The Stacks project

Proof. By Lemma 29.32.3 we may replace $X$ and $S$ by $U$ and $V$. Thus we may assume $X = \mathop{\mathrm{Spec}}(A)$ and $S = \mathop{\mathrm{Spec}}(R)$ and we have to show the lemma with $U = X$ and $V = S$. Consider the $A$-module $M = \Gamma (X, \Omega _{X/S})$ together with the $R$-derivation $\text{d}_{X/S} : A \to M$. Let $N$ be another $A$-module and denote $\widetilde{N}$ the quasi-coherent $\mathcal{O}_ X$-module associated to $N$, see Schemes, Section 26.7. Precomposing by $\text{d}_{X/S} : A \to M$ we get an arrow

Using Lemmas 29.32.2 and 29.32.4 we get identifications

Taking global sections determines an arrow $\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\Omega _{X/S}, \widetilde{N}) \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N)$. Combining this arrow and the identifications above we get an arrow

Checking what happens on global sections, we find that $\alpha$ and $\beta$ are each others inverse. Hence we see that $\text{d}_{X/S} : A \to M$ satisfies the same universal property as $\text{d} : A \to \Omega _{A/R}$, see Algebra, Lemma 10.131.3. Thus the Yoneda lemma (Categories, Lemma 4.3.5) implies there is a unique isomorphism of $A$-modules $M \cong \Omega _{A/R}$ compatible with derivations. $\square$