Suppose $S$ is a closed, connected, oriented surface of genus at least two and $G$ is any group. Suppose further that $\Gamma$ is any group that fits into the following short exact sequence:
$$ 1 \to \pi\_1 S \to \Gamma \to G \to 1. $$


The claim is that groups $\Gamma$ correspond exactly to fundamental groups of $S$-bundles, but I am a little unclear as to why. I'm a little rusty on my algebraic topology, so can someone explain why this is, or point me to a resources that explains this?



Surface bundles associated to a short exact sequence of groups
(1) It is not true that these groups are precisely the fundamental groups of $S$-bundles. The correct statement is that these groups are precisely the fundamental groups of *$S$-bundles over a base space with $\pi\_2 = 0$.* (For instance, $U(2)$ is a torus bundle over $S^2$, but its fundamental group is $\Bbb Z$, which certainly does not contain $\pi\_1 T^2$.)


(2) If $S \to E \to B$ is a fiber bundle and $\pi\_2 B = 0$ (and $S$ is a connected surface) then the long exact sequence of a fibration shows that $\pi\_1 E$ fits into the desired short exact sequence.


(3) Given an extension as above, construct the classifying space $BG=EG/G$ and $B\Gamma=E\Gamma/\Gamma$ by whatever procedure you like (here $EG$ is a contactible space so that $p: EG \to BG$ is a covering map.) Write $f: \pi\_1 S \to \Gamma$ for the group homomorphism outlined above. Since $f: \pi\_1 S \to \Gamma$ is injective, consider $$X = \widetilde S \times\_{\pi\_1 S} (E\Gamma \times EG),$$ acting on the latter spaces via $f$ (and so the last space trivially); this retains an action of $G = \Gamma/\pi\_1 S$. Now consider $Y = X/G$. This has a natural projection map to $(E\Gamma \times EG)/G \simeq BG$, and fiber $\widetilde S/\pi\_1 S = S.$ We have thus constructed an $S$-bundle over a space with $\pi\_1 = G$.


Intuitively the point is that whenever you have a short exact sequence of groups you can deloop this to a fibration of classifying spaces. $S$ is a classifying space for $\pi\_1 S$. This is only well-defined up to homotopy equivalence (whereas "bundle" is a notion about raw topological spaces) so we have to do some fiddling to replace what might look like a fiber bundle with fiber a "very large, but homotopy equivalent" version of S with an actual honest copy of S. (The naive approach would have given you fiber $\widetilde S \times\_{\pi\_1 S} E\Gamma$.)


