exterior derivation on 1-forms

Let us define an $A$ -module homomorphism

$$\psi:B\otimes_A B \to \Omega^2_{B/A}$$

by the following formula.

$$\psi(b_1\otimes b_2)=d b_1 \wedge d b_2.$$

Then we may easily see that

$$\psi((1\otimes f -f \otimes 1)(1\otimes g -g \otimes 1))=0.$$

Thus $\psi$ defines a unique $A$ -module homomorphism

$$d_1:\Omega^1_{B/A}=I_\Delta/I_\Delta^2 \ni f d g\mapsto d f \wedge d g \in \Omega^2_{B/A}$$