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