The factor set satisfies the two-cocycle equation

Relation [F1] at $q=1$ reads $P(1)P(r)=\alpha (1,r)P(r)$, that is $P(r)=\alpha (1,r)P(r)$ by [A1]; multiplying by the inverse of $P(r)$ gives $\alpha (1,r)=1$. Symmetrically, [F1] at $r=1$ gives $P(q)P(1)=\alpha (q,1)P(q)$, so $\alpha (q,1)=1$.

Multiplying the relation of [F1] on the left by $P(q)$ and applying it twice, $P(q)(P(r)P(s))=\alpha (r,s)P(q)P(rs)=\alpha (r,s)\alpha (q,rs)P(qrs)$. Applying it in the other order, $(P(q)P(r))P(s)=\alpha (q,r)P(qr)P(s)=\alpha (q,r)\alpha (qr,s)P(qrs)$.

By associativity in $\mathrm{GL}(V)$, the two expressions of step 1.2 are equal; since $P(qrs)$ is invertible, cancelling it gives $\alpha (q,r)\alpha (qr,s)=\alpha (r,s)\alpha (q,rs)$.

Reading the additive data of [F2] in multiplicative notation for the abelian group ${\mathbb{C}}^{\times }$ with trivial $Q$-action — sums become products, negatives become inverses and $0$ becomes $1$ — the cocycle equation $g\cdot f(h,k)-f(gh,k)+f(g,hk)-f(g,h)=0$ becomes $\alpha (h,k)\alpha (g,hk)=\alpha (gh,k)\alpha (g,h)$, which after renaming $(g,h,k)$ as $(q,r,s)$ is the equation of step 2.1, and $f(1,g)=f(g,1)=0$ becomes $\alpha (1,q)=\alpha (q,1)=1$. Thus $\alpha$ is a normalized two-cocycle on $Q$ with values in ${\mathbb{C}}^{\times }$ in the multiplicative form of the published convention.