Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $BG$?

share|cite|improve this question
Re: edit. In this description, it follows that $[X, EG]$ classifies isomorphism classes of principal $G$-bundles over $X$ with a section. But these are all 1) trivial and 2) canonically isomorphic, so this is the one point set. This holds for any CW-complex $X$, so if your $EG$ is a CW-complex then $\mathrm{Id} = * \in [EG, EG]$ and it is contractible. If it is not a CW-complex you can take $X=S^n$ and deduce it is weakly contractible (there is a little basepoint subtlety here). – Oscar Randal-Williams Sep 16 '11 at 16:19
@Oscar : That's also a great answer. If it were an answer instead of a comment, I would be torn as to whether to accept your answer or Neil's answer. – Will B Sep 16 '11 at 16:25

3 Answers 3

up vote 4 down vote accepted

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.

share|cite|improve this answer
That's a beautiful answer. Thanks! – Will B Sep 16 '11 at 16:24

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.

share|cite|improve this answer
I don't think it is part of the definition of classifying spaces that $EG$ is contractible. All we know is that $[X,BG]$ is naturally in bijection with the set of principal $G$ bundles on $X$ via the map that takes $\phi : X \rightarrow BG$ to $\phi^{\ast}(EG)$. Of course, it is true that $EG$ is contractible, but I don't know how to see this except via one of the explicit constructions of $BG$. Do you know how to do this? – Will B Sep 16 '11 at 15:52
It is, indeed – Fernando Muro Sep 16 '11 at 18:07
@Fernando : Two points. First, I think that the wikipedia article is flawed in that you also have to assume that the action by $G$ is proper (otherwise, the map $EG \rightarrow BG$ might not be a principal $G$-bundle). Second, if you define it like in the wikipedia article, then it is a theorem that if it exists, then it classifies the functor "principal $G$-bundles", and thus it is not completely obvious that the object that the Brown representability theorem gives you actually satisfies the conditions of the theorem. You need some argument like that given by Oscar or Johannes above. – Will B Sep 16 '11 at 20:47
Not really, the fact that it is a principal bundle is superfluous. It's enough that EG be contractible and the action of G be free. This is what I usually take as the definition of classifying space. Now I see that you take a different one. – Fernando Muro Sep 16 '11 at 22:16
@Fernando Muro : I think the OP is correct that properness is needed. Otherwise, you would get pathological behavior. For instance, $\mathbb{Z}^2$ acts freely on $\mathbb{R}$ with $(1,0)$ acting as translation by $1$ and $(0,1)$ acting as translation by $\pi$. Of course, if $G$ is a compact group then properness is automatic. – Andy Putman Sep 17 '11 at 0:26

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.

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.