A $k$-algebra over a field $k$ is quasi-free if for any bimodule $M$ the second Hochschild cohomology $H^2(A,M)$ vanishes or, equivalently, if for any square-zero extension $R/M=A$ there is a lifting homomorphism $A\to R$.
A (seemingly) useful criterion of quasi-freeness is this: $A$ is quasi-free if for some epresentation $F/I$ of $A$ as a quotient of a free algebra the square-zero extension $$0\to I\to F/I^2\to F/I\to0$$ admits a lifting homomorphism $\zeta: F/I\to F/I^2$.
It's easy to see that a product of a quasi-free algebras is quasi free by working in terms of bimodule resolutions of the factots. But is there a way to see this through the above criterion?
Specifically, let $A=k\langle x_1,\ldots,x_n\rangle/I,\ B=k\langle y_1,\ldots,y_m\rangle/J$. Then we can get a presentation of the product $A\times B$ as a quotient of $F=k\langle x_1,\ldots,x_n,y_1,\ldots, y_m\rangle$ by the ideal $K$ generated by the relations of $I$, relations of $J$ and the relations of the form $x_iy_j,\ y_ix_j$.
If both $A$ and $B$ are quasi-free, there must be a lifting $\zeta: F/K\to F/K^2$. Is there a way to canonically construct it given the liftings $k\langle x_1,\ldots,x_n\rangle/I\to k\langle x_1,\ldots,x_n\rangle/I^2,$ $k\langle y_1,\ldots,y_m\rangle/J\to k\langle y_1,\ldots,y_m\rangle/J^2?$