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Define the classifying space $BG$ of a well-pointed topological group $G$ as the fat realization of the nerve of $G$.

Let $G$, $H$ be well-pointed topological groups. Assume that there is a continuous group homomorphism $f:G\rightarrow H$ that induces a homotopy equivalence on the underlying topological spaces. It is claimed in an answer to this question that the induced map $BG\rightarrow BH$ is a homotopy equivalence. It is important here that we have a map between groups that simultaneously respects algebra and topology.

The question is basically what happens if we do not have such a map. Let $G$, $H$ be well-pointed topological groups. Assume that their underlying abstract groups are isomorphic and that their underlying topological spaces are homeomorphic. This by itself does not mean that $G$ and $H$ are isomorphic as topological groups (a pro-choice example is given by $p$-adic rationals for different $p$).

But what happens if we assume in addition that the classifying spaces of $G$ and $H$ are homotopy equivalent?

Note that a theorem of Notbohm says that two compact Lie groups are isomorphic as Lie groups iff their classifying spaces are homotopy equivalent so at least in some contexts, this question has a positive answer.

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    $\begingroup$ What do you mean with "strongly homotopy equivalent"? $\endgroup$ Commented Mar 6, 2019 at 9:27
  • $\begingroup$ @DenisNardin ncatlab says that a notion with this name exists. But now that I think of it that's not what I had in mind $\endgroup$
    – rori
    Commented Mar 6, 2019 at 9:32

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