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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.

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My understanding is that homotopic maps induce the same pullback for any fiber bundles at all. I don't have a clue as to the first question though. – Jason DeVito Apr 24 '11 at 17:28
I'm not entirely sure what the exact question is. My guess is this: We know for principal G-bundles that homotopic maps induce isomorphisms on the pullback bundles, to what extent can we remove the "principal G-bundle" aspect and just replace it with an arbitrary fiber bundle? I'm pretty sure this is always true (for reasonable X). I've seen a proof for vector bundles on manifolds, but I don't think it used the "vector" part or "manifold" part. – Matt Apr 24 '11 at 17:31
@Matt: in the rank $k$ vector bundle case the base space is the Grassmannian of $k$-planes in $\mathbb R^\infty$. Does this really generalize in the proof you saw to give the base space when the fiber is arbitrary? – Eric O. Korman Apr 24 '11 at 18:18
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. – Matt Apr 24 '11 at 20:57

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.

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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).

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I should have mentioned in my question that I know what I said holds true for $G$ a Lie group. But does this still hold if $G$ is any topological group? Also, if $F$ is a CW complex is $Aut(F)$ also? – Eric O. Korman Apr 24 '11 at 18:17
@Eric Sure. (I don't think there is a specific proof for Lie groups -- so if you know a proof for Lie groups, perhaps it works for arbitrary topological groups as well.) – Grigory M Apr 24 '11 at 18:23
@Eric: As far as $Aut(F)$, I think these are pretty poorly understood. Of course by saying "automorphism" we're not really saying what we mean -- homeomorphism, or diffeomorphism, or biholomorphism? Anyways, I don't think it's at all obvious that the group of homeomorphisms of a CW complex should be another CW complex. – Aaron Mazel-Gee Apr 25 '11 at 5:57
@Aaron Oh, I missed the second question (whether Aut(F) is a CW complex -- I don't know the answer and it's certainly not obvious). – Grigory M Apr 25 '11 at 17:28
@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.) – Grigory M Jan 10 '14 at 16:18

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