The proof is exactly the same as the proof of Proposition \ref{proposition-poincare-duality}. \medskip\noindent By the relative Hodge-to-de Rham spectral sequence $$ E_1^{p, q} = R^qf_*\Omega^p_{X/S} \Rightarrow R^{p + q}f_*\Omega^\bullet_{X/S} $$ (Section \ref{section-hodge-to-de-rham}), the vanishing of $\Omega^i_{X/S}$ for $i > d$, the vanishing in for example Limits, Lemma \ref{limits-lemma-higher-direct-images-zero-above-dimension-fibre} and the results of Lemmas \ref{lemma-relative-bottom-part-degenerates} and \ref{lemma-relative-top-part-degenerates} we see that $R^0f_*\Omega_{X/S} = R^0f_*\mathcal{O}_X$ and $R^df_*\Omega^d_{X/S} = R^{2d}f_*\Omega^\bullet_{X/S}$. More precisesly, these identifications come from the maps of complexes $$ \Omega^\bullet_{X/S} \to \mathcal{O}_X[0] \quad\text{and}\quad \Omega^d_{X/S}[-d] \to \Omega^\bullet_{X/S} $$ Let us choose $t : R^{2d}f_*\Omega_{X/S} \to \mathcal{O}_S$ which via this identification corresponds to a $t$ as in Lemma \ref{lemma-relative-duality-hodge}. \medskip\noindent Let us abbreviate $\Omega^\bullet = \Omega^\bullet_{X/S}$. Consider the map (\ref{equation-wedge}) which in our situation reads $$ \wedge : \text{Tot}(\Omega^\bullet \otimes_{f^{-1}\mathcal{O}_S} \Omega^\bullet) \longrightarrow \Omega^\bullet $$ For every integer $p = 0, 1, \ldots, d$ this map annihilates the subcomplex $\text{Tot}(\sigma_{> p} \Omega^\bullet \otimes_{f^{-1}\mathcal{O}_S} \sigma_{\geq d - p} \Omega^\bullet)$ for degree reasons. Hence we find that the restriction of $\wedge$ to the subcomplex $\text{Tot}(\Omega^\bullet \otimes_{f^{-1}\mathcal{O}_S} \geq_{d - p}\Omega^\bullet)$ factors through a map of complexes $$ \gamma_p : \text{Tot}(\sigma_{\leq p} \Omega^\bullet \otimes_{f^{-1}\mathcal{O}_S} \sigma_{\geq d - p} \Omega^\bullet) \longrightarrow \Omega^\bullet $$ Using the same procedure as in Section \ref{section-cup-product} we obtain relative cup products $$ Rf_*\sigma_{\leq p} \Omega^\bullet \otimes_{\mathcal{O}_S}^\mathbf{L} Rf_*\sigma_{\geq d - p}\Omega^\bullet \longrightarrow Rf_*\Omega^\bullet $$ We will prove by induction on $p$ that these cup products via $t$ induce perfect pairings between $Rf_*\sigma_{\leq p} \Omega^\bullet$ and $Rf_*\sigma_{\geq d - p}\Omega^\bullet[2d]$. For $p = d$ this is the assertion of the proposition. \medskip\noindent The base case is $p = 0$. In this case we have $$ Rf_*\sigma_{\leq p}\Omega^\bullet = Rf_*\mathcal{O}_X \quad\text{and}\quad Rf_*\sigma_{\geq d - p}\Omega^\bullet[2d] = Rf_*(\Omega^d[-d])[2d] = Rf_*\Omega^d[d] $$ In this case we simply obtain the pairing between $Rf_*\mathcal{O}_X$ and $Rf_*\Omega^d[d]$ of Lemma \ref{lemma-relative-duality-hodge} and the result is true. \medskip\noindent Induction step. Say we know the result is true for $p$. Then we consider the distinguished triangle $$ \Omega^{p + 1}[-p - 1] \to \sigma_{\leq p + 1}\Omega^\bullet \to \sigma_{\leq p}\Omega^\bullet \to \Omega^{p + 1}[-p] $$ and the distinguished triangle $$ \sigma_{\geq d - p}\Omega^\bullet \to \sigma_{\geq d - p - 1}\Omega^\bullet \to \Omega^{d - p - 1}[-d + p + 1] \to (\sigma_{\geq d - p}\Omega^\bullet)[1] $$ Observe that both are distinguished triangles in the homotopy category of complexes of sheaves of $f^{-1}\mathcal{O}_S$-modules; in particular the maps $\sigma_{\leq p}\Omega^\bullet \to \Omega^{p + 1}[-p]$ and $\Omega^{d - p - 1}[-d + p + 1] \to (\sigma_{\geq d - p}\Omega^\bullet)[1]$ are given by actual maps of complexes, namely using the differential $\Omega^p \to \Omega^{p + 1}$ and the differential $\Omega^{d - p - 1} \to \Omega^{d - p}$. Consider the distinguished triangles associated gotten from these distinguished triangles