Consider a short exact sequence of Abelian groups — I'm happy to assume they're finite as a toy example:
$$
0 \to H \to G \to G/H \to 0\ .
$$
I want to understand the classifying space of $G$. Since $BH \cong EG/H$, $G/H$ acts on $BH$ and we can write $BG \cong E(G/H) \times_{G/H} EG/H$. Thus, we have a fiber bundle (which I'll write horizontally)
$$
BH \to BG \to B(G/H)
$$
On the other hand, the central extension is classified by an element of the group cohomology $H^2(G/H,H)$ which is the same as $H^2(B(G/H),H)$. The latter is an element in the homotopy class of maps $[B(G/H),K(H,2)]$ and $K(H,2)\cong BBH$. This map looks like it classifies a principal $BH$ bundle over $B(G/H)$. I find it hard to imagine that this 'principal' $BH$ bundle is not 'the same' as the bundle above, so the question is, how do you see that? From this construction, it's not even obvious to me that the above bundle is a principal bundle.
I would guess (and being a poor physicist, I'm not so up on my homotopy theory), there's a sense that the classifying space of an Abelian group is an 'Abelian group', and taking classifying spaces of an exact sequence gives you back an 'exact sequence'. That gets you a 'principal bundle' (aren't quotation marks fun?), but even then I'm not sure how to see that the classifying map of this bundle is the same as the class in group cohomology.
Any references to the needed background would also be greatly appreciated.
Best Answer
Yes. The principal bundles are the same and your guess that $BA$ is an abelian group is exactly right. A good reference for this story, and of Segal's result that David Roberts quotes, is Segal's paper:
G. Segal. Cohomology of topological groups, Symposia Mathematica IV (1970), 377- 387.
The functors $E$ and $B$ can be described in two steps. First you form a simplicial topological space, and then you realize this space. It is easy to see directly that $EG$ is always a group and that there is an inclusion $G \rightarrow EG$, which induces the action. The quotient is $BG$. Under suitable conditions, for example if $G$ is locally contractible (which includes the discrete case), the map $EG\rightarrow BG$ will admit local sections and so $EG$ will be a $G$-principal bundle over $BG$. This is proven in the appendix of Segal's paper, above. There are other conditions (well pointedness) which will do a similar thing.
The inclusion of $G$ into $EG$ is a normal subgroup precisely when $G$ is abelian, and so in this case $BG$ is again an abelian group.
I believe your question was implicitly in the discrete setting, but the non-discrete setting is relevant and is the subject of Segal's paper. Roughly here is the answer: Given an abelian (topological) group $H$, the $BH$-princical bundles over a space $X$ are classified by the homotopy classes of maps $[X, BBH]$. When $H$ is discrete, $BBH = K(H,2)$. If $X = K(G,1)$ for a discrete group $G$, these correspond to (central) group extensions:
$$H \rightarrow E \rightarrow G$$
If $G$ has topology, then the group extensions can be more interesting. For example there can be non-trivial group extensions which are trivial as principal bundles. Easy example exist when H is a contractible group. However Segal developed a cohomology theory which classifies all these extensions. That is the subject of his paper.