Fiber product of schemes: when is $X \times_S Y \cong X \times_U i^{-1}(U)$?

Question

Let $j \colon X \rightarrow S$ and $i \colon Y \rightarrow S$ be morphisms of schemes. Let $U \subset S$ be an open subscheme of $S$ such that $j(X) \subset U$. Under which assumptions do we get an isomorphism of fiber products $X \times_S Y \cong X \times_U i^{-1}(U)$?

Answer 1

Always. In fact, $i^{-1}(U)$ is $U \times_S Y$. Now use the general rule $$X \times_U (U \times_S Y) \cong X \times_S Y.$$