Localization, base change and functoriality of differentials

For (1), let $D\text{\hspace{0.17em}⁣}:B^{\mathrm{\prime }}\to B^{\mathrm{\prime }}{\otimes }_B{\mathrm{\Omega }}_{B\mathrm{/}A}$ be given on the generating tensors of $B^{\mathrm{\prime }}=B{\otimes }_A{A}^{\mathrm{\prime }}$ by $D(b\otimes a^{\mathrm{\prime }}):=a^{\mathrm{\prime }}\cdot (1\otimes \mathrm{d}b)$. This is a well-defined additive map by the defining property of the tensor product of $A$-modules, since $(b,a^{\mathrm{\prime }})\mapsto a^{\mathrm{\prime }}(1\otimes \mathrm{d}b)$ is $A$-balanced, and it satisfies Leibniz because $(b\otimes a^{\mathrm{\prime }})(b^{\mathrm{\prime }\mathrm{\prime }}\otimes a^{\mathrm{\prime }\mathrm{\prime }})=b{b}^{\mathrm{\prime }\mathrm{\prime }}\otimes a^{\mathrm{\prime }}{a}^{\mathrm{\prime }\mathrm{\prime }}$ and $\mathrm{d}(b{b}^{\mathrm{\prime }\mathrm{\prime }})=b\mathrm{d}{b}^{\mathrm{\prime }\mathrm{\prime }}+b^{\mathrm{\prime }\mathrm{\prime }}\mathrm{d}b$; it is $A^{\mathrm{\prime }}$-constant since $D(1\otimes a^{\mathrm{\prime }})=a^{\mathrm{\prime }}(1\otimes \mathrm{d}1)=0$ and $A^{\mathrm{\prime }}$-linear as the written scalar action shows, the $A^{\mathrm{\prime }}$-algebra structure on $B^{\mathrm{\prime }}$ being the one of [F2]. By [F1] for the $A^{\mathrm{\prime }}$-algebra $B^{\mathrm{\prime }}$ there is a $B^{\mathrm{\prime }}$-linear $\psi \text{\hspace{0.17em}⁣}:{\mathrm{\Omega }}_{B^{\mathrm{\prime }}\mathrm{/}{A}^{\mathrm{\prime }}}\to B^{\mathrm{\prime }}{\otimes }_B{\mathrm{\Omega }}_{B\mathrm{/}A}$ with $\psi (\mathrm{d}(b\otimes a^{\mathrm{\prime }}))=a^{\mathrm{\prime }}(1\otimes \mathrm{d}b)$. In the other direction $b\mapsto \mathrm{d}(b\otimes 1)$ is an $A$-derivation of $B$ into the $B$-module ${\mathrm{\Omega }}_{B^{\mathrm{\prime }}\mathrm{/}{A}^{\mathrm{\prime }}}$, so [F1] gives a $B$-linear ${\mathrm{\Omega }}_{B\mathrm{/}A}\to {\mathrm{\Omega }}_{B^{\mathrm{\prime }}\mathrm{/}{A}^{\mathrm{\prime }}}$, and extension of scalars gives the $B^{\mathrm{\prime }}$-linear $\phi \text{\hspace{0.17em}⁣}:B^{\mathrm{\prime }}{\otimes }_B{\mathrm{\Omega }}_{B\mathrm{/}A}\to {\mathrm{\Omega }}_{B^{\mathrm{\prime }}\mathrm{/}{A}^{\mathrm{\prime }}}$ displayed in the statement. Both composites are $B^{\mathrm{\prime }}$-linear and fix the generators: $\phi \psi (\mathrm{d}(b\otimes a^{\mathrm{\prime }}))=\phi (a^{\mathrm{\prime }}(1\otimes \mathrm{d}b))=a^{\mathrm{\prime }}\mathrm{d}(b\otimes 1)=\mathrm{d}(b\otimes a^{\mathrm{\prime }})$, using $\mathrm{d}(1\otimes a^{\mathrm{\prime }})=0$ and Leibniz; and $\psi \phi ((b\otimes a^{\mathrm{\prime }})\otimes \mathrm{d}{b}^{\mathrm{\prime }\mathrm{\prime }})=\psi ((b\otimes a^{\mathrm{\prime }})\mathrm{d}(b^{\mathrm{\prime }\mathrm{\prime }}\otimes 1))=(b\otimes a^{\mathrm{\prime }})(1\otimes \mathrm{d}{b}^{\mathrm{\prime }\mathrm{\prime }})=(b\otimes a^{\mathrm{\prime }})\otimes \mathrm{d}{b}^{\mathrm{\prime }\mathrm{\prime }}$. Hence $\phi$ and $\psi$ are mutually inverse and $\phi$ is the isomorphism of (1).

