# 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.

-
If all the groups involved are discrete, then $BG = K(G, 1)$ and you can always replace the induced map $BG \rightarrow B(G/H)$ into a fibration. The long exact sequence of homotopy groups implies that the fibre will be $BH$. –  Piotr Pstrągowski Aug 28 '13 at 23:13
@PiotrPstragowski Yes, I mean all the groups are discrete. Is there anything we can say about the manifold part then? –  Zuriel Aug 29 '13 at 3:38
If you wanted the map to be a fibre bundle, you would need that the automorphisms of $BH$ in the bundle should all be realizable by diffeomorphisms of $BH$. So this is the question of if the homomorphism $\pi_0 Diff(BH) \to \pi_0 HomEq(BH)$ admits a section. –  Ryan Budney Aug 29 '13 at 4:53
Thanks @RyanBudney! Are there any references to this topic? Would like to read more about it. –  Zuriel Aug 29 '13 at 5:22

## 1 Answer

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 $\pi_1 B \to \pi_0 Diff(F)$.

-