Lemma 18.33.8. In Lemma 18.33.7 suppose that $\mathcal{O}_2 \to \mathcal{O}'_2$ is surjective with kernel $\mathcal{I} \subset \mathcal{O}_2$ and assume that $\mathcal{O}_1 = \mathcal{O}'_1$. Then there is a canonical exact sequence of $\mathcal{O}'_2$-modules

\[ \mathcal{I}/\mathcal{I}^2 \longrightarrow \Omega _{\mathcal{O}_2/\mathcal{O}_1} \otimes _{\mathcal{O}_2} \mathcal{O}'_2 \longrightarrow \Omega _{\mathcal{O}'_2/\mathcal{O}_1} \longrightarrow 0 \]

The leftmost map is characterized by the rule that a local section $f$ of $\mathcal{I}$ maps to $\text{d}f \otimes 1$.

**Proof.**
For a local section $f$ of $\mathcal{I}$ denote $\overline{f}$ the image of $f$ in $\mathcal{I}/\mathcal{I}^2$. To show that the map $\overline{f} \mapsto \text{d}f \otimes 1$ is well defined we just have to check that $\text{d} f_1f_2 \otimes 1 = 0$ if $f_1, f_2$ are local sections of $\mathcal{I}$. And this is clear from the Leibniz rule $\text{d} f_1f_2 \otimes 1 = (f_1 \text{d}f_2 + f_2 \text{d} f_1 )\otimes 1 = \text{d}f_2 \otimes f_1 + \text{d}f_2 \otimes f_1 = 0$. A similar computation show this map is $\mathcal{O}'_2 = \mathcal{O}_2/\mathcal{I}$-linear. The map on the right is the one from Lemma 18.33.7.

To see that the sequence is exact, we argue as follows. Let $\mathcal{O}''_2 \subset \mathcal{O}'_2$ be the presheaf of $\mathcal{O}_1$-algebras whose value on $U$ is the image of $\mathcal{O}_2(U) \to \mathcal{O}'_2(U)$. By Algebra, Lemma 10.131.9 the sequences

\[ \mathcal{I}(U)/\mathcal{I}(U)^2 \longrightarrow \Omega _{\mathcal{O}_2(U)/\mathcal{O}_1(U)} \otimes _{\mathcal{O}_2(U)} \mathcal{O}''_2(U) \longrightarrow \Omega _{\mathcal{O}''_2(U)/\mathcal{O}_1(U)} \longrightarrow 0 \]

are exact for all objects $U$ of $\mathcal{C}$. Since sheafification is exact this gives an exact sequence of sheaves of $(\mathcal{O}'_2)^\# $-modules. By Lemma 18.33.4 and the fact that $(\mathcal{O}''_2)^\# = \mathcal{O}'_2$ we conclude.
$\square$

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