The Stacks project

Proof. We first assume $S$ is a Noetherian scheme. Let $\omega ^\bullet _{X/S}$ be the relative dualizing complex of $X$ over $S$ as in Duality for Schemes, Remark 48.12.5 and let $Rf_*\omega _{X/S}^\bullet \to \mathcal{O}_ S$ be its trace map. By Duality for Schemes, Lemma 48.15.7 (here we use that $S$ is Noetherian) there exists an isomorphism $\omega ^\bullet _{X/S} \cong \Omega ^ n_{X/S}[n]$ and using this isomorphism we obtain $t$. The complexes $Rf_*\Omega ^ p_{X/S}$ are perfect by Lemma 50.3.5. Since $\Omega ^ p_{X/S}$ is locally free and since $\Omega ^ p_{X/S} \otimes _{\mathcal{O}_ X} \Omega ^{n - p}_{X/S} \to \Omega ^ n_{X/S}$ exhibits an isomorphism $\Omega ^ p_{X/S} \cong \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\Omega ^{n - p}_{X/S}, \Omega ^ n_{X/S})$ we see that the pairing induced by the relative cup product is perfect by Duality for Schemes, Remark 48.12.6.

Proof of existence in the general case. By absolute Noetherian approximation we can find a cartesian diagram

with $S_0$ Noetherian and $f_0$ a proper and smooth morphism all of whose fibres are nonempty and equidimensional of dimension $n$. See Limits, Lemmas 32.10.1, 32.8.9, 32.18.4, 32.8.15, and 32.13.1. Choose a map

as in the statement of the lemma. By cohomology and base change the complexes $Rf_*(\Omega ^ p_{X/S})$ are equal to $Lg^*(Rf_{0, *}\Omega ^ p_{X_0/S_0})$ for all $p \geq 0$, see Lemma 50.3.5. In particular the pullback of $t_0$ by $g$ gives us a map $t$ as in the statement of the lemmma. Namely, by Cohomology, Lemma 20.31.8, the cup product map

is the pullback by $g$ of the corresponding cup product map for $f_0$. Hence the fact that $t_0$ produces perfect pairings, pulls back to the same property for $t$ on $S$.

Uniqueness of $t$. Choose a distinguished triangle $f_*\mathcal{O}_ X \to Rf_*\mathcal{O}_ X \to P \to f_*\mathcal{O}_ X[1]$. By Lemma 50.24.2 the object $P$ is perfect of tor amplitude in $[1, \infty )$ and the triangle is locally on $S$ split. Thus $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(P, \mathcal{O}_ X)$ is perfect of tor amplitude in $(-\infty , -1]$. Hence duality (above) shows that locally on $S$ we have

This shows that $R^ nf_*\Omega ^ n_{X/S}$ is finite locally free and that we obtain a perfect $\mathcal{O}_ S$-bilinear pairing

using $t$. This implies that any $\mathcal{O}_ S$-linear map $t' : R^ nf_*\Omega ^ n_{X/S} \to \mathcal{O}_ S$ is of the form $t' = t \circ g$ for some $g \in \Gamma (S, f_*\mathcal{O}_ X) = \Gamma (X, \mathcal{O}_ X)$. In order for $t'$ to still determine a perfect pairing $g$ will have to be a unit. This finishes the proof. $\square$