Let $G$ be a nice topological group and $E\to B$ a universal $G$-bundle. I'm interested in a proof of contractibility of $E$ using only the universal property of it. I also know that if there is a contractible $G$-bundle, then all of others are also contractible, but does there exist a direct proof not using a special construction of contractible universal $G$-bundles?

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    $\begingroup$ If $B$ represents the functor «$X$ → principal $G$-bundles on $X$», then $E$ represents the functor «$X$ → principal $G$-bundles on $X$ with a fixed section». This functor is trivial, so $E$ is contractible. $\endgroup$ – Grigory M Jun 4 '15 at 21:57
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    $\begingroup$ @GrigoryM I know that a map to $E$ gives a section of a $G$-bundle but I'm having trouble to prove your correspondence for homotopy classes of maps. Could you please write a complete answer? thanks! $\endgroup$ – Mostafa Jun 5 '15 at 18:43
  • $\begingroup$ Nobody has an answer for this important fact? $\endgroup$ – Eduardo Longa Jun 10 at 18:49

First note that the projection $\pi: E \rightarrow B$ is nullhomotopic by the universal property of $B$ since the pullback of a principal bundle by its projection is always trivial.

It follows from clutching and the loop-suspension adjunction that both $\Omega B$ and $G$ represent $G$-bundles over the suspension of your space.

This means that we can use the following rectangle which is commutative up to homotopy (formatting taken from https://mathoverflow.net/a/132340/134512) . $$\begin{matrix} G & \to & E & \to & B \\ \downarrow \simeq && \downarrow * && \downarrow = \\ \Omega B & \to & PB & \to & B \end{matrix}$$

This square induces a morphism of long exact sequences, which by the five lemma is an isomorphism from the homotopy groups of $B$ to the trivial group, as desired.


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