This question is from Qing Liu Algebraic Geometry and Arithmetic Curves chapter 3 exercise 2.5 on pg. 96. I've approached this is the same spirit as the answer given to this highly related question.
Let $S$ be a locally Noetherian scheme, $X$, $Y$ be $S$-schemes of finite type. Fix $s\in S$, and let $\varphi:X\times_S\operatorname{Spec}\mathcal{O}_{S,s}\to Y\times_S\operatorname{Spec}\mathcal{O}_{S,s}$ be an $S$-morphism of schemes. Show that there exists an open set $U\ni s$ and a morphism of $S$-schemes $f:X\times_SU\to Y\times_SU$ such that $\varphi$ is obtained from $f$ by the base change $\operatorname{Spec}\mathcal{O}_{S,s}\to U$. If $\varphi$ is an isomorphism, show that there exists such an $f$ which is moreover an isomorphism.
As a first step, I'd like to argue that this problem is affine in $S$, $X$, and $Y$. For $S$ this is trivial, as we may take $U$ to be an open affine. For $X$ and $Y$, I'd like to argue along the following lines: if for every open affine $V\subset Y$, and every open affine $W\subset X$ for which $\varphi(W\times_S\operatorname{Spec}\mathcal{O}_{S,s})\subset V\times_S\operatorname{Spec}\mathcal{O}_{S,s}$, there is an $f_W$ extending $\varphi|_{W\times_S\operatorname{Spec}\mathcal{O}_{S,s}}$ by base change, then the $f_W$ agree on overlaps and glue to a morphism $f$ extending $\varphi$.
Question 1: Is this argument correct, and how do I make it more formal?
Reducing to the affine case, I may assume $S=\operatorname{Spec}R$ with $R$ Noetherian (since $S$ is locally Noetherian), and that $X = \operatorname{Spec} B$ and $Y = \operatorname{Spec} A$ for finite-type $R$-algebras $A$ and $B$. Then if $\mathfrak{p}$ is the prime corresponding to $s\in S$, $\varphi^\#(Y):A\otimes_R R_\mathfrak{p}\to B\otimes_RR_\mathfrak{p}$, so $\varphi$ extends to a neighborhood of $s$ if and only if there is some $h\in R-\mathfrak{p}$ such that there is a homomorphism $$\varphi^\#(Y)\otimes \pi:A\otimes_R R_h\to B\otimes_R R_h$$ where $\pi:R_h\to R_\mathfrak{p}$ is the canonical map.
To find such an $h$, write $A = R_\mathfrak{p}[T_1,\dots,T_n]/(f_1,\dots,f_r)$ and $B=R_\mathfrak{p}[S_1,\dots,S_m]/(g_1,\dots,g_s)$ where the $f_i\in R[T_1,\dots,T_n]$ and $g_j\in R[S_1,\dots,S_m]$. Define $$\Phi:R_\mathfrak{p}[T_1,\dots,T_n]\to R_\mathfrak{p}[S_1,\dots,S_m]/(g_1,\dots,g_s)$$ where $\Phi(p(T))$ is a coset representative of $\varphi(p(T)+(f_1,\dots,f_r)$. Then $\Phi(T_i) = q_i(S)\in R_\mathfrak{p}[S_1,\dots,S_m]$, and $\Phi(f_j) \in (g_1,\dots,g_s)$, so $$\Phi(f_j) = \sum_{k=1}^s \frac{a_{jk}}{s_{jk}}g_k.$$ where the $a_{jk}\in R$ and $s_{jk}\in R-\mathfrak{p}$ and $$\Phi(T_i) = \sum_{k_i\geq 0}\frac{a_{k_1,\dots,k_m,i}}{s_{k_1,\dots,d_m,i}}S^{k_1}\cdots S^{k_m}$$ where $a_{k_1,\dots,k_m,i}\in R$ and $s_{k_1,\dots,k_m,i}\in R-\mathfrak{p}$. I'm tempted to define $h$ to be the product of all such $s$ and argue that $\Phi\otimes \pi$ is well-defined as a mapping $A\otimes_RR_h\to B\otimes_RR_h$, but the notation has become extremely cumbersome.
Question 2: Am I on the right track/is there a more elegant approach?
