# Topological groups with homeomorphic underlying spaces, isomorphic abstract groups and homotopy equivalent classifying spaces

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.

• What do you mean with "strongly homotopy equivalent"? – Denis Nardin Mar 6 at 9:27
• @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 – rori Mar 6 at 9:32