The Heisenberg transgression and its sign

For triples with first two coordinates (a,b), (u,v), (x,y), the extra central terms in the two associative products are $av+(a+u)y$ and $uy+a(v+y)$, which are equal. The identity is (0,0,0) and the inverse of (a,b,c) is $(-a,-b,-c+ab)$ by multiplication on both sides. Projection onto the first two coordinates is an onto homomorphism with central kernel $N=\{(0,0,c)\}$. Thus these formulas really give the stated group extension.

The action on A is trivial and N is central, so d is a crossed homomorphism and conjugation fixes it. Choose $\alpha (a,b)=(a,b,0)$ and $\eta (a,b)=0$. They are normalized and $\alpha (q)\ast d-d=0=\delta \eta (q)$. Multiplication gives $\alpha (a,b)\alpha (u,v)=(a+u,b+v,av)$, so the factor set, as an element of N identified with $\mathbb{Z}$, is $f((a,b),(u,v))=av$. F1 yields $F=-f$, not f. These explicit maps supply every choice required for this instance of F1 and F2.

The cochain F vanishes if either input is zero. Its cocycle identity is $-uy+(a+u)y-a(v+y)+av=0$ for the three inputs in step 1.1. With trivial action on the abelian quotient, every coboundary has the form $b(q)+b(r)-b(q+r)$ and is symmetric in q,r. But $F((1,0),(0,1))=-1$, whereas $F((0,1),(1,0))=0$. Thus F is not a coboundary and its class is nonzero.