Transitivity sequence for differentials

The second map exists by [F1]: ${\mathrm{d}}_{C\mathrm{/}B}\text{\hspace{0.17em}⁣}:C\to {\mathrm{\Omega }}_{C\mathrm{/}B}$ is in particular an $A$-derivation, so it induces a $C$-linear $\pi \text{\hspace{0.17em}⁣}:{\mathrm{\Omega }}_{C\mathrm{/}A}\to {\mathrm{\Omega }}_{C\mathrm{/}B}$ with $\pi ({\mathrm{d}}_{C\mathrm{/}A}c)={\mathrm{d}}_{C\mathrm{/}B}c$. It is surjective because the elements ${\mathrm{d}}_{C\mathrm{/}B}c$ generate ${\mathrm{\Omega }}_{C\mathrm{/}B}$. The composite $\pi \circ (\text{first map})$ is zero: on the generators $c\otimes \mathrm{d}b$ of $C{\otimes }_B{\mathrm{\Omega }}_{B\mathrm{/}A}$ the first map sends $c\otimes \mathrm{d}b$ to $c{\mathrm{d}}_{C\mathrm{/}A}b$, and $\pi$ sends that to $c{\mathrm{d}}_{C\mathrm{/}B}b=0$, since $b$ lies in the image of $B$ so that $b$ is killed by the universal $B$-derivation of $C$. Hence $\mathrm{im}(\text{first map})\subseteq \ker \pi$.

For the reverse inclusion put $Q:=\mathrm{coker}(C{\otimes }_B{\mathrm{\Omega }}_{B\mathrm{/}A}\to {\mathrm{\Omega }}_{C\mathrm{/}A})$, with quotient map $q\text{\hspace{0.17em}⁣}:{\mathrm{\Omega }}_{C\mathrm{/}A}\to Q$. The $A$-derivation $C\to Q$, $c\mapsto q({\mathrm{d}}_{C\mathrm{/}A}c)$, kills $B$, since $q$ kills the image of the first map; hence by [F1] it induces a $C$-linear $\beta \text{\hspace{0.17em}⁣}:{\mathrm{\Omega }}_{C\mathrm{/}B}\to Q$ with $\beta ({\mathrm{d}}_{C\mathrm{/}B}c)=q({\mathrm{d}}_{C\mathrm{/}A}c)$. The map $\pi$ of step 1.1 kills the image of the first map, so it factors as $\pi =\gamma \circ q$ for a $C$-linear $\gamma \text{\hspace{0.17em}⁣}:Q\to {\mathrm{\Omega }}_{C\mathrm{/}B}$, and then $\gamma \beta$ is a $C$-linear endomorphism of ${\mathrm{\Omega }}_{C\mathrm{/}B}$ fixing the generators ${\mathrm{d}}_{C\mathrm{/}B}c$, so $\gamma \beta ={\mathrm{i}\mathrm{d}}_{{\mathrm{\Omega }}_{C\mathrm{/}B}}$. Symmetrically $\beta \gamma$ is a $C$-linear endomorphism of $Q$, and the elements $q({\mathrm{d}}_{C\mathrm{/}A}c)$ generate $Q$ because the elements ${\mathrm{d}}_{C\mathrm{/}A}c$ generate ${\mathrm{\Omega }}_{C\mathrm{/}A}$ and $q$ is onto, so from $\beta \gamma (q({\mathrm{d}}_{C\mathrm{/}A}c))=\beta (\pi ({\mathrm{d}}_{C\mathrm{/}A}c))=\beta ({\mathrm{d}}_{C\mathrm{/}B}c)=q({\mathrm{d}}_{C\mathrm{/}A}c)$ we get $\beta \gamma ={\mathrm{i}\mathrm{d}}_Q$. Thus $\gamma$ is injective, and for $\omega \in \ker \pi$ we get $\gamma (q(\omega ))=\pi (\omega )=0$, hence $q(\omega )=0$ and $\omega \in \ker q=\mathrm{im}(\text{first map})$. Hence $\ker \pi \subseteq \mathrm{im}(\text{first map})$ and, with step 1.1, $\ker \pi =\mathrm{im}(\text{first map})$; moreover $Q\cong {\mathrm{\Omega }}_{C\mathrm{/}B}$.

The first arrow is not injective in general: take $A=k$ a field, $B=k[x]$, $C=k$ with $x\mapsto 0$. By [F3] the source is $C{\otimes }_B{\mathrm{\Omega }}_{B\mathrm{/}A}\cong (k[x]\mathrm{/}(x)){\otimes }_{k[x]}k[x]\mathrm{d}x\cong k\mathrm{d}x\mathrm{\ne }0$, while the target is ${\mathrm{\Omega }}_{C\mathrm{/}A}={\mathrm{\Omega }}_{k\mathrm{/}k}=0$, the universal derivation of a ring over itself being zero. So the first map is zero on a nonzero module, and with steps 1.1 and 2.1 the asserted sequence is exact with a noninjective first arrow.