Let $A$ be a $k$-algebra with underlying vector space $V$ and let $F_1:V\times V\to V$ be a bilinear map. Let $A:V\times V\times V\to V$ be the associator of $F_1$, i.e.
$A(a,b,c)=F_1(F_1(a,b),c)-F_1(a,F_1(b,c))$.
$F_1$ can be thought as an element of the Hochschild complex, in particular, $F_1\in C^2(A,A)=\hom_k(A\otimes A,A)$. According to this paper (Chapter I, section 2, page 64), the associator of $F_1$ is an element of $Z^3(A,A)$ whenever $F_1\in Z^2(A,A)$. This means that if $\delta F_1=0$, then $\delta A=0$, where
$\delta F_1 (a,b,c)=aF_1(b,c)-F_1(ab,c)+F_1(a,bc)-F_1(a,b)c$
and
$\delta A(a,b,c,d)=aA(b,c,d)-A(ab,c,d)+A(a,bc,d)-A(a,b,cd)+A(a,b,c)d$.
But by definition
$\delta A(a,b,c,d)=a[F_1(F_1(b,c),d)-F_1(b,F_1(c,d))]-[F_1(F_1(ab,c),d)-F_1(ab,c),d)]+[F_1(F_1(a,bc),d)-F_1(a,F_1(bc,d))]-[F_1(F_1(a,b),cd)-F_1(a,F_1(b,cd))]+[F_1(F_1(a,b),c)-F_1(a,F_1(b,c))]d$
If $F_1$ was $A$-bilinear and not only $k$-bilinear, then I could simplify to get $\delta A(a,b,c,d)=F_1(\delta F_1(a,b,c),d)+A(a,b,c)d$. But I still have that term $A(a,c,d)d$ which doesn't depend on $\delta F_1$. And this is asuming more than I can. I have no idea what to do if $F_1$ is only $k$-bilinear.
What am I missing?