I am going through the proof of Proposition 4.17 in Liu's Algebraic Geometry and Arithmetic Curves:
First question (SOLVED):
I can't verify why $V(f\vert_U)$ contains $W\cap U$. If we write $\phi:W\cong\operatorname{Spec}A$ and $\psi:U\cong\operatorname{Spec}B$, then formally $W=V(f\vert_W)$ means that $$ \phi(W)=V({\phi^\#}^{-1}f\vert_W)=V(\mathfrak p). $$ where for notational convenience $a={\phi^\#}^{-1}f\vert_W$. And $W\cap U\subset V(f\vert_U)$ means that $$ \psi(W\cap U)=\psi(W)\cap\operatorname{Spec}B=V({\psi^\#}^{-1}f\vert_V)=V(b) $$ with $b={\psi^\#}^{-1}f\vert_V$. So let $x\in W$. We need to show that $b\in\psi(x)$. We know that $$ a\in\phi(x). $$
Attempt (SOLVED):
Let's say for notational convenience that $W=\operatorname{Spec}A$ (equality as opposed to an isomorphism). We still work with an explicit isomorphism $\psi:U\cong\operatorname{Spec}(B)$. So we want to show that $$ \psi(\psi^{-1}(U)\cap W)=V(b) $$ where $b=(\psi^{-1})^\#f\vert_W$. Let $x\in \psi^{-1}(U)\cap W$. Then $x$ is given by a prime $\mathfrak p$ of $A$, and write $f\vert_W=a\in A$. Then we know that $a\in\mathfrak p$. Write $\mathfrak q=\psi(\mathfrak p)$. Choose some $g\in A$ such that $$ \mathfrak p\in\operatorname{Spec}(A_g)\subset \psi^{-1}(U)\cap W. $$ Note that we have a sheaf map $$ \Psi=\psi\vert_{\operatorname{Spec}(A_g)}:\operatorname{Spec} A_g\to\psi(\operatorname{Spec}A_g)\subset\operatorname{Spec}B, $$ which corresponds to a map of rings $$ G:B\to A_g, $$ i.e., $\Psi=\operatorname{Spec}(G)$. So we have $$ G^{-1}(\mathfrak p)=\mathfrak q. $$ To conclude that $b\in\mathfrak q$, I would need that $G(b)=a$. This is true, if we remember that $b=(\psi^{-1})^\# f\vert_U$ and $a=f\vert_W$ and that $G$ corresponds to $\psi^\#$ (along with restriction). More precisely, we have $$ a/1\in A_g=\mathcal O_{\operatorname{Spec}A}(\operatorname{Spec}A_g), $$ and $$ G:\sigma\mapsto \sigma\vert_{\psi(\operatorname{Spec}A_g)}\mapsto \psi^\#\sigma\vert_{\psi(\operatorname{Spec}A_g)}. $$ so recalling the definition of $b$, we get indeed $$ G:b\mapsto f\vert_{\operatorname{Spec}A_g}=a/1\in A_g. $$ It's clear to me how to adapt this proof for the 'general' case where we have an isomorphism $\phi:U\cong\operatorname{Spec}A$ (instead of equality), which I won't write down.
Second question (SOLVED):
How can I show that $\xi_1$ and $\xi_2$ would be generic points of $U_1\cap U_2$?
Attempt: (SOLVED)
Let $Z_i=\overline{\{\xi_i\}}$. I know that $U_1\cap U_2$ is irreducible, and hence contains a generic point $\xi$ (since $U$ is a scheme), i.e. $$ \operatorname{Cl}_{U_1\cap U_2}(\xi)=U_1\cap U_2. $$ It follows that $$ \operatorname{Cl}_{U_1}(\{\xi\})=\operatorname{Cl}_{U_1}\operatorname{Cl}_{U_1\cap U_2}(\{\xi\})=\operatorname{Cl}_{U_1}(U_1\cap U_2)=U_1, $$ where in the last step I used that $U_1\cap U_2$ is dense in $U_1$ (which follows from irreducibility of $U_1$). Therefore $$ \operatorname{Cl}_{Z_1\cap U_1}(\{\xi\})=\operatorname{Cl}_{U_1}(\{\xi\})\cap U_1\cap Z_1=U_1\cap Z_1. $$ Now using that $U_1\cap Z_1$ is dense in $Z_1$ (by irreducibility of $Z_1$), we conclude that $$ \operatorname{Cl}_{Z_1}(\{\xi_i\})=\operatorname{Cl}_{Z_1}(\operatorname{Cl}_{Z_1\cap U_1}(\{\xi\}))=Z_1. $$ and hence $\xi=\xi_1$ by uniqueness of the generic point. The case for $Z_2$ is similar.
