Let $G$, $H$ be topological groups and $f:G\rightarrow H$ a continuous group homomorphism which happens to be a homotopy equivalence of the underlying topological spaces. Let us assume that $G$, $H$ are well-pointed compactly generated Hausdorff as topological spaces, where well-pointedness means that the inclusions of basepoints are closed cofibrations. (By well-pointedness, there is no difference between a homotopy equivalence and a based homotopy equivalence.) Let $B$ be the classifying space functor.

My question is: Is $Bf: BG \rightarrow BH$ a homotopy equivalence? (Again, there is no difference between based and non-based homotopy equivalence since $BG$ is well-pointed if $G$ is.)

I understand that $Bf$ is a weak homotopy equivalence even without the assumptions made above on the topologies of $G$ and $H$, by this post. I would like it to be a homotopy equivalence with those extra assumptions. Can we show $BG$ and $BH$ have the homotopy type of CW complexes or something?

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    $\begingroup$ If $BG$ has the homotopy type of a CW complex then $G=\Omega BG$ has it too. $\endgroup$ Mar 29, 2015 at 0:01
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    $\begingroup$ You might try looking at Segal's paper Classifying Spaces and Spectral Sequences, where he explains that Milnor's infinite join construction of $BG$ can be built as the classifying space of a certain topological category $G_N$. Then induced functor $G_N \to H_N$ will induce a level-wise homotopy equivalence between the nerves of these cateogies (whose realizations give $BG$ and $BH$)... $\endgroup$
    – Dan Ramras
    Mar 29, 2015 at 0:46
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    $\begingroup$ (cont'd) I think your conditions on $G$ and $H$ ought to imply that these nerves are good simplicial spaces. But I'm not certain if that's ensure that a level-wise homotopy equivalence induces a homotopy equivalence of realizations. Goerss and Jardine's book would be one place to look. $\endgroup$
    – Dan Ramras
    Mar 29, 2015 at 0:48
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    $\begingroup$ @user46652: you need to specify which classifying space functor $B$ you are using for the question to make any sense. $\endgroup$
    – John Klein
    Mar 29, 2015 at 1:49
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    $\begingroup$ $BG$ is fundamentally an object which only makes sense up to weak homotopy equivalence; asking questions like this means getting bogged down in the technicalities of spaces not having the homotopy type of a CW complex and there's just no reason to torture yourself like that. $\endgroup$ Mar 29, 2015 at 3:31

1 Answer 1


As John Klein remarked, the answer to this question will depend on the classifying space functor $B$ one uses.

Let me present one case for which the question can be answered positive which is basically the case Dan Ramras mentioned.

We define $BG$ for a topological group $G$ to be the fat realization of the simplicial space obtained by applying the topological nerve construction to the topological category one obtains by regarding $G$ as a category in the usual way and including its topology. (See Segal's 'Classifying Spaces and Spectral Sequences' §3) A continuous group homomorphism $G\rightarrow H$ which is a homotopy equivalence will induce a morphism of the corresponding simplicial spaces which is a degreewise homotopy equivalence. (This is easy to check, right from the definitions.) Since we have chosen the fat realization the induced map $BG\rightarrow BH$ will be a homotopy equivalence by Proposition A.1 in Appendix A in Segal's 'Catgeories and Cohomology Theories'.

Up here, no point-set topological restrictions are needed, it even all works if the spaces are not compactly generated.

In the last paper cited, you'll also find information about the question, in which cases the fat realization is homotopy equivalent to the usual one. In the case of well pointed compactly generated groups, their simplicial nerves will be "good" in the sense of Segal. A survey on the results of that manner is given here.

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    $\begingroup$ I think your discussion of the relationship with Milnor's construction is a little off. Segal has to use a modified ("unraveled" he calls it) version of the category associated to G. This is what I called G_N in my comment on the question. For the record, Oscar Randall-Williams once corrected a similar mis-statement that I made (see the comments here mathoverflow.net/questions/104406/…). $\endgroup$
    – Dan Ramras
    Apr 1, 2015 at 3:44
  • $\begingroup$ This implies that $EG \to BG$ is always a principal bundle if I use the fat geometric realization for $EG$ and $BG$. Where can I find this statement in the literature? $\endgroup$ Jul 7, 2015 at 15:39

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