A fibration of classifying spaces Let $G$ be a Lie group, $N$ a closed connected normal subgroup. Let $BG$, $BN$, $B(G/N)$ be the classifying spaces of $G,N$ and $G/N$. Is there a fibration $BN\to BG\to B(G/N)$ ?
It seems that such a construction is used in Atiyah, Bott: Yang-Mills equations on Riemann surfaces, formula (9.2), but I can't see how it works.
Thanks a lot!
 A: By functoriality there is a map $BG\to B(G/N)$. Let $X$ to be its homotopy fiber. Then you can get a fiber sequence
$\Omega X \to G \to G/N \to X \to BG \to B(G/N)$
using the fact that $\Omega BG = G$ for every (nice enough) topological group $G$. Then, since $N$ is the fiber of $G\to G/N$ we have that $\Omega X = N$. So we only need to show that $X$ is connected to conclude that $X=BN$. Consider then the long exact sequence of homotopy groups induced by $X\to BG\to B(G/N)$. It is
$\pi_1(BG)\to \pi_1(B(G/N))\to \pi_0(X)\to \pi_0(BG)=*$
So we just need to show that the map $\pi_1(BG)\to \pi_1(B(G/N))$ is onto. But this is obvious, since this coincides with $\pi_0(G)\to \pi_0(G/N)$, which is certainly onto.
As a final note, I think that a reference for these matters can be the book by May "Classifying spaces and fibrations".
A: This is an edited extract from a book in preparation (Bruner, Catanzaro, May)
tentatively titled Characteristic Classes and is therefore overlong for an
answer. This is similar to Denis Nardin's answer, but more bundle theoretic; he did refer to an old Memoir of mine, so I thought I'd give an answer.  Let $N$ be a closed normal subgroup
of a topological group $G$ with quotient group $K$. Let $i\colon N\longrightarrow G$
be the inclusion and $j\colon G\longrightarrow K$ the quotient map.  Let 
$EG \longrightarrow BG$ and $E K \longrightarrow B K$ be universal bundles for $G$ 
and $K$ and take $EG \longrightarrow EG/ N = B N$ to be the universal
bundle for $ N$, so that inclusion of orbits gives $Bi\colon BN \longrightarrow BG$ as a bundle with fiber $K$.  There is a map $Ej\colon  EG \longrightarrow E K$ such that
$(Ej)(yg) = (Ej)(y)j(g)$, either by a functorial construction or general principles. Then passage to orbits gives $Bj\colon BG \longrightarrow BK$. By construction this gives a bundle map [this toy does not
seem to tex \xymatrix] from the bundle  $Bi$ to the bundle $EK\longrightarrow BK$, so that $BN$ is the pullback along $Bj$ of the universal bundle for $K$. Since $EK$ is contractible, this implies that $BN\longrightarrow BG \longrightarrow BK$ is a fibration sequence.  [Further details are standard, but I'll supply if wanted.]
