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 fibration of classifying spaces
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.]


