Lemma 91.7.1. With notation as in (91.7.0.2) set

$\begin{matrix} \Omega _1 = \Omega _{\mathcal{O}/A} \otimes _\mathcal {O} \underline{B} \text{ on }\mathcal{C}_{B/A} \\ \Omega _2 = \Omega _{\mathcal{O}/A} \otimes _\mathcal {O} \underline{C} \text{ on }\mathcal{C}_{C/A} \\ \Omega _3 = \Omega _{\mathcal{O}/B} \otimes _\mathcal {O} \underline{C} \text{ on }\mathcal{C}_{C/B} \end{matrix}$

Then we have a canonical short exact sequence of sheaves of $\underline{C}$-modules

$0 \to g_1^{-1}\Omega _1 \otimes _{\underline{B}} \underline{C} \to g_2^{-1}\Omega _2 \to g_3^{-1}\Omega _3 \to 0$

on $\mathcal{C}_{C/B/A}$.

Proof. Recall that $g_ i^{-1}$ is gotten by simply precomposing with $u_ i$. Given an object $U = (P \to B, Q \to C)$ we have a split short exact sequence

$0 \to \Omega _{P/A} \otimes Q \to \Omega _{Q/A} \to \Omega _{Q/P} \to 0$

for example by Algebra, Lemma 10.138.9. Tensoring with $C$ over $Q$ we obtain a short exact sequence

$0 \to \Omega _{P/A} \otimes C \to \Omega _{Q/A} \otimes C \to \Omega _{Q/P} \otimes C \to 0$

We have $\Omega _{P/A} \otimes C = \Omega _{P/A} \otimes B \otimes C$ whence this is the value of $g_1^{-1}\Omega _1 \otimes _{\underline{B}} \underline{C}$ on $U$. The module $\Omega _{Q/A} \otimes C$ is the value of $g_2^{-1}\Omega _2$ on $U$. We have $\Omega _{Q/P} \otimes C = \Omega _{Q \otimes _ P B/B} \otimes C$ by Algebra, Lemma 10.131.12 hence this is the value of $g_3^{-1}\Omega _3$ on $U$. Thus the short exact sequence of the lemma comes from assigning to $U$ the last displayed short exact sequence. $\square$

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