The diagonal ideal modulo its square is Omega

The class map is a derivation. Put $j_b:=1\otimes b-b\otimes 1\in J$ and $D(b):=[j_b]\in J\mathrm{/}{J}^2$. For $b,c\in B$, expansion in $B{\otimes }_{A}B$ gives $j_{bc}-[(b\otimes 1)j_c+(c\otimes 1)j_b]=j_b{j}_c\in J^2$, so $D(bc)=bD(c)+cD(b)$. The factors $b\otimes 1$ and $1\otimes b$ act identically on $J\mathrm{/}{J}^2$, since their difference lies in $J$ and $J(J\mathrm{/}{J}^2)=0$. Also $D$ is additive and $D(a)=0$ for $a\in A$, since $1\otimes a=a\otimes 1$. Hence $D$ is an $A$-derivation into the $B$-module $J\mathrm{/}{J}^2$, and [F1] gives a unique $B$-linear $\alpha \text{\hspace{0.17em}⁣}:{\mathrm{\Omega }}_{B\mathrm{/}A}\to J\mathrm{/}{J}^2$ with $\alpha (\mathrm{d}b)=D(b)$.

$\alpha$ is surjective. Every $x={\sum }_i{b}_i\otimes c_i\in J$ satisfies ${\sum }_i{b}_i{c}_i=0$, hence $x={\sum }_i(b_i\otimes 1)(1\otimes c_i-c_i\otimes 1)={\sum }_i(b_i\otimes 1)j_{c_i}$ in $B{\otimes }_{A}B$, using $(b_i\otimes 1)(c_i\otimes 1)=b_i{c}_i\otimes 1$ and ${\sum }_i{b}_i{c}_i\otimes 1=0$. Thus $J$ is generated as a left $B$-module by the $j_c$, and $J\mathrm{/}{J}^2$ is generated by their classes $D(c)$, which lie in the image of $\alpha$. Hence $\alpha$ is surjective.

A left inverse. The assignment $(b,c)\mapsto b\mathrm{d}c$ is $A$-bilinear, so [F2] defines an $A$-linear map $\mathrm{\Psi }\text{\hspace{0.17em}⁣}:B{\otimes }_{A}B\to {\mathrm{\Omega }}_{B\mathrm{/}A}$ with $\mathrm{\Psi }(b\otimes c)=b\mathrm{d}c$. It is linear for the left $B$-action $a\cdot (b\otimes c)=(ab)\otimes c$. For $b,c\in B$, expand $j_b{j}_c=1\otimes bc-c\otimes b-b\otimes c+bc\otimes 1$. Then $\mathrm{\Psi }(j_b{j}_c)=\mathrm{d}(bc)-c\mathrm{d}b-b\mathrm{d}c+bc\mathrm{d}1=0$ by the Leibniz rule. By step 2.1, $J$ is generated as a left $B$-module by the $j_b$, so $J^2$ is generated as a left $B$-module by their pairwise products; left $B$-linearity of $\mathrm{\Psi }$ therefore gives $\mathrm{\Psi }(J^2)=0$. Restricting $\mathrm{\Psi }$ to $J$ and passing to the quotient gives a $B$-linear map $\beta \text{\hspace{0.17em}⁣}:J\mathrm{/}{J}^2\to {\mathrm{\Omega }}_{B\mathrm{/}A}$ with $\beta ([j_b])=\mathrm{d}b$.

The maps are inverse. For $b\in B$ one has $\beta (\alpha (\mathrm{d}b))=\mathrm{\Psi }(j_b)=\mathrm{\Psi }(1\otimes b-b\otimes 1)=\mathrm{d}b-b\mathrm{d}1=\mathrm{d}b$. The differentials $\mathrm{d}b$ generate ${\mathrm{\Omega }}_{B\mathrm{/}A}$, so $\beta \circ \alpha =\mathrm{i}\mathrm{d}$. Since $\alpha$ is surjective by step 2.1, it follows also that $\alpha \circ \beta =\mathrm{i}\mathrm{d}$. Thus $\alpha$ is an isomorphism. The formulas defining $j_b$ and $\mathrm{\Psi }$ commute with maps of ring homomorphisms $A\to B$, so the isomorphism is natural.