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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.2011-04-24
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    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.2011-04-24
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    @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?2011-04-24
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    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

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