The Stacks project

Proof. The question is local on $X$ hence we may assume $X$, $Y$, $S$ are affine. Say $S = \mathop{\mathrm{Spec}}(R)$, $X = \mathop{\mathrm{Spec}}(A)$, $Y = \mathop{\mathrm{Spec}}(B)$. Write $B = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$. Let $\mathfrak p \subset A$ be the prime ideal corresponding to $x$. The local ring $\mathcal{O}_{X, x} = A_\mathfrak p$ is a G-ring by More on Algebra, Proposition 15.50.10. The map $\varphi $ is a map

where $\mathfrak q \subset B$ is the prime corresponding to $y$. Let $a_1, \ldots , a_ n \in A_\mathfrak p^\wedge $ be the images of $x_1, \ldots , x_ n$ via $R[x_1, \ldots , x_ n] \to B \to B_\mathfrak q^\wedge \to A_\mathfrak p^\wedge $. Then we can apply Smoothing Ring Maps, Lemma 16.13.4 to get an étale ring map $A \to A'$ and a prime ideal $\mathfrak p' \subset A'$ and $b_1, \ldots , b_ n \in A'$ such that $\kappa (\mathfrak p) = \kappa (\mathfrak p')$, $a_ i - b_ i \in (\mathfrak p')^ N(A'_{\mathfrak p'})^\wedge $, and $f_ j(b_1, \ldots , b_ n) = 0$ for $j = 1, \ldots , n$. This determines an $R$-algebra map $B \to A'$ by sending the class of $x_ i$ to $b_ i \in A'$. This finishes the proof by taking $U = \mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(B)$ as the morphism $f$ and $u = \mathfrak p'$. $\square$