Group Extension and Classifying Space 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. 
 A: 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 $\pi_1 B \to \pi_0 Diff(F)$.    
