Conormal exact sequence for an algebra quotient

The second map exists and is surjective. Regard ${\mathrm{\Omega }}_{B\mathrm{/}A}$ as a $P$-module along $\pi$. The composite $P\stackrel{\pi }{\to }B\stackrel{{\mathrm{d}}_{B\mathrm{/}A}}{\to }{\mathrm{\Omega }}_{B\mathrm{/}A}$ is an $A$-derivation of $P$ into ${\mathrm{\Omega }}_{B\mathrm{/}A}$: it is additive, kills $A$, and satisfies Leibniz because $\pi$ is a ring map and ${\mathrm{d}}_{B\mathrm{/}A}$ is a derivation. By [F1] it corresponds to a $P$-linear map $u\text{\hspace{0.17em}⁣}:{\mathrm{\Omega }}_{P\mathrm{/}A}\to {\mathrm{\Omega }}_{B\mathrm{/}A}$ with $u(\mathrm{d}p)={\mathrm{d}}_{B\mathrm{/}A}(\pi (p))$. For $i\in I$ we have $u(\mathrm{d}i)={\mathrm{d}}_{B\mathrm{/}A}(0)=0$, and $P$-linearity gives $u(i\omega )=\pi (i)u(\omega )=0$, so $u$ kills the submodule $I{\mathrm{\Omega }}_{P\mathrm{/}A}\subseteq {\mathrm{\Omega }}_{P\mathrm{/}A}$. By [F3] applied to $I\to P\to B\to 0$ tensored with ${\mathrm{\Omega }}_{P\mathrm{/}A}$ we have $B{\otimes }_P{\mathrm{\Omega }}_{P\mathrm{/}A}\cong {\mathrm{\Omega }}_{P\mathrm{/}A}\mathrm{/}I{\mathrm{\Omega }}_{P\mathrm{/}A}$, so $u$ induces a $B$-linear map $\beta \text{\hspace{0.17em}⁣}:B{\otimes }_P{\mathrm{\Omega }}_{P\mathrm{/}A}\to {\mathrm{\Omega }}_{B\mathrm{/}A}$ with $\beta (1\otimes \mathrm{d}p)={\mathrm{d}}_{B\mathrm{/}A}(\pi (p))$. It is surjective: every $b\in B$ is $\pi (p)$ for some $p\in P$, and the elements ${\mathrm{d}}_{B\mathrm{/}A}(b)$ generate ${\mathrm{\Omega }}_{B\mathrm{/}A}$ over $B$ by [F2].

The first map is well defined. The assignment $i\mapsto 1\otimes \mathrm{d}i$ defines a $P$-linear map $I\to B{\otimes }_P{\mathrm{\Omega }}_{P\mathrm{/}A}$, and it kills $I^2$: for $i,j\in I$, $1\otimes \mathrm{d}(ij)=1\otimes (i\mathrm{d}j+j\mathrm{d}i)=i(1\otimes \mathrm{d}j)+j(1\otimes \mathrm{d}i)=0$ in the $B$-module $B{\otimes }_P{\mathrm{\Omega }}_{P\mathrm{/}A}$, because the classes of $i$ and $j$ in $B$ are zero. Hence it induces a $B$-linear map $\alpha \text{\hspace{0.17em}⁣}:I\mathrm{/}{I}^2\to B{\otimes }_P{\mathrm{\Omega }}_{P\mathrm{/}A}$ with $\alpha ([i])=1\otimes \mathrm{d}i$.

The composite vanishes. For $i\in I$, $\beta (\alpha ([i]))=\beta (1\otimes \mathrm{d}i)={\mathrm{d}}_{B\mathrm{/}A}(\pi (i))={\mathrm{d}}_{B\mathrm{/}A}(0)=0$; thus $\beta$ factors through the cokernel $Q:=\mathrm{coker}\alpha =(B{\otimes }_P{\mathrm{\Omega }}_{P\mathrm{/}A})\mathrm{/}\alpha (I\mathrm{/}{I}^2)$, giving a surjective $B$-linear map $\bar{\beta }\text{\hspace{0.17em}⁣}:Q\to {\mathrm{\Omega }}_{B\mathrm{/}A}$.

A left inverse for $\bar{\beta }$. Let $D\text{\hspace{0.17em}⁣}:P\to Q$ send $p$ to the class of $1\otimes \mathrm{d}p$; it is the composite of the $A$-derivation $p\mapsto 1\otimes \mathrm{d}p$ with the $B$-linear quotient map, hence an $A$-derivation, and it kills $I$ because the class of $1\otimes \mathrm{d}i$ is $\alpha ([i])=0$ for $i\in I$. Since $Q$ is a $B$-module, $D$ is constant on cosets of $I$ and satisfies Leibniz, so it descends to an $A$-derivation $\bar{D}\text{\hspace{0.17em}⁣}:B\to Q$: any $b$ has a lift $p$, and $\bar{D}(b):=D(p)$ is well defined because $D$ kills $I$. Applying [F1] to the ring map $A\to B$ gives a $B$-linear map $\mathrm{\ell }\text{\hspace{0.17em}⁣}:{\mathrm{\Omega }}_{B\mathrm{/}A}\to Q$ with $\mathrm{\ell }({\mathrm{d}}_{B\mathrm{/}A}(b))=\bar{D}(b)$ for all $b\in B$.

$\mathrm{\ell }$ is inverse to $\bar{\beta }$. For $p\in P$ we have $\mathrm{\ell }(\bar{\beta }([1\otimes \mathrm{d}p]))=\mathrm{\ell }({\mathrm{d}}_{B\mathrm{/}A}(\pi (p)))=\bar{D}(\pi (p))=D(p)=[1\otimes \mathrm{d}p]$, and the classes $[1\otimes \mathrm{d}p]$ generate $Q$ over $B$ because the $\mathrm{d}p$ generate ${\mathrm{\Omega }}_{P\mathrm{/}A}$, so $\mathrm{\ell }\circ \bar{\beta }={\mathrm{i}\mathrm{d}}_Q$. Conversely, for $b\in B$ with lift $p$, $\bar{\beta }(\mathrm{\ell }({\mathrm{d}}_{B\mathrm{/}A}(b)))=\bar{\beta }([1\otimes \mathrm{d}p])={\mathrm{d}}_{B\mathrm{/}A}(b)$, and the ${\mathrm{d}}_{B\mathrm{/}A}(b)$ generate ${\mathrm{\Omega }}_{B\mathrm{/}A}$ by [F2], so $\bar{\beta }\circ \mathrm{\ell }=\mathrm{i}\mathrm{d}$. Hence $\bar{\beta }$ is an isomorphism, $\ker \beta =\mathrm{im}\alpha$, and with $\beta$ surjective the displayed sequence is exact.