# What is relation between the path fibration of classifying map and the Borel construction

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 $$\require{AMScd} \begin{CD} P_f @>>> BG^I \\ @VVV @VVV \\ B @>{f}>> BG \end{CD}$$

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.

• Excellent answer! Thank you very much! – Aolong Li Apr 28 at 1:36
• I have some questions: 1. Should "the top left-hand square is a pullback" be "the right-hand square"? Otherwise why can we say the inclusion is $G$-bundle? 2. since we use the inverse of $q$, why can we use this when $q$ is a weak homotopy equivalence, not a homotopy equivalence? 3. why $\pi\circ q^{-1}$ is the classifying map? I cannot draw the corresponding pull-back diagram... – Aolong Li Apr 28 at 2:47
• The top left-hand square in 1 is a pullback. The pull back along an identity map is an idenitity, so the top right-hand square cannot be a pullback. The inclusion is a $G$-bundle because diagram 2 is a pulllback. In diagram 2 you have that the canonical pullback consists of elements $([x,e],y)\in (EG\times_GE)\times EG$ with $xG=yG\in BG$. The second point $y$ gives you a representative for the coset so you can uniquely write $[x,e]=[y,e']$ for some $e'\in E$, and in this way get a map $EG\times_GE\rightarrow E$, $([x,e],y)\mapsto e'$, which is a homeomorphism. Hence the square is a pullback. – Tyrone Apr 28 at 9:27
• Of course the inverse to $q$ may not exist if $q$ is not a true homotopy equivalence, which is the reason I indicated that $q$ is only a weak homotopy equivalence without further assumption. Since classifying spaces are only defined up to (weak) homotopy equivalence anyway, see Theorem 9.15, where the existence of $EG$ is characterised by it being weakly contractible. For reasonable groups you can assume that EG (and hence $B$ also) is CW. I introduce $q^{-1}$ really as an extra observation in this special case and you'll note that I don't refer to it at any other point. – Tyrone Apr 28 at 9:39
• $\pi\circ q^{-1}$ is a classifying map. These objects really only make sense up to homotopy, so it may not be a strict classifying map in the sense that $E$ actually is $(\pi\circ q^{-1})^*EG$, but rather $E$ is only bundle-isomorphic to this latter pullback. To see that it indeed is, you have that $\pi^*EG\cong EG\times_BE$ by square 2. Then since homotopy equivalences give you bundle isomorphisms you have $q^*((q^{-1})^*E)\cong E$ over $B$. – Tyrone Apr 28 at 9:44