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1. • CommentRowNumber1.
   • CommentAuthordomenico_fiorenza
   • CommentTimeFeb 24th 2014
   • PermaLink
   Author: domenico_fiorenza
   Format: MarkdownItexLet $K\hookrightarrow G$ be a normal subgroup. Then we have a [[quotient group]] $G/K$ and a short exact sequence of groups $1\to K\to G\to G/K\to 1$. This sequence deloops giving the [[fiber sequence]] $\mathbf{B}K\to \mathbf{B}G\to \mathbf{B}(G/K)$. Is this also a [[cofiber sequence]]? I suspect so, but I'm not sure (for sure it is true for abelian groups). Yet the nLab page on [[cofiber sequences]] it seems this example is not discussed, so maybe I'm wrong here. The reason I'm interested in $\mathbf{B}K\to \mathbf{B}G\to \mathbf{B}(G/K)$ being a cofiber sequence is that this would give a natural description of representations of the quotient $G/K$ as representations of $G$ together with the datum of a trivialization of their restriction to $K$ (actually proving by hand this equivalence would be a proof that the sequence is a cofiber sequence, but if the result is already known (either as true or as false) I'd avoid working this out in detail myself).

   Let $K\hookrightarrow G$ be a normal subgroup. Then we have a quotient group $G/K$ and a short exact sequence of groups $1\to K\to G\to G/K\to 1$. This sequence deloops giving the fiber sequence $\mathbf{B}K\to \mathbf{B}G\to \mathbf{B}(G/K)$. Is this also a cofiber sequence? I suspect so, but I’m not sure (for sure it is true for abelian groups). Yet the nLab page on cofiber sequences it seems this example is not discussed, so maybe I’m wrong here.

   The reason I’m interested in $\mathbf{B}K\to \mathbf{B}G\to \mathbf{B}(G/K)$ being a cofiber sequence is that this would give a natural description of representations of the quotient $G/K$ as representations of $G$ together with the datum of a trivialization of their restriction to $K$ (actually proving by hand this equivalence would be a proof that the sequence is a cofiber sequence, but if the result is already known (either as true or as false) I’d avoid working this out in detail myself).

2. • CommentRowNumber2.
   • CommentAuthorDavidRoberts
   • CommentTimeFeb 24th 2014
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexI think it may not be. $K\to G$ is a crossed module, and $G/K$ acts by outer automorphisms on $K$, and so at least in the topological classifying case sense, I think $BG\to B(G/K)$ is a fibre bundle with fibre $BK$ and structure group $Out(K)$. Or at least, something like that. This is related somewhat to figuring out what the nonabelian version of a lifting bundle gerbe is.

   I think it may not be. $K\to G$ is a crossed module, and $G/K$ acts by outer automorphisms on $K$, and so at least in the topological classifying case sense, I think $BG\to B(G/K)$ is a fibre bundle with fibre $BK$ and structure group $Out(K)$. Or at least, something like that.

   This is related somewhat to figuring out what the nonabelian version of a lifting bundle gerbe is.

3. • CommentRowNumber3.
   • CommentAuthorMike Shulman
   • CommentTimeFeb 24th 2014
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   Author: Mike Shulman
   Format: MarkdownItexIs it really true even for abelian groups if you're talking about spaces rather than spectra?

   Is it really true even for abelian groups if you’re talking about spaces rather than spectra?

4. • CommentRowNumber4.
   • CommentAuthorDavidRoberts
   • CommentTimeFeb 24th 2014
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexWell, for central extensions, yes.

   Well, for central extensions, yes.

5. • CommentRowNumber5.
   • CommentAuthorDylan Wilson
   • CommentTimeFeb 25th 2014
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   Author: Dylan Wilson
   Format: MarkdownItexUnless I'm misinterpreting the notation, this seems like it would almost never happen just by inspection of cohomology groups. As a fiber sequence the relationship between the cohomology groups is a spectral sequence; in a cofiber sequence you get a long exact sequence.

   Unless I’m misinterpreting the notation, this seems like it would almost never happen just by inspection of cohomology groups. As a fiber sequence the relationship between the cohomology groups is a spectral sequence; in a cofiber sequence you get a long exact sequence.

6. • CommentRowNumber6.
   • CommentAuthorDavidRoberts
   • CommentTimeFeb 25th 2014
   • (edited Feb 25th 2014)
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   Author: DavidRoberts
   Format: MarkdownItexGah, I misread the question, so just ignore what I wrote in #2 and #4.

   Gah, I misread the question, so just ignore what I wrote in #2 and #4.

7. • CommentRowNumber7.
   • CommentAuthorChris SchommerPries
   • CommentTimeFeb 25th 2014
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   Author: Chris SchommerPries
   Format: MarkdownItexThis is generally false. For example consider the central extension of abelian groups $$ \mathbb{Z}\to \mathbb{Z} \to \mathbb{Z}/2 $$ after taking classifying spaces you get $$ S^1 \to S^1 \to \mathbb{RP}^\infty $$ but the cofiber of the first map (which is a double covering map) is only $\mathbb{RP}^2$.

   This is generally false. For example consider the central extension of abelian groups

   $$\mathbb{Z}\to \mathbb{Z} \to \mathbb{Z}/2$$

   after taking classifying spaces you get

   $$S^1 \to S^1 \to \mathbb{RP}^\infty$$

   but the cofiber of the first map (which is a double covering map) is only $\mathbb{RP}^2$.

8. • CommentRowNumber8.
   • CommentAuthorMike Shulman
   • CommentTimeFeb 25th 2014
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexI've added this counterexample to the page [[cofiber sequence]].

   I’ve added this counterexample to the page cofiber sequence.

9. • CommentRowNumber9.
   • CommentAuthordomenico_fiorenza
   • CommentTimeFeb 26th 2014
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   Author: domenico_fiorenza
   Format: MarkdownItexHi Chris, Thanks, that's a beautiful example! By the way, the natural morphism $cofiber(B\mathbb{Z}\to B\mathbb{Z})\to B(\mathbb{Z}/2)$ is homotopy equivalent the canonical inclusion $\mathbb{R}\mathbb{R}^2\hookrightarrow \mathbb{R}\mathbb{P}^\infty$ in this case and so it induces in particular an isomorphism on fundamental groups up to $\pi_1$. This I guess should be a general phenomenon, i.e., one should have $cofiber(B K\to B G)\to B(G/K)$ inducing an iso on $\pi_0$ and $\pi_1$ in general (the $\pi_0$ case being obvious, and the $\pi_1$ being Seifert-van Kampen theorem), am I right?

   Hi Chris,

   Thanks, that’s a beautiful example!

   By the way, the natural morphism $cofiber(B\mathbb{Z}\to B\mathbb{Z})\to B(\mathbb{Z}/2)$ is homotopy equivalent the canonical inclusion $\mathbb{R}\mathbb{R}^2\hookrightarrow \mathbb{R}\mathbb{P}^\infty$ in this case and so it induces in particular an isomorphism on fundamental groups up to $\pi_1$. This I guess should be a general phenomenon, i.e., one should have $cofiber(B K\to B G)\to B(G/K)$ inducing an iso on $\pi_0$ and $\pi_1$ in general (the $\pi_0$ case being obvious, and the $\pi_1$ being Seifert-van Kampen theorem), am I right?

10. • CommentRowNumber10.
    • CommentAuthorMike Shulman
    • CommentTimeFeb 27th 2014
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexYes, that seems right to me.

    Yes, that seems right to me.