Proof.
Say $Y = \mathop{\mathrm{Spec}}(A)$ and $S = \mathop{\mathrm{Spec}}(R)$. In this case $\Omega ^\bullet _{A/R}$ computes $R\Gamma (Y, \Omega ^\bullet _{Y/S})$ by Lemma 50.3.1. Choose a finite affine open covering $\mathcal{U} : X = \bigcup _{i \in I} U_ i$. Consider the complex
\[ K^\bullet = \text{Tot}(\check{\mathcal{C}}^\bullet (\mathcal{U}, \Omega _{X/S}^\bullet )) \]
as in Cohomology, Section 20.25. Let us collect some facts about this complex most of which can be found in the reference just given:
$K^\bullet $ is a complex of $R$-modules whose terms are $A$-modules,
$K^\bullet $ represents $R\Gamma (X, \Omega ^\bullet _{X/S})$ in $D(R)$ (Cohomology of Schemes, Lemma 30.2.2 and Cohomology, Lemma 20.25.2),
there is a natural map $\Omega ^\bullet _{A/R} \to K^\bullet $ of complexes of $R$-modules which is $A$-linear on terms and induces the pullback map $H^*_{dR}(Y/S) \to H^*_{dR}(X/S)$ on cohomology,
$K^\bullet $ has a multiplication denoted $\wedge $ which turns it into a differential graded $R$-algebra,
the multiplication on $K^\bullet $ induces the cup product on $H^*_{dR}(X/S)$ (Cohomology, Section 20.31),
the filtration $F$ on $\Omega ^*_{X/S}$ induces a filtration
\[ K^\bullet = F^0K^\bullet \supset F^1K^\bullet \supset F^2K^\bullet \supset \ldots \]
by subcomplexes on $K^\bullet $ such that
$F^ kK^ n \subset K^ n$ is an $A$-submmodule,
$F^ kK^\bullet \wedge F^ lK^\bullet \subset F^{k + l}K^\bullet $,
$\text{gr}^ kK^\bullet $ is a complex of $A$-modules,
$\text{gr}^0K^\bullet = \text{Tot}(\check{\mathcal{C}}^\bullet (\mathcal{U}, \Omega _{X/Y}^\bullet ))$ and represents $R\Gamma (X, \Omega ^\bullet _{X/Y})$ in $D(A)$,
multiplication induces an isomorphism $\Omega ^ k_{A/R}[-k] \otimes _ A \text{gr}^0K^\bullet \to \text{gr}^ kK^\bullet $
We omit the detailed proofs of these statements; please see discussion leading up to the construction of the spectral sequence in Lemma 50.12.1.
For every $i = 1, \ldots , N$ we choose a cocycle $x_ i \in K^{n_ i}$ representing $\xi _ i$. Next, we look at the map of complexes
\[ \tilde x : M^\bullet = \bigoplus \nolimits _{i = 1, \ldots , N} \Omega ^\bullet _{A/R}[-n_ i] \longrightarrow K^\bullet \]
which sends $\omega $ in the $i$th summand to $x_ i \wedge \omega $. All that remains is to show that this map is a quasi-isomorphism. We endow $M^\bullet $ with the structure of a filtered complex by the rule
\[ F^ kM^\bullet = \bigoplus \nolimits _{i = 1, \ldots , N} (\sigma _{\geq k}\Omega ^\bullet _{A/R})[-n_ i] \]
With this choice the map $\tilde x$ is a morphism of filtered complexes. Observe that $\text{gr}^0M^\bullet = \bigoplus A[-n_ i]$ and multiplication induces an isomorphism $\Omega ^ k_{A/R}[-k] \otimes _ A \text{gr}^0M^\bullet \to \text{gr}^ kM^\bullet $. By construction and Lemma 50.13.1 we see that
\[ \text{gr}^0\tilde x : \text{gr}^0M^\bullet \longrightarrow \text{gr}^0K^\bullet \]
is an isomorphism in $D(A)$. It follows that for all $k \geq 0$ we obtain isomorphisms
\[ \text{gr}^ k \tilde x : \text{gr}^ kM^\bullet = \Omega ^ k_{A/R}[-k] \otimes _ A \text{gr}^0M^\bullet \longrightarrow \Omega ^ k_{A/R}[-k] \otimes _ A \text{gr}^0K^\bullet = \text{gr}^ kK^\bullet \]
in $D(A)$. Namely, the complex $\text{gr}^0K^\bullet = \text{Tot}(\check{\mathcal{C}}^\bullet (\mathcal{U}, \Omega _{X/Y}^\bullet ))$ is K-flat as a complex of $A$-modules by Derived Categories of Schemes, Lemma 36.23.3. Hence the tensor product on the right hand side is the derived tensor product as is true by inspection on the left hand side. Finally, taking the derived tensor product $\Omega ^ k_{A/R}[-k] \otimes _ A^\mathbf {L} -$ is a functor on $D(A)$ and therefore sends isomorphisms to isomorphisms. Arguing by induction on $k$ we deduce that
\[ \tilde x : M^\bullet /F^ kM^\bullet \to K^\bullet /F^ kK^\bullet \]
is an isomorphism in $D(R)$ since we have the short exact sequences
\[ 0 \to F^ kM^\bullet /F^{k + 1}M^\bullet \to M^\bullet /F^{k + 1}M^\bullet \to \text{gr}^ kM^\bullet \to 0 \]
and similarly for $K^\bullet $. This proves that $\tilde x$ is a quasi-isomorphism as the filtrations are finite in any given degree.
$\square$
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