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.



An element in the product of schemes
Let $Z=\{z\}$. Then $p$ and $q$ induce morphisms $Z\to U$ and $Z\to V$ compatible with the structure maps $U\to W$ and $V\to W$. Then by the definition of a product there exists a *unique* morphism $Z\to U\times\_W V$ compatible with these morphisms. You can repeat the same argument for $X,Y,S$ and get a unique morphism compatible with the appropriate morphisms. This has to equal the one coming from the original embedding $Z\hookrightarrow X\times\_SY$. Now clearly the first one combined with the natural morphism $U\times\_W V\to X\times\_S Y$ gives another such morphism so by uniqueness it has to be equal to the original embedding, so in other words the latter has to factor through $U\times\_W V\to X\times\_S Y$.


