I have a question about the proof of the existence of the exterior derivative that is found in Grothendieck's EGA IV.16.6. Let me say that I know how to construct it (there is an excellent answer in this question: The algebraic de Rham complex) but I don't understand the construction in EGA.
Let $A$ be a ring and $B$ an $A$-algebra. The existence of the exterior derivative follows if I can construct an $A$-linear homomorphism of modules $$ u : \Omega_{B/A}^1 \rightarrow \Omega_{B/A}^2 $$ such that $u(g\cdot df) = dg \wedge df$.
EGA then claims (this is the problematic part) that it suffices to construct an $A$-linear homomorphism $$ v : B \otimes_A \Omega_{B/A}^1 \rightarrow \Omega_{B/A}^2 $$ such that $ v(g \cdot \omega) = dg \wedge \omega$, where $g \in B$ and $\omega \in \Omega_{B/A}^1$.
The rest of the argument is clear, however I don't see how contructing $v$ implies the existence of $u$. Furthermore, it looks like $v$ is missing a term. I would expect that $$ v(g \cdot \omega) = dg \wedge \omega + g \cdot d\omega.$$
If you want to see the original then here is an English translation: https://fppf.site/ega/ega4-auto.pdf page 27, equation (16.6.2.4).
Any help is very appreciated.
