Relation between the path fibration of classifying map and the Borel construction

algebraic-topologyclassifying-spacesfiber-bundleshomotopy-theory

I am trying to solve exercise 179 on Davis-Kirk:

  1. Show that given a principal $G$-bundle $E\to B$, there is a fibration
    $$E\hookrightarrow EG\times_G E\to BG$$
    where $EG\times_G E$ denotes the Borel construction.

  2. How is this fibration related to the path fibration $$P_f\to BG$$
    where $f:B\to BG$ is the classifying map of the bundle $E\to B$?

The first part is trivial since $BG$ is a CW complex and we obtain a fiber bundle by Borel construction. The statement follows from the fact any fiber bundle over paracompact space is a fibration.

However, I have no idea what exactly does the second question mean. Should we construct a morphism between these two fibrations?

P.S.

The path fibration is given by the pull-back
$$\require{AMScd}
\begin{CD}
P_f @>>> BG^I \\
@VVV @VVV \\
B @>{f}>> BG
\end{CD}
$$

Best Answer

Let $p:E\rightarrow B$ be a principal $G$-bundle and consider the Borel construction $EG\times_G E$. As you have found there is an induced map $\pi:EG\times_GE\rightarrow BG=EG/G$ with fibre $E$ obtained by projecting to the first factor. On the other hand, you can also project to the second factor to get a map

$$q:EG\times_GE\rightarrow B=E/G.$$

Since $p$ is a locally-trivial $G$-bundle the map $q$ has the structure of a locally-trivial fibre bundle with fibre $EG$. Moreover, since $EG$ is contractible, $q$ is a (weak) homotopy equivalence, which you can check by studying its long-exact sequence of homotoyp groups.

Therefore consider the following diagram $\require{AMScd}$ \begin{CD} G@>>> E @>p>> B\\ @VV V @VV V@VV=V\\ EG @>>> EG\times_GE@>q>\simeq >B\\ @VVV @VV\pi V\\ BG@>>=>BG \end{CD} which you can check commutes strictly. The top left-hand square is a pullback, which tells you that the fibre inclusion $E\hookrightarrow EG\times_GE$ of $\pi$ is a principal $G$-bundle. Since we know that $EG\times_GE\simeq B$, what we have achieved here is to construct an explicit classifying map for the bundle $p$, and at the same time have arranged for it to be a fibration. Namely the composite $\pi\circ q^{-1}:B\rightarrow BG$ is a classifying map for $E$, with $\pi$ a fibration. The point is that we have used only 'geometric' information to achieve this. Had we started with a map $f:B\rightarrow BG$ classifying $p$ and asked to understand its homotopy fibre $F_f$, this is exactly what we would have got.

To compare these two results directly choose a homotopy equivalence $\xi:EG\xrightarrow{\simeq}PBG$ and it together with the map $q$ to construct a map from the square $\require{AMScd}$ \begin{CD} E@>>> EG \\ @VV V @VV V\\ EG\times_GE @>\pi>> BG\\ \end{CD} to the square $\require{AMScd}$ \begin{CD} F_f@>>> PBG \\ @VV V @VV V\\ B @>f>> BG\\ \end{CD} Since both $q$ and $\xi$ are homotopy equivalences, the induced map of pullbacks $E\rightarrow F_f$ is a homotopy equivalence.

Related Question