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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 $BG$?

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    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?2011-08-13
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    @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."2011-08-13
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    what do you mean by the "same"? are $G$ and $G'$ CW complexes?2011-08-13
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    @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.2011-08-13
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    you want to avoid $BG$ but $H^1(X;G) \cong [X, BG]$, right?2011-08-13
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    @Sean: Sure. Well, by $H^1$ I mean the Cech cohomology group, which can be defined in an elementary way (before you know that $BG$ exists).2011-08-13
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    @AkhilMathew let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/1070/discussion-between-sean-tilson-and-akhil-mathew)2011-08-13

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