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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