My original question was:
Is there a long exact sequence for classifying spaces of topological groups?
It is solved by a comment from @JHF , since $BA$ is not usually a group. The sequence cannot go long enough in general.
The following are still puzzling to me.
I am reading the book Basic Bundle Theory and K-Cohomology Invariants (springer webpage). On page 140, there goes a sequence $$\mathbb{Z/2}\rightarrow Spin(n) \rightarrow SO(n) \xrightarrow{\sigma} B(\mathbb{Z/2})\, ,$$ where $B$ is for taking classifying sapce of a topological group, $\mathbb{Z/2}$ is a $2$-element group. I guess the map $\sigma$ is a classifying map here.
The author then claims that $\sigma$ is a group homomorphism and apply the fiber sequence again to produce $$B(\mathbb{Z/2})\rightarrow BSpin(n) \rightarrow BSO(n) \xrightarrow{B\sigma} B^2(\mathbb{Z/2}) \, . $$
The group structure of $B(\mathbb{Z/2})$ comes from the functoriality of $B$ and the Abelian group structure of $\mathbb{Z/2}$.
A similar sequence is later refered to as a classifying space sequence in that book at page 284.
There are some related posts like this, which shows that for good cases, $B$ is exact in the sense that it sends a short exact sequence to a fibration. My question is why $\sigma$ here is a group homomorphism and how it "connects" the two short sequences.
In the book Spin Geometry by Lawson and Michelsohn, a statement for the "exactness" of $B$ (page 84) seems to come from the exact sequence of cohomology groups. I also wonder why it is true and whether it gives the "connecting map" between two sequences mentioned above. I know a result that cohomology classes can be represented by mappings into Eilenberg-MacLane spaces up to homotopy.
