Lemma 49.8.1. In the situation above there is a canonical isomorphism

$\text{Tot}(p^{-1}\Omega ^\bullet _{X/S} \otimes _{f^{-1}\mathcal{O}_ S} q^{-1}\Omega ^\bullet _{Y/S}) \longrightarrow \Omega ^\bullet _{X \times _ S Y/S}$

of complexes of $f^{-1}\mathcal{O}_ S$-modules.

Proof. By Derived Categories of Schemes, Remark 35.22.2 we have

$p^{-1}\Omega ^ i_{X/S} \otimes _{f^{-1}\mathcal{O}_ S} q^{-1}\Omega ^ j_{Y/S} = p^*\Omega ^ i_{X/S} \otimes _{\mathcal{O}_{X \times _ S Y}} q^*\Omega ^ j_{Y/S}$

for all $i, j$. On the other hand, we know that $\Omega _{X \times _ S Y/S} = p^*\Omega _{X/S} \oplus q^*\Omega _{Y/S}$ by Morphisms, Lemma 28.31.11. Taking exterior powers we obtain

$\Omega ^ n_{X \times _ S Y/S} = \bigoplus \nolimits _{i + j = n} p^*\Omega ^ i_{X/S} \otimes _{\mathcal{O}_{X \times _ S Y}} q^*\Omega ^ j_{Y/S} = \bigoplus \nolimits _{i + j = n} p^{-1}\Omega ^ i_{X/S} \otimes _{f^{-1}\mathcal{O}_ S} q^{-1}\Omega ^ j_{Y/S}$

by elementary properties of exterior powers. This finishes the proof. $\square$

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