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.

Are 0,8 and 9 homeomorphic topological space?


Consider the topological spaces “0”,”8″ and “9” in $ \mathbb{R}^{2}$ . Are they homeomorphic?

I have an approach that doesnt look very rigorous to me. I wanted to know how to formalize this if its correct.

  • 0 and 8 are not homeomorphic since excluding one point of 0 the space is still connected, but excluding the “tangent point” of 8, we have a disconnected space.

  • Same idea for 8 and 9.

  • The space 9 is union of one circle and one arc. The arc is homeomorphic to the circle, so we can view 9 as a union of two circles, then 8 and 9 are homeomorphic

PS: the topology of the spaces is induced by topology of $ \mathbb{R}^{2}$ .