}\mathbf{B}^2 G \to \mathbf{B}^2 G$$
11. • CommentRowNumber11.
    • CommentAuthorMike Shulman
    • CommentTimeJul 23rd 2010
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexOk. I think I'm not going to be able to help you without spending a lot more time to understand your situation; sorry.

    Ok. I think I’m not going to be able to help you without spending a lot more time to understand your situation; sorry.