Lemma 37.61.2. In Situation 37.61.1 then for $y \in Y$ there exist affine opens $U \subset X$ and $V \subset Y$ with $i^{-1}(U) = j^{-1}(V)$ and $y \in V$.

**Proof.**
Let $y \in Y$. Choose an affine open $U \subset X$ such that $j^{-1}(\{ y\} ) \subset i^{-1}(U)$ (possible by assumption). Choose an affine open $V \subset Y$ neighbourhood of $y$ such that $j^{-1}(V) \subset i^{-1}(U)$. This is possible because $j : Z \to Y$ is a closed morphism (Morphisms, Lemma 29.44.7) and $i^{-1}(U)$ contains the fibre over $y$. Since $j$ is integral, the scheme theoretic fibre $Z_ y$ is the spectrum of an algebra integral over a field. By Limits, Lemma 32.11.6 we can find an $\overline{f} \in \Gamma (i^{-1}(U), \mathcal{O}_{i^{-1}(U)})$ such that $Z_ y \subset D(\overline{f}) \subset j^{-1}(V)$. Since $i|_{i^{-1}(U)} : i^{-1}(U) \to U$ is a closed immersion of affines, we can choose an $f \in \Gamma (U, \mathcal{O}_ U)$ whose restriction to $i^{-1}(U)$ is $\overline{f}$. After replacing $U$ by the principal open $D(f) \subset U$ we find affine opens $y \in V \subset Y$ and $U \subset X$ with

Now we (in some sense) repeat the argument. Namely, we choose $g \in \Gamma (V, \mathcal{O}_ V)$ such that $y \in D(g)$ and $j^{-1}(D(g)) \subset i^{-1}(U)$ (possible by the same argument as above). Then we can pick $f \in \Gamma (U, \mathcal{O}_ U)$ whose restriction to $i^{-1}(U)$ is the pullback of $g$ by $i^{-1}(U) \to V$ (again possible by the same reason as above). Then we finally have affine opens $y \in V' = D(g) \subset V \subset Y$ and $U' = D(f) \subset U \subset X$ with $j^{-1}(V') = i^{-1}(V')$. $\square$

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