# Homotopy equivalence and universal $G$-bundle (Showing total space is contractible then the bundle is universal)

I am trying to show if the total space of a principal $$G$$-bundle is contractible, then the bundle is universal. I am across this problem:

$$\require{AMScd}\begin{CD} E@<\simeq<\simeq>> EG\\ @VVV@VVV @VVV\\ B@<\simeq<<(E\times EG)/G@>\simeq>> BG \end{CD}$$\

If we have such a commutative diagram, such that $$EG\to BG$$ is a universal $$G$$-bundle and $$E\times EG\to (E\times EG)/G$$, $$E\to B$$ are principal $$G$$-bundles with contractible fibers, and the horizontal arrows are homotopy equivalence, then can we show $$E\times EG\to (E\times EG)/G$$, $$E\to B$$ are universal $$G$$-bundles?

First let's show, that if $$W \simeq BG$$ through $$\phi$$, the function $$[X,W] \rightarrow P$$ given by $$[f] \rightarrow f^*(\phi^*(EG))$$ is bijective and natural, where $$P$$ is the set of all principal G-bundles.

This should be a general categorical fact:

Imagine we have a category $$C$$ and above each object $$c \in C$$ there is a set $$P_C$$ (think isomorphism classes of principal bundles over a space). Further, a map $$c \rightarrow c'$$ induces a map $$P_{c'} \rightarrow P_c$$. There is an obvious functor $$F$$ taking $$c \rightarrow P(c)$$. Finally, we have that an object $$s$$ represents $$F$$ by way of pulling back a distinguished element $$\gamma$$ of $$P(s)$$.

We would like to say that given an isomorphism $$\Phi:s' \rightarrow s$$, $$\Phi^*(\gamma)$$ is a distinguished element of $$P(s')$$ that pulling back by yields a functor naturally isomorphic to $$F$$.

Surjectivity is easy: suppose $$x \in P(c)$$. Let $$f: c \rightarrow s$$ be the map that gives rise to $$x$$. The composition $$\Phi^{-1} f$$ is the map we are after.

Injectivity is also easy: suppose $$g^*(s')=h^*(s')$$ then writing out what this means we see $$g^*(\Phi^*(s))=h^*(\Phi^*(s))$$ which implies $$\Phi g = \Phi h$$ which implies the $$g=h$$.

Naturality also comes out easily. Note, this is really just the Yoneda lemma when one recognizes that the universal bundle is the pullback of the identity on $$BG$$.

Now note that the isomorphism type of the principal bundle depends only on the homotopy type of the map you pull back by. This means we can use the above fact to show that any space homotopy equivalent to a classifying space is another classifying space with classifying bundle the pull back.

The only thing left to check is that given given a commutative square of principle bundles where the top maps are equivariant, the pullback of the map between orbit space is isomorphic to the bundle you started out with. If the map between principal bundles is $$L$$ an isomorphism is given by $$x \rightarrow ([x],L(x))$$, this does not actually depend on the map being a homotopy equivalence.

For a bundle to universal, you just need the total space to be contractible, if you assume that $$E\times EG$$ is contactible, so is $$E$$ therefore, all the bundles designated above are universal.

https://ncatlab.org/nlab/show/universal+principal+bundle

• I am actually proving the iff condition, i.e. i am trying to show if the total space is contractible, then it's universal. So I can't apply the result you mentioned
– 6666
Commented Jul 20, 2019 at 20:24
• There is a short proof in "Notes on principal bundles and classifying spaces" by Stephen Mitchell on page 13. Since we are dealing with CW complexes, it is also enough to assume weakly contractible. His proof is pretty simple- it relies on some basic properties of Serre fibrations and the interaction between sections of a certain bundle and morphisms from a principal G-bundle into a G-space. Commented Jul 21, 2019 at 0:40
• @ConnorMalin In the proof of the notes you mentioned why does weakly contractible fibre imply it admits a section?
– 6666
Commented Jul 21, 2019 at 4:48
• Weakly contractible fiber implies a (weak) homotopy equivalence that is also a fibration. These have lifts because you can draw a lifting diagram with the identity on the bottom and the acyclic fibration on the top and properties of cofibrant objects and acyclic fibrations implies you can lift the map (which means a section). Commented Jul 21, 2019 at 6:29
• @Connor Malin for the diagram you described, did you mean the pullback of the identity along the fibration(fiber bundle map)? And if so, what’s your cofibrant object in the diagram?
– 6666
Commented Jul 26, 2019 at 23:55