If $$ 0 \to H \to G \to G/H \to 0\ $$ is a group extension, under what conditions do we have a fibration of the form $$ BH \to BG \to B(G/H), $$ where $BG$ is a classifying space of $G$? Suppose further that we know in the fibration $$ BH \to BG \to B(G/H), $$ if $BH$ and $B(G/H)$ are closed manifolds, can we conclude that $BG$ is also a closed manifold? If not, what condition(s) can we add to make it true?
Many thanks in advance for your time.
My answer involves nothing flashy. You just have to compare the homotopy long exact sequence of a fibration with the locally-trivial fibre bundle condition.
In particular, if
$$ F \to E \to B $$
is a fibration, there is a map
$\pi_1 B \to \pi_0 HomEq(F)$. This comes from the definition of a Serre Fibration, just like how the homotopy long-exact sequence follows from the definition of a fibration. Specifically, let $p : E \to B$ be the bundle map. Consider the inclusion of $F$ in $E$, and the homotopy $H : F \times [0,1] \to B$ given by $(x,t) \longmapsto \gamma(t)$. This is a homotopy of $p$ applied to the above inclusion, provided $\gamma : [0,1] \to B$ is a path starting at the basepoint of $B$. By the homotopy extension property there exists a map $\tilde H : F \times [0,1] \to E$ such that $\tilde H(x,0) = x$ for all $x \in F$ and $p(F(x,t)) \in p^{-1}\{\gamma(t)\}$ for all $t$. So if the path $\gamma$ is a closed loop, you can check this is a homotopy-equivalence of the fibre.
Now just check that if your fibration is a smooth fibre bundle, this map is actually a map $\pi_1 B \to \pi_0 Diff(F)$.