Lemma 50.23.9. Let $X \to S$, $i : Z \to X$, and $c \geq 0$ be as in Lemma 50.23.3. Assume $X \to S$ smooth and $Z \to X$ Koszul regular. Given $\alpha \in H^ q(X, \Omega ^ p_{X/S})$ we have $\gamma ^{p, q}(\alpha |_ Z) = \alpha \cup \gamma ^{0, 0}(1)$ in $H^{q + c}(X, \Omega ^{p + c}_{X/S})$ with $\gamma ^{a, b}$ as in Remark 50.23.7.

**Proof.**
This lemma follows from Lemma 50.23.5 and Cohomology, Lemma 20.34.11. We suggest the reader skip over the more detailed discussion below.

We will use without further mention that $R\mathcal{H}_ Z(\Omega ^ j_{X/S}) = \mathcal{H}^ c_ Z(\Omega ^ j_{X/S})[-c]$ for all $j$ as pointed out in Remark 50.23.7. We will also silently use the identifications $H^{q + c}_ Z(X, \Omega ^ j_{X/S}) = H^{q + c}(Z, R\mathcal{H}_ Z(\Omega ^ j_{X/S}) = H^ q(Z, \mathcal{H}^ c_ Z(\Omega ^ j_{X/S}))$, see Cohomology, Lemma 20.34.4 for the first one. With these identifications

$\gamma ^0(1) \in H^ c_ Z(X, \Omega ^ c_{X/S})$ maps to $\gamma ^{0, 0}(1)$ in $H^ c(X, \Omega ^ c_{X/S})$,

the right hand side $i^{-1}\alpha \wedge \gamma ^0(1)$ of the equality in Lemma 50.23.5 is the (image by wedge product of the) cup product of Cohomology, Remark 20.34.10 of the elements $\alpha $ and $\gamma ^0(1)$, in other words, the constructions in the proof of Lemma 50.23.5 and in Cohomology, Remark 20.34.10 match,

by Cohomology, Lemma 20.34.11 this maps to $\alpha \cup \gamma ^{0, 0}(1)$ in $H^{q + c}(X, \Omega ^ p_{X/S} \otimes \Omega ^ c_{X/S})$, and

the left hand side $\gamma ^ p(\alpha |_ Z)$ of the equality in Lemma 50.23.5 maps to $\gamma ^{p, q}(\alpha |_ Z)$.

This finishes the proof. $\square$

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