Kähler differentials commute with localization

The canonical map. Since $\phi (V)\subseteq U$, [F4] extends $A\to U^{-1}B$ uniquely to a ring map $V^{-1}A\to U^{-1}B$, and $\lambda \text{\hspace{0.17em}⁣}:B\to U^{-1}B$ is an $A$-algebra homomorphism, and the composite $B\stackrel{\lambda }{\to }{U}^{-1}B\stackrel{{\mathrm{d}}^{\mathrm{\prime }}}{\to }{\mathrm{\Omega }}_{U^{-1}B\mathrm{/}{V}^{-1}A}$ is an $A$-derivation of $B$ into ${\mathrm{\Omega }}_{U^{-1}B\mathrm{/}{V}^{-1}A}$: it is additive, kills $\phi (A)$, and satisfies Leibniz. By [F1] it corresponds to a $B$-linear $\rho \text{\hspace{0.17em}⁣}:{\mathrm{\Omega }}_{B\mathrm{/}A}\to {\mathrm{\Omega }}_{U^{-1}B\mathrm{/}{V}^{-1}A}$ with $\rho (\mathrm{d}b)={\mathrm{d}}^{\mathrm{\prime }}(b\mathrm{/}1)$, and by [F3] applied to the canonical map ${\lambda }_{\mathrm{\Omega }}\text{\hspace{0.17em}⁣}:{\mathrm{\Omega }}_{B\mathrm{/}A}\to U^{-1}{\mathrm{\Omega }}_{B\mathrm{/}A}$ the map $\rho$ factors uniquely through a $U^{-1}B$-linear map $\alpha \text{\hspace{0.17em}⁣}:U^{-1}{\mathrm{\Omega }}_{B\mathrm{/}A}\to {\mathrm{\Omega }}_{U^{-1}B\mathrm{/}{V}^{-1}A}$ with $\alpha (\omega \mathrm{/}u)=(1\mathrm{/}u)\rho (\omega )$; in particular $\alpha (\mathrm{d}b\mathrm{/}u)={\mathrm{d}}^{\mathrm{\prime }}(b\mathrm{/}1)\mathrm{/}u$. This is the canonical map of the statement.

A derivation of the localization. Let $E:=U^{-1}B\oplus U^{-1}{\mathrm{\Omega }}_{B\mathrm{/}A}$ with the product $(x,\omega )(x^{\mathrm{\prime }},{\omega }^{\mathrm{\prime }})=(x{x}^{\mathrm{\prime }},x{\omega }^{\mathrm{\prime }}+x^{\mathrm{\prime }}\omega )$, a commutative ring in which the second summand is an ideal of square zero, and let $s\text{\hspace{0.17em}⁣}:B\to E$, $s(b):=(b\mathrm{/}1,\mathrm{d}b)$. Then $s$ is a unital ring homomorphism: it is additive, and multiplicativity is exactly the Leibniz rule $\mathrm{d}(b{b}^{\mathrm{\prime }})=b\mathrm{d}{b}^{\mathrm{\prime }}+b^{\mathrm{\prime }}\mathrm{d}b$ of [F5]. For $u\in U$ the element $s(u)=(u\mathrm{/}1,\mathrm{d}u)$ is a unit of $E$ with inverse $(1\mathrm{/}u,-\mathrm{d}u\mathrm{/}{u}^2)$, since $(u\mathrm{/}1)(-\mathrm{d}u\mathrm{/}{u}^2)+(1\mathrm{/}u)\mathrm{d}u=-\mathrm{d}u\mathrm{/}u+\mathrm{d}u\mathrm{/}u=0$. By [F4] there is a unique unital ring homomorphism $\tilde{s}\text{\hspace{0.17em}⁣}:U^{-1}B\to E$ with $\tilde{s}\circ \lambda =s$; writing $\tilde{s}(x)=(s_1(x),D(x))$, the first coordinate $s_1$ is a unital ring homomorphism $U^{-1}B\to U^{-1}B$ with $s_1(\lambda (b))=b\mathrm{/}1$, so $s_1=\mathrm{i}\mathrm{d}$ by the uniqueness clause of [F4] applied to the identity. Hence $\tilde{s}(x)=(x,D(x))$.

