By the Hodge-to-de Rham spectral sequence (Section \ref{section-hodge-to-de-rham}), the vanishing of $\Omega^i_{X/k}$ for $i > d$, the vanishing in Cohomology, Proposition \ref{cohomology-proposition-vanishing-Noetherian} and the results of Lemmas \ref{lemma-bottom-part-degenerates} and \ref{lemma-top-part-degenerates} we see that $H^0_{dR}(X/k) = H^0(X, \mathcal{O}_X)$ and $H^d(X, \Omega^d_{X/k}) = H_{dR}^{2d}(X/k)$. More precisesly, these identifications come from the maps of complexes $$ \Omega^\bullet_{X/k} \to \mathcal{O}_X[0] \quad\text{and}\quad \Omega^d_{X/k}[-d] \to \Omega^\bullet_{X/k} $$ Let us choose $t : H_{dR}^{2d}(X/k) \to k$ which via this identification corresponds to a $t$ as in Lemma \ref{lemma-duality-hodge}. Then in any case we see that the pairing displayed in the lemma is perfect for $i = 0$. \medskip\noindent Denote $\underline{k}$ the constant sheaf with value $k$ on $X$. Let us abbreviate $\Omega^\bullet = \Omega^\bullet_{X/k}$. Consider the map (\ref{equation-wedge}) which in our situation reads $$ \wedge : \text{Tot}(\Omega^\bullet \otimes_{\underline{k}} \Omega^\bullet) \longrightarrow \Omega^\bullet $$ For every integer $p = 0, 1, \ldots, d$ this map annihilates the subcomplex $\text{Tot}(\sigma_{> p} \Omega^\bullet \otimes_{\underline{k}} \sigma_{\geq d - p} \Omega^\bullet)$ for degree reasons. Hence we find that the restriction of $\wedge$ to the subcomplex $\text{Tot}(\Omega^\bullet \otimes_{\underline{k}} \sigma_{\geq d - p}\Omega^\bullet)$ factors through a map of complexes $$ \gamma_p : \text{Tot}(\sigma_{\leq p} \Omega^\bullet \otimes_{\underline{k}} \sigma_{\geq d - p} \Omega^\bullet) \longrightarrow \Omega^\bullet $$ Using the same procedure as in Section \ref{section-cup-product} we obtain cup products $$ H^i(X, \sigma_{\leq p} \Omega^\bullet) \times H^{2d - i}(X, \sigma_{\geq d - p}\Omega^\bullet) \longrightarrow H_{dR}^{2d}(X, \Omega^\bullet) $$ We will prove by induction on $p$ that these cup products via $t$ induce perfect pairings between $H^i(X, \sigma_{\leq p} \Omega^\bullet)$ and $H^{2d - i}(X, \sigma_{\geq d - p}\Omega^\bullet)$. For $p = d$ this is the assertion of the proposition. \medskip\noindent The base case is $p = 0$. In this case we simply obtain the pairing between $H^i(X, \mathcal{O}_X)$ and $H^{d - i}(X, \Omega^d)$ of Lemma \ref{lemma-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 $\underline{k}$-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 long exact cohomology sequences associated to these distinguished triangles $$ \xymatrix{ H^{i - 1}(X, \sigma_{\leq p}\Omega^\bullet) \ar[d]_a \\ H^i(X, \Omega^{p + 1}[-p - 1]) \ar[d]_b \\ H^i(X, \sigma_{\leq p + 1}\Omega^\bullet) \ar[d]_c \\ H^i(X, \sigma_{\leq p}\Omega^\bullet) \ar[d]_d \\ H^{i + 1}(X, \Omega^{p + 1}[-p - 1]) } \quad\quad \xymatrix{ H^{2d - i  + 1}(X, \sigma_{\geq d - p}\Omega^\bullet) \\ H^{2d - i}(X, \Omega^{d - p - 1}[-d + p + 1]) \ar[u]_{a'} \\ H^{2d - i}(X, \sigma_{\geq d - p - 1}\Omega^\bullet) \ar[u]_{b'} \\ H^{2d - i}(X, \sigma_{\geq d - p}\Omega^\bullet) \ar[u]_{c'} \\ H^{2d - i - 1}(X, \Omega^{d - p - 1}[-d + p + 1]) \ar[u]_{d'} } $$ By induction and Lemma \ref{lemma-duality-hodge} we know that the pairings constructed above between the $k$-vectorspaces on the first, second, fourth, and fifth rows are perfect. By the $5$-lemma, in order to show that the pairing between the cohomology groups in the middle row is perfect, it suffices to show that the pairs $(a, a')$, $(b, b')$, $(c, c')$, and $(d, d')$ are compatible with the given pairings (see below). \medskip\noindent Let us prove this for the pair $(c, c')$. Here we observe simply that we have a commutative diagram $$ \xymatrix{ \text{Tot}(\sigma_{\leq p} \Omega^\bullet \otimes_{\underline{k}} \sigma_{\geq d - p} \Omega^\bullet) \ar[d]_{\gamma_p} & \text{Tot}(\sigma_{\leq p + 1} \Omega^\bullet \otimes_{\underline{k}} \sigma_{\geq d - p} \Omega^\bullet) \ar[l] \ar[d] \\ \Omega^\bullet & \text{Tot}(\sigma_{\leq p + 1} \Omega^\bullet \otimes_{\underline{k}} \sigma_{\geq d - p - 1} \Omega^\bullet) \ar[l]_-{\gamma_{p + 1}} } $$ Hence if we have $\alpha \in H^i(X, \sigma_{\leq p + 1}\Omega^\bullet)$ and $\beta \in H^{2d - i}(X, \sigma_{\geq d - p}\Omega^\bullet)$ then we get $\gamma_p(\alpha \cup c'(\beta)) = \gamma_{p + 1}(c(\alpha) \cup \beta)$ by functoriality of the cup product. \medskip\noindent Similarly for the pair $(b, b')$ we use the commutative diagram $$ \xymatrix{ \text{Tot}(\sigma_{\leq p + 1} \Omega^\bullet \otimes_{\underline{k}} \sigma_{\geq d - p - 1} \Omega^\bullet) \ar[d]_{\gamma_{p + 1}} & \text{Tot}(\Omega^{p + 1}[-p - 1] \otimes_{\underline{k}} \sigma_{\geq d - p - 1} \Omega^\bullet) \ar[l] \ar[d] \\ \Omega^\bullet & \Omega^{p + 1}[-p - 1] \otimes_{\underline{k}} \Omega^{d - p - 1}[-d + p + 1] \ar[l]_-\wedge } $$ and argue in the same manner. \medskip\noindent For the pair $(d, d')$ we use the commutative diagram $$ \xymatrix{ \Omega^{p + 1}[-p] \otimes_{\underline{k}} \Omega^{d - p - 1}[-d + p] \ar[d] & \text{Tot}(\sigma_{\leq p}\Omega^\bullet \otimes_{\underline{k}} \Omega^{d - p - 1}[-d + p]) \ar[l] \ar[d] \\ \Omega^\bullet & \text{Tot}(\sigma_{\leq p}\Omega^\bullet \otimes_{\underline{k}} \sigma_{\geq d - p}\Omega^\bullet) \ar[l] } $$ and we look at cohomology classes in $H^i(X, \sigma_{\leq p}\Omega^\bullet)$ and $H^{2d - i}(X, \Omega^{d - p - 1}[-d + p])$. Changing $i$ to $i - 1$ we get the result for the pair $(a, a')$ thereby finishing the proof that our pairings are perfect. \medskip\noindent We omit the argument showing the uniqueness of $t$ up to precomposing by multiplication by a unit in $H^0(X, \mathcal{O}_X)$.