I'm new to homological algebra. I try to state what I know, and ask the question at the end. I know that given a group extension $0\to K\to G\to Q\to 1$, it may have many liftings. Each lifting produce a factor set: if there are exactly 10 liftings, then there are exactly 10 factor sets. And each two factor set $f,~f'$ satisfy the relation $f-f'\in B^2(Q,K)$, so the really "meaningful"(or representative) factor set of the group extension given at the begin is only one.
On the other hand, given another extension $0\to K\to G'\to Q\to 1$ which is equivalent to the previous one (i.e. $\exists\phi:G\to G'$ such that the diagram commutes), then it can be shown that there is a factor set $f$ of $G$ and a factor set $f'$ of $G'$ such that $f-f'\in B^2(Q,K)$. But it is not true that the difference of every factor sets $f$ from $G$ and $f'$ from $G'$ will belong to $B^2(Q,K)$ right?
So how to characterize the "representative uniqueness" of the group extensions by the language of, say $f-f'$, $B^2(Q,K)$? If we define $H^2(Q,K)=Z^2(Q,K)/B^2(Q,K)$, we may hence view two factor sets that satisfying $f-f'\in B^2(Q,K)$ as the same one, but it is not naturally to use this to reflect the equivalence of group extensions. I feel weird here.