Let $ f : X \to S, g : Y \to S$ be the scheme morphisms, and $ X \times_S Y$ be the product of shemes. Let $ z \in X \times_S Y$ , and $ x=p(z) , y=q(z) ,$ where $ p: X \times_S Y \to X, q: X \times_S Y\to Y $ are projections. Suppose $ s=f(x)=g(y) \in S$ and $x\in U \subset X, y\in V \subset Y, s\in W \subset S, f(U) \subset W, g(V) \subset W$ where $U, V, W$ are arbitrary open sets. Is it ture $ z\in U \times_W V$ ?

(The question should be read as : Is it ture that $ z$ is in the image of open immersion $U \times_W V \to X \times_S Y$ )

Let $ f : X \to S, g : Y \to S$ be the scheme morphisms, and $ X \times_S Y$ be the product of shemes. Let $ z \in X \times_S Y$ , and $ x=p(z) , y=q(z) ,$ where $ p: X \times_S Y \to X, q: X \times_S Y\to Y $ are projections. Suppose $ s=f(x)=g(y) \in S$ and $x\in U \subset X, y\in V \subset Y, s\in W \subset S, f(U) \subset W, g(V) \subset W$ where $U, V, W$ are arbitrary open sets. Is it ture $ z\in U \times_W V$ ?

(The question should be read as : Is it ture that $ z$ is in the image of open immersion $U \times_W V \to X \times_S Y$ )

Just a little comment: this can be used to prove monomorphism (particularly open and closed immersions) is seperated. The reason is, by the above result, one can pick up an open cover of $ X \times_Y X$ as $ U \times_V U$, because of the monomorphism property, two projections are the same, so one can choose the same open set $U$, and by the morphism of affines is seperated, we are done.