Classifying spaces, Brown representability, and homotopy equivalences

Question

Let $G_1$ and $G_2$ be topological groups. Assume that there exists a continuous homomorphism $f : G_1 \rightarrow G_2$ which (ignoring the group structure) is a homotopy equivalence. If $BG_i$ is a classifying space for $G_i$, then we get an induced map $f_{\ast} : BG_1 \rightarrow BG_2$ which is a homotopy equivalence.

The fact that $f_{\ast}$ is a homotopy equivalence is clear if you construct $BG_i$ via Milnor's construction of classifying spaces. However, the existence of $BG_i$ is also an easy consequence of the (unpointed) Brown representability theorem. This leads me to my question : is there an "abstract nonsense" proof that $f_{\ast}$ is a homotopy equivalence which avoids having to know an explicit description of $BG_i$?

EDIT : As Fernando Muro pointed out, the desired result is equivalent to the fact that if $EG \rightarrow BG$ is the universal $G$-bundle over $BG$, then the space $EG$ is contractible. Of course, this is true, but I don't see how to prove it without knowing an explicit construction of $BG$. My definition of $BG$ and the universal bundle $EG \rightarrow BG$ is that $BG$ is a space such that for any CW complex $X$, the set of principal $G$-bundles on $X$ is naturally in bijection with $[X,BG]$ via the map that takes $\phi : X \rightarrow BG$ to $\phi^{\ast}(EG)$. Is there a way to see that $EG$ has to be contractible without knowing a construction of $BG$?

Answer 1

The space $\Omega BG$ is, by categorical nonsense, a classifying space for principal $G$-bundles over $\Sigma X$ with a chosen trivialisation at the basepoint. Any such bundle can be trivialised over the upper and lower halves of $\Sigma X$, and the difference between the two trivialisations gives a map $X\to G$. One can show using this that $G$ represents the same functor as $\Omega BG$, so they are homotopy equivalent. (We probably want to assume here that $G$ has the homotopy type of a CW complex). In fact, this line of argument gives a map $G\to\Omega BG$ that is well-defined and natural up to homotopy. This should be enough to show that $\Omega(f_*)$ is a homotopy equivalence, and thus that $\pi_*(f_*)$ is an isomorphism, so $f_*$ is a weak equivalence.

Answer 2

First of all let me point out that, to my knowledge, your statement is true for weak equivalences, and therefore for homotopy equivalences if your topological groups have the homotopy type of a CW-complex.

Now, I'll answer your question. By the very definition of classifying space there is a principal $G$-bundle $$G\longrightarrow EG\longrightarrow BG$$ where $EG$ is contractible. Therefore $BG$ is connected and there are natural isomorphisms $\pi_n(G)\cong\pi_{n+1}(BG)$. Since weak equivalences are detected by homotopy groups, we are done.

Answer 3

Theorem: ''Let $EG \to BG$ be a principal bundle such that for each finite CW $X$, the transformation $\eta_X$ from $[X;BG]$ to the set of isomorphism classes of principal bundles is a bijection. Then $EG$ is weakly contractible.''

Proof: Let $P \to X$ be a principal bundle. A bundle map $P \to EG$ is the same as an equivariant map. Put the compactly generated compact-open topology on $map_G (P;EG)$. The assumption says that $map_G(P;EG)$ is nonempty and path-connected (the first property expresses the surjectivity of $\eta_X$, the second one the injectivity). If $P=X \times G$ is a trivial bundle, then $map_G (P;EG)=map(X;EG)$ and you get that $[X;EG]$ is a point for each $X$. If the set of all free homotopy classes $[S^n;X]$ is trivial, then all homotopy groups are trivial).

You see that it is enough to check the criterion for spheres; you get honest contractibility if $EG$ belongs to the class of admissible base spaces.