by applying $Rf_*$ $$ \xymatrix{ Rf_*\sigma_{\leq p}\Omega^\bullet \ar[d]_a \\ Rf_*\Omega^{p + 1}[-p - 1] \ar[d]_b \\ Rf_*\sigma_{\leq p + 1}\Omega^\bullet \ar[d]_c \\ Rf_*\sigma_{\leq p}\Omega^\bullet \ar[d]_d \\ Rf_*\Omega^{p + 1}[-p - 1] } \quad\quad \xymatrix{ Rf_*\sigma_{\geq d - p}\Omega^\bullet \\ Rf_*\Omega^{d - p - 1}[-d + p + 1] \ar[u]_{a'} \\ Rf_*\sigma_{\geq d - p - 1}\Omega^\bullet \ar[u]_{b'} \\ Rf_*\sigma_{\geq d - p}\Omega^\bullet \ar[u]_{c'} \\ Rf_*\Omega^{d - p - 1}[-d + p + 1] \ar[u]_{d'} } $$ We will show below that the pairs $(a, a')$, $(b, b')$, $(c, c')$, and $(d, d')$ are compatible with the given pairings. This means we obtain a map from the distinguished triangle on the left to the distuiguished triangle obtained by applying $R\SheafHom(-, \mathcal{O}_S)$ to the distinguished triangle on the right. By induction and Lemma \ref{lemma-duality-hodge} we know that the pairings constructed above between the complexes on the first, second, fourth, and fifth rows are perfect, i.e., determine isomorphisms after taking duals. By Derived Categories, Lemma \ref{derived-lemma-third-isomorphism-triangle} we conclude the pairing between the complexes in the middle row is perfect as desired. \medskip\noindent Let $e : K \to K'$ and $e' : M' \to M$ be maps of objects of $D(\mathcal{O}_S)$ and let $K \otimes_{\mathcal{O}_S}^\mathbf{L} M \to \mathcal{O}_S$ and $K' \otimes_{\mathcal{O}_S}^\mathbf{L} M' \to \mathcal{O}_S$ be pairings. Then we say these pairings are compatible if the diagram $$ \xymatrix{ K' \otimes_{\mathcal{O}_S}^\mathbf{L} M' \ar[d] & K \otimes_{\mathcal{O}_S}^\mathbf{L} M' \ar[l]^{e \otimes 1} \ar[d]^{1 \otimes e'} \\ \mathcal{O}_S & K \otimes_{\mathcal{O}_S}^\mathbf{L} M \ar[l] } $$ commutes. This indeed means that the diagram $$ \xymatrix{ K \ar[r] \ar[d]_e & R\SheafHom(M, \mathcal{O}_S) \ar[d]^{R\SheafHom(e', -)} \\ K' \ar[r] & R\SheafHom(M', \mathcal{O}_S) } $$ commutes and hence is sufficient for our purposes. \medskip\noindent Let us prove this for the pair $(c, c')$. Here we observe simply that we have a commutative diagram $$ \xymatrix{ \text{Tot}(\sigma_{\leq p} \Omega^\bullet \otimes_{f^{-1}\mathcal{O}_S} \sigma_{\geq d - p} \Omega^\bullet) \ar[d]_{\gamma_p} & \text{Tot}(\sigma_{\leq p + 1} \Omega^\bullet \otimes_{f^{-1}\mathcal{O}_S} \sigma_{\geq d - p} \Omega^\bullet) \ar[l] \ar[d] \\ \Omega^\bullet & \text{Tot}(\sigma_{\leq p + 1} \Omega^\bullet \otimes_{f^{-1}\mathcal{O}_S} \sigma_{\geq d - p - 1} \Omega^\bullet) \ar[l]_-{\gamma_{p + 1}} } $$ By functoriality of the cup product we obtain commutativity of the desired diagram. \medskip\noindent Similarly for the pair $(b, b')$ we use the commutative diagram $$ \xymatrix{ \text{Tot}(\sigma_{\leq p + 1} \Omega^\bullet \otimes_{f^{-1}\mathcal{O}_S} \sigma_{\geq d - p - 1} \Omega^\bullet) \ar[d]_{\gamma_{p + 1}} & \text{Tot}(\Omega^{p + 1}[-p - 1] \otimes_{f^{-1}\mathcal{O}_S} \sigma_{\geq d - p - 1} \Omega^\bullet) \ar[l] \ar[d] \\ \Omega^\bullet & \Omega^{p + 1}[-p - 1] \otimes_{f^{-1}\mathcal{O}_S} \Omega^{d - p - 1}[-d + p + 1] \ar[l]_-\wedge } $$ \medskip\noindent For the pairs $(d, d')$ and $(a, a')$ we use the commutative diagram $$ \xymatrix{ \Omega^{p + 1}[-p] \otimes_{f^{-1}\mathcal{O}_S} \Omega^{d - p - 1}[-d + p] \ar[d] & \text{Tot}(\sigma_{\leq p}\Omega^\bullet \otimes_{f^{-1}\mathcal{O}_S} \Omega^{d - p - 1}[-d + p]) \ar[l] \ar[d] \\ \Omega^\bullet & \text{Tot}(\sigma_{\leq p}\Omega^\bullet \otimes_{f^{-1}\mathcal{O}_S} \sigma_{\geq d - p}\Omega^\bullet) \ar[l] } $$ \medskip\noindent We omit the argument showing the uniqueness of $t$ up to precomposing by multiplication by a unit in $H^0(X, \mathcal{O}_X)$.