4
$\begingroup$

For principal $G$-bundles with $G$ a Lie group there exists a principal $G$-bundle $EG \to BG$ such that we have a bijection [X,BG] \leftrightarrow \text{(principal $G$-bundles over X)}
$ f \mapsto f^* EG $ where $[X,BG]$ is the set of homotopy classes of maps from $X$ to $BG$. As a result of this, homotopic maps induce the same pullback maps of bundles.

My question is the following: for what class of spaces $F$ does there exist $F \to EF \to BF$ that gives a correspondence as above. I am also interested in knowing for what type of $F$ homotopic maps induce the same pullback.

Let's also assume all spaces are (countable) CW complexes.

  • 0
    I just looked it up, and the proof only requires the maps $f, g : A\to B$ that are homotopic to have the property that $A$ is paracompact. The fiber bundle over $B$ that you are pulling back can be anything.2011-04-24

2 Answers 2

3

This seems to be the only answer I ever give these days... Peter May wrote a memoir called Classifying spaces and Fibrations that might be what you are looking for. He writes down spaces that play the role of $BG$ but for "arbitrary" fibrations! (arbitrary is in quotes because there are topological restrictions which I think you will be fine accepting, I believe that if the homotopy fiber has the homotopy type of a non-degenerately based CW complex the results go through). You don't need local triviality or anything.

1

Pullback gives the bijection between $[X,BG]$ and isomorphism classes of bundles on X with structure group G. In particular, all bundles on X with fiber F is the same thing as [X,BAut(F)] (where Aut is the group of autohomeomorphisms of F).

  • 0
    @archipelago Indeed, the action of $\operatorname{Aut}F$ on $F$ is not (in general) free. And?.. (Example: certainly action of $GL(V)$ on $V$ is far from being free, but categories of principle $GL$-bundles and vector bundles are equivalent etc.)2014-01-10