For (2), write $B_U:=U^{-1}B$ and let $D\text{\hspace{0.17em}⁣}:B_U\to U^{-1}{\mathrm{\Omega }}_{B\mathrm{/}A}$ be $D(b\mathrm{/}u):=(1\mathrm{/}u)\mathrm{d}b-(b\mathrm{/}{u}^2)\mathrm{d}u$, an element of $U^{-1}{\mathrm{\Omega }}_{B\mathrm{/}A}$. This is well defined: if $b\mathrm{/}u=b^{\mathrm{\prime }}\mathrm{/}{u}^{\mathrm{\prime }}$ in $B_U$, there is $w\in U$ with $wx=0$ for $x:=b{u}^{\mathrm{\prime }}-b^{\mathrm{\prime }}u$, and applying $\mathrm{d}$ to $wx=0$ gives $w\mathrm{d}x=-x\mathrm{d}w$; multiplying by $u^2{u}^{\mathrm{\prime }2}$ and writing $G:=u^{\mathrm{\prime }2}(u\mathrm{d}b-b\mathrm{d}u)-u^2(u^{\mathrm{\prime }}\mathrm{d}{b}^{\mathrm{\prime }}-b^{\mathrm{\prime }}\mathrm{d}{u}^{\mathrm{\prime }})$, this yields $wG=-x(u{u}^{\mathrm{\prime }}\mathrm{d}w+wu\mathrm{d}{u}^{\mathrm{\prime }}+w{u}^{\mathrm{\prime }}\mathrm{d}u)$. In $U^{-1}{\mathrm{\Omega }}_{B\mathrm{/}A}$ the class of $x$ is zero, because $wx=0$ and $w$ becomes invertible, so the class of $wG$ is zero; since $w$ also becomes invertible as a scalar, the class of $G$ is zero, which is exactly $D(b\mathrm{/}u)=D(b^{\mathrm{\prime }}\mathrm{/}{u}^{\mathrm{\prime }})$. The map $D$ is additive and kills $V^{-1}A$; for the Leibniz rule one uses the relations implied by $\mathrm{d}(u{u}^{-1})=0$, namely $\mathrm{d}(u^{-1})=-u^{-2}\mathrm{d}u$ inside $U^{-1}{\mathrm{\Omega }}_{B\mathrm{/}A}$, after which the product rule for fractions follows from the product rule for $\mathrm{d}$. Hence $D$ is a $V^{-1}A$-derivation and [F1] for the $V^{-1}A$-algebra $B_U$ gives a $B_U$-linear $\psi \text{\hspace{0.17em}⁣}:{\mathrm{\Omega }}_{B_U\mathrm{/}{V}^{-1}A}\to U^{-1}{\mathrm{\Omega }}_{B\mathrm{/}A}$ with $\psi (\mathrm{d}(b\mathrm{/}u))=D(b\mathrm{/}u)$. Conversely $b\mapsto \mathrm{d}(b\mathrm{/}1)$ is an $A$-derivation of $B$ into ${\mathrm{\Omega }}_{B_U\mathrm{/}{V}^{-1}A}$, so [F1] gives a $B$-linear ${\mathrm{\Omega }}_{B\mathrm{/}A}\to {\mathrm{\Omega }}_{B_U\mathrm{/}{V}^{-1}A}$ with $\mathrm{d}b\mapsto \mathrm{d}(b\mathrm{/}1)$, and under the identification $U^{-1}{\mathrm{\Omega }}_{B\mathrm{/}A}\cong (U^{-1}B){\otimes }_B{\mathrm{\Omega }}_{B\mathrm{/}A}$ of [F3] the balanced assignment $(c,\omega )\mapsto c\cdot (\text{image of }\omega )$ yields the $B_U$-linear $\phi \text{\hspace{0.17em}⁣}:U^{-1}{\mathrm{\Omega }}_{B\mathrm{/}A}\to {\mathrm{\Omega }}_{B_U\mathrm{/}{V}^{-1}A}$ with $\phi ((b\mathrm{/}u)\mathrm{d}{b}^{\mathrm{\prime }})=(b\mathrm{/}u)\mathrm{d}(b^{\mathrm{\prime }}\mathrm{/}1)$. On generators, $\phi (\psi (\mathrm{d}(b\mathrm{/}u)))=\phi ((1\mathrm{/}u)\mathrm{d}b-(b\mathrm{/}{u}^2)\mathrm{d}u)=(1\mathrm{/}u)\mathrm{d}(b\mathrm{/}1)-(b\mathrm{/}{u}^2)\mathrm{d}(u\mathrm{/}1)=\mathrm{d}(b\mathrm{/}u)$, the last equality being the Leibniz rule applied to $b\mathrm{/}u=(b\mathrm{/}1)(u\mathrm{/}1{)}^{-1}$ together with $\mathrm{d}((u\mathrm{/}1{)}^{-1})=-(u\mathrm{/}1{)}^{-2}\mathrm{d}(u\mathrm{/}1)$; and $\psi (\phi ((1\mathrm{/}u)\mathrm{d}b))=\psi ((1\mathrm{/}u)\mathrm{d}(b\mathrm{/}1))=(1\mathrm{/}u)D(b\mathrm{/}1)=(1\mathrm{/}u)\mathrm{d}b$. Both composites are module maps fixing generating sets, so they are inverse and $\phi$ is the isomorphism of (2).

For (3) let $B\to C$ be an $A$-algebra map. The map $b\mapsto \mathrm{d}(1\otimes b)$ from $B$ to the $C$-module ${\mathrm{\Omega }}_{C\mathrm{/}A}$ is additive and $A$-constant and satisfies Leibniz, since $b\mapsto 1\otimes b$ is a ring homomorphism and $\mathrm{d}$ is an $A$-derivation; here $1\otimes b$ denotes the image of $b$ in $C$ when $C$ is written as a $B$-algebra. So it is an $A$-derivation, [F1] turns it into a $B$-linear map ${\mathrm{\Omega }}_{B\mathrm{/}A}\to {\mathrm{\Omega }}_{C\mathrm{/}A}$, and extension of scalars along $B\to C$ gives the $C$-linear map of (3). For $C=B{\otimes }_A{A}^{\mathrm{\prime }}$, composing this map with the canonical map ${\mathrm{\Omega }}_{C\mathrm{/}A}\to {\mathrm{\Omega }}_{C\mathrm{/}{A}^{\mathrm{\prime }}}$ gives the base-change isomorphism of step 1.1: both send $1\otimes \mathrm{d}b$ to $\mathrm{d}(b\otimes 1)$. The map before this composition need not be an isomorphism. For $C=U^{-1}B$, step 1.2 with $V=\{1\}$ does identify the functorial map with a localization isomorphism. No injectivity or surjectivity is claimed for a general $B\to C$.