I believe there is a theorem like if $G$ and $H$ are homotopy equivalent topological groups, then $G$-bundles can have their structure group reduced to $H$ or equivalently (?) the classifying space $BG$ is homotopy equivalent to $BH$.
A more specific version of this fact is that real (complex) vector bundles can have their structure groups reduced from $GL_n(\mathbb{R})$ ($GL_n(\mathbb{C})$) to $O(n)$ ($U(n)$), which is proved via linear algebra techniques like Gram-Schmidt or $QR$-decomposition.
But for the more general fact, I thought I once knew a purely homotopy-theoretic proof, using maybe fiber sequences or functoriality of the classifying space functor $B$. Can someone give the correct statement of the theorem and a nice homotopy-theoretic proof?
Best Answer
What is true is that if $f:G\to H$ is a group homomorphism that is also a (weak) homotopy equivalence, then it induces a weak homotopy equivalence $BG\simeq BH$. This is essentially immediate from the functoriality of $BG$ and the naturality of the isomorphisms $\pi_n(G)\cong \pi_{n+1}(BG)$ coming from the natural fiber sequence $G\to EG\to BG$: if $f:G\to H$ is a continuous homomorphism that induces isomorphisms $\pi_n(G)\to \pi_n(H)$, then $Bf:BG\to BH$ induces isomorphisms $\pi_{n+1}(BG)\to\pi_{n+1}(BH)$ since these groups are naturally isomorphic to $\pi_n(G)$ and $\pi_n(H)$.
Note that it is crucial that $f$ is not just a homotopy equivalence but also a homomorphism, since only homomorphisms induce maps $BG\to BH$. Indeed, $BG$ and $BH$ may not be homotopy equivalent if $G$ and $H$ are merely homotopy equivalent. For instance, $G$ and $H$ could be non-isomorphic discrete groups of the same cardinality.