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Let $G$ be a group, either a Lie group or a discrete group. Let a principal $G$-bundle $$ G\to E\to B,$$ then $B=E/G$, the orbit space under action of $G$.

Let $BG$ be the classifying space of $G$.

My question:

How to get the fiber sequence

$G\simeq \Omega BG\to E\to B\to BG$?

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1 Answer 1

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The bundle $E \rightarrow B$ is classified by a map $B \rightarrow BG$, which means that we have a pullback $$ \require{AMScd} \begin{CD} E @>{}>> EG\\ @VVV @VVV \\ B @>{}>> BG \end{CD} $$ As $EG$ is contractible, this gives us a fiber sequence $E \rightarrow B \rightarrow BG$. As $E \rightarrow B$ is a $G$-principal bundle, its fiber is $G$ which is homotopy equivalent to $\Omega BG$. This can be seen from the fibrations $G \rightarrow EG \rightarrow BG$ and $\Omega BG \rightarrow PBG \rightarrow BG$ where $PBG$ is contractible.

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