# Why universal G-bundles are contractible?

by Mostafa   Last Updated August 14, 2019 06:20 AM

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?

Tags :

Here are two different sketches-- I believe the second is easily completed:

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.

This implies that the long exact sequence of the fibration decomposes into short exact sequences $$0 \rightarrow \pi_{n+1}(B) \rightarrow \pi_n(G) \rightarrow \pi_n(E) \rightarrow 0$$.

The map $$\pi_{n+1}(B) \rightarrow \pi_n(G)$$ being surjective implies the total space is weakly contractible, as desired.

This map is surjective thanks to the following construction:

Given a map $$S^n \rightarrow G$$ we may use it to construct a $$G$$-bundle over $$S^{n+1}$$ by clutching together two copies of the space $$D^{n+1} \times G$$ along $$S^n \times G$$ according to our map. By the universal property, this bundle corresponds to a map $$S^{n+1}$$ into our space, which in a just world would be the element we are after.

Here is the second:

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 implies that they represent $$G$$-bundles over the spheres which by the cofiber sequence/clutching implies they both represent $$G$$-bundles over the suspension for all CW complexes.

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

This square induces a morphism of long exact sequences which by the five lemma is an isomorphism, which implies that $$E$$ is weakly contractible.

Connor Malin
August 14, 2019 05:46 AM

## Related Questions

Updated July 25, 2018 03:20 AM

Updated May 10, 2019 21:20 PM

Updated July 16, 2019 14:20 PM

Updated February 24, 2016 04:08 AM