The second coordinate is a derivation. Multiplicativity of $\tilde{s}$ in the square-zero extension gives $D(x{x}^{\mathrm{\prime }})=xD(x^{\mathrm{\prime }})+x^{\mathrm{\prime }}D(x)$, and additivity of $\tilde{s}$ gives $D(x+x^{\mathrm{\prime }})=D(x)+D(x^{\mathrm{\prime }})$. For $a\in A$ we have $\tilde{s}(\lambda (\phi (a)))=s(\phi (a))=(\phi (a)\mathrm{/}1,0)$, so $D$ kills $\lambda \circ \phi (A)$; since $D$ also kills $\lambda (\phi (v))$ for $v\in V$ and $D$ satisfies Leibniz with $D(1)=0$, it kills the inverse of each such unit, hence the image of $V^{-1}A\to U^{-1}B$. So $D$ is a $V^{-1}A$-derivation of $U^{-1}B$ into the $U^{-1}B$-module $U^{-1}{\mathrm{\Omega }}_{B\mathrm{/}A}$, and by [F1] it corresponds to a $U^{-1}B$-linear map $\beta \text{\hspace{0.17em}⁣}:{\mathrm{\Omega }}_{U^{-1}B\mathrm{/}{V}^{-1}A}\to U^{-1}{\mathrm{\Omega }}_{B\mathrm{/}A}$ with $\beta ({\mathrm{d}}^{\mathrm{\prime }}x)=D(x)$.

The two maps are inverse. For $b\in B$ and $u\in U$, multiplicativity of $\tilde{s}$ gives $D(b\mathrm{/}u)=D(\lambda (b)\cdot (1\mathrm{/}u))=(1\mathrm{/}u)D(b)+(b\mathrm{/}1)D(1\mathrm{/}u)$, and $D(1\mathrm{/}u)=-\mathrm{d}u\mathrm{/}{u}^2$ because $0=D(1)=D((u\mathrm{/}1)(1\mathrm{/}u))=(1\mathrm{/}u)D(u)+(u\mathrm{/}1)D(1\mathrm{/}u)$ with $D(u)=\mathrm{d}u$ from $\tilde{s}(\lambda (u))=(u\mathrm{/}1,\mathrm{d}u)$; hence $D(b\mathrm{/}u)=(u\mathrm{d}b-b\mathrm{d}u)\mathrm{/}{u}^2$. Therefore $\alpha (\beta ({\mathrm{d}}^{\mathrm{\prime }}(b\mathrm{/}u)))=\alpha ((u\mathrm{d}b-b\mathrm{d}u)\mathrm{/}{u}^2)=(u{\mathrm{d}}^{\mathrm{\prime }}(b\mathrm{/}1)-b{\mathrm{d}}^{\mathrm{\prime }}(u\mathrm{/}1))\mathrm{/}{u}^2={\mathrm{d}}^{\mathrm{\prime }}(b\mathrm{/}u)$, the last equality being the derivation identity for the fraction $b\mathrm{/}u$ with $u$ invertible, obtained from the Leibniz rule and ${\mathrm{d}}^{\mathrm{\prime }}(u\cdot (1\mathrm{/}u))=0$. Since the elements ${\mathrm{d}}^{\mathrm{\prime }}(b\mathrm{/}u)$ generate ${\mathrm{\Omega }}_{U^{-1}B\mathrm{/}{V}^{-1}A}$ over $U^{-1}B$, this gives $\alpha \circ \beta =\mathrm{i}\mathrm{d}$. Conversely $\beta (\alpha (\mathrm{d}b\mathrm{/}u))=\beta ({\mathrm{d}}^{\mathrm{\prime }}(b\mathrm{/}1)\mathrm{/}u)=(1\mathrm{/}u)D(b\mathrm{/}1)=\mathrm{d}b\mathrm{/}u$ for all $b\in B$, $u\in U$, and the elements $\mathrm{d}b\mathrm{/}u$ generate $U^{-1}{\mathrm{\Omega }}_{B\mathrm{/}A}$ over $U^{-1}B$ by [F2], so $\beta \circ \alpha =\mathrm{i}\mathrm{d}$. Hence $\alpha$ is an isomorphism. Taking $V=\{1\}$ gives $V^{-1}A=A$ and taking $U=\{1\}$ makes $\lambda$ the identity, so both degenerate cases are covered by the same computation.