Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ be a topological group. Then the classifying space $BG$'s homotopy type depends on the "homotopy type" of the topological group $G$: that is, if $G \to G'$ is a morphism of topological groups (i.e., a continuous homomorphism) which is a weak equivalence, then $BG \to BG'$ is a weak equivalence (note that $BG$ is functorial by the usual construction, and its homotopy groups are those of $G$ with a shift). This means that on a CW complex, giving a principal $G$-bundle is the same as giving a principal $G'$-bundle.

Is there a direct proof of this (e.g. using $H^1$) that does not resort to $BG$?

share|cite|improve this question
you want a proof that a G-bundle is the same as a G'-bundle without using the classifying spaces? (just asking for clarification I misread it the first time). Also, why do you want an alternative proof? – Sean Tilson Aug 13 '11 at 0:27
@Sean: Yes. For one thing, (say we assume $G, G'$ homotopy equivalent), I'd like to know that principal $G$-bundles are the same as principal $G'$-bundles under weaker hypotheses than "$X$ is a CW complex." – Akhil Mathew Aug 13 '11 at 1:11
what do you mean by the "same"? are $G$ and $G'$ CW complexes? – Sean Tilson Aug 13 '11 at 17:16
@Sean: I mean that $H^1(X, G) \to H^1(X, G')$ is an isomorphism, so isomorphism classes of $G$ and $G'$-bundles are the same. I don't necessarily want to assume $G, G'$ CW complexes, but if it makes it easier, then I'd be curious about a result in that case. – Akhil Mathew Aug 13 '11 at 17:35
you want to avoid $BG$ but $H^1(X;G) \cong [X, BG]$, right? – Sean Tilson Aug 13 '11 at 17:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.