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  1. Let KGK\hookrightarrow G be a normal subgroup. Then we have a quotient group G/KG/K and a short exact sequence of groups 1KGG/K11\to K\to G\to G/K\to 1. This sequence deloops giving the fiber sequence BKBGB(G/K)\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 BKBGB(G/K)\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/KG/K as representations of GG together with the datum of a trivialization of their restriction to KK (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).

    • CommentRowNumber2.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 24th 2014

    I think it may not be. KGK\to G is a crossed module, and G/KG/K acts by outer automorphisms on KK, and so at least in the topological classifying case sense, I think BGB(G/K)BG\to B(G/K) is a fibre bundle with fibre BKBK and structure group Out(K)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.

    • CommentRowNumber3.
    • CommentAuthorMike Shulman
    • CommentTimeFeb 24th 2014

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

    • CommentRowNumber4.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 24th 2014

    Well, for central extensions, yes.

    • CommentRowNumber5.
    • CommentAuthorDylan Wilson
    • CommentTimeFeb 25th 2014

    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.

    • CommentRowNumber6.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 25th 2014
    • (edited Feb 25th 2014)

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

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

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

    after taking classifying spaces you get

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

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

    • CommentRowNumber8.
    • CommentAuthorMike Shulman
    • CommentTimeFeb 25th 2014

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

  3. Hi Chris,

    Thanks, that’s a beautiful example!

    By the way, the natural morphism cofiber(BB)B(/2)cofiber(B\mathbb{Z}\to B\mathbb{Z})\to B(\mathbb{Z}/2) is homotopy equivalent the canonical inclusion 2 \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 π 1\pi_1. This I guess should be a general phenomenon, i.e., one should have cofiber(BKBG)B(G/K)cofiber(B K\to B G)\to B(G/K) inducing an iso on π 0\pi_0 and π 1\pi_1 in general (the π 0\pi_0 case being obvious, and the π 1\pi_1 being Seifert-van Kampen theorem), am I right?

    • CommentRowNumber10.
    • CommentAuthorMike Shulman
    • CommentTimeFeb 27th 2014

    Yes, that seems right to me.