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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