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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?

Mostafa
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    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. – Grigory M Jun 04 '15 at 21:57
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    @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! – Mostafa Jun 05 '15 at 18:43
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    Nobody has an answer for this important fact? – Eduardo Longa Jun 10 '19 at 18:49
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    Worth mentioning there is a nice note that I found which includes a proof of this fact... https://math.uchicago.edu/~dannyc/courses/alg_topol_2013/F_bundle_notes.pdf – ah-- Sep 19 '19 at 05:05
  • @chikurin nice reference! thanks! – Mostafa Sep 19 '19 at 08:20

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