Transitivity sequence for differential modules

The first map. The composite $B\to C\stackrel{{\mathrm{d}}_{C\mathrm{/}A}}{\to }{\mathrm{\Omega }}_{C\mathrm{/}A}$ is an $A$-derivation of $B$ into the $C$-module ${\mathrm{\Omega }}_{C\mathrm{/}A}$; by [F1] it corresponds to a $B$-linear map ${\mathrm{\Omega }}_{B\mathrm{/}A}\to {\mathrm{\Omega }}_{C\mathrm{/}A}$ with $\mathrm{d}b\mapsto {\mathrm{d}}_{C\mathrm{/}A}(b)$. Its extension of scalars along $B\to C$ is the $C$-linear map $\gamma \text{\hspace{0.17em}⁣}:C{\otimes }_B{\mathrm{\Omega }}_{B\mathrm{/}A}\to {\mathrm{\Omega }}_{C\mathrm{/}A}$ with $\gamma (c\otimes \mathrm{d}b)=c{\mathrm{d}}_{C\mathrm{/}A}(b)$.

The second map. The universal $B$-derivation ${\mathrm{d}}_{C\mathrm{/}B}\text{\hspace{0.17em}⁣}:C\to {\mathrm{\Omega }}_{C\mathrm{/}B}$ is also an $A$-derivation, so [F1] applied to $A\to C$ gives a $C$-linear map $\delta \text{\hspace{0.17em}⁣}:{\mathrm{\Omega }}_{C\mathrm{/}A}\to {\mathrm{\Omega }}_{C\mathrm{/}B}$ with $\delta ({\mathrm{d}}_{C\mathrm{/}A}(c))={\mathrm{d}}_{C\mathrm{/}B}(c)$. It is surjective because the elements ${\mathrm{d}}_{C\mathrm{/}B}(c)$ generate ${\mathrm{\Omega }}_{C\mathrm{/}B}$ over $C$ by [F2].

The composite vanishes. For $c\in C$ and $b\in B$, $\delta (\gamma (c\otimes \mathrm{d}b))=c{\mathrm{d}}_{C\mathrm{/}B}(b)=0$, since ${\mathrm{d}}_{C\mathrm{/}B}$ is $B$-constant: $b$ is the image of an element of $B$, so ${\mathrm{d}}_{C\mathrm{/}B}(b)=0$ in the definition of a $B$-derivation of $C$. Hence there is an induced $C$-linear map $\bar{\delta }\text{\hspace{0.17em}⁣}:Q\to {\mathrm{\Omega }}_{C\mathrm{/}B}$ out of $Q:=\mathrm{coker}\gamma$, and it is surjective by step 1.2.

A left inverse for $\bar{\delta }$. The map $D\text{\hspace{0.17em}⁣}:C\to Q$ sending $c$ to the class of ${\mathrm{d}}_{C\mathrm{/}A}(c)$ is the composite of the $A$-derivation ${\mathrm{d}}_{C\mathrm{/}A}$ with the $C$-linear quotient map, hence an $A$-derivation, and it kills $B$ because ${\mathrm{d}}_{C\mathrm{/}A}(b)$ is the class of $\gamma (1\otimes \mathrm{d}b)$, which is zero in $Q$. As $D$ is $A$-linear and kills $B$, it satisfies $D(bc)=bD(c)$ for $b\in B$, $c\in C$ by the Leibniz rule, so $D$ is a $B$-derivation; [F1] applied to the ring map $B\to C$ gives a $C$-linear map $\mathrm{\ell }\text{\hspace{0.17em}⁣}:{\mathrm{\Omega }}_{C\mathrm{/}B}\to Q$ with $\mathrm{\ell }({\mathrm{d}}_{C\mathrm{/}B}(c))=[{\mathrm{d}}_{C\mathrm{/}A}(c)]$.

$\mathrm{\ell }$ is inverse to $\bar{\delta }$. For all $c\in C$ we have $\bar{\delta }(\mathrm{\ell }({\mathrm{d}}_{C\mathrm{/}B}(c)))=\bar{\delta }([{\mathrm{d}}_{C\mathrm{/}A}(c)])={\mathrm{d}}_{C\mathrm{/}B}(c)$ and $\mathrm{\ell }(\bar{\delta }([{\mathrm{d}}_{C\mathrm{/}A}(c)]))=\mathrm{\ell }({\mathrm{d}}_{C\mathrm{/}B}(c))=[{\mathrm{d}}_{C\mathrm{/}A}(c)]$ by the defining property of $\mathrm{\ell }$ in step 3.1. The elements ${\mathrm{d}}_{C\mathrm{/}B}(c)$ generate ${\mathrm{\Omega }}_{C\mathrm{/}B}$ and the classes $[{\mathrm{d}}_{C\mathrm{/}A}(c)]$ generate $Q$ over $C$ by [F2], so $\bar{\delta }\circ \mathrm{\ell }=\mathrm{i}\mathrm{d}$ and $\mathrm{\ell }\circ \bar{\delta }=\mathrm{i}\mathrm{d}$. Hence $\bar{\delta }$ is an isomorphism, $\ker \delta =\mathrm{im}\gamma$, and with $\delta$ surjective the displayed sequence is exact.