I am currently learning some algebraic geometry. I try to find out whether the projection morphisms of a fiber product are always surjective (as maps of the underlying sets). So suppose that $X_1,X_2$ are schemes over some scheme $S$. Are the morphisms $p_i: X_1 \times_S X_2 \to X_i$ then necessarily surjective? Are they surjective if $X_1,X_2$ are separated?
I tried to show that this is indeed the case, but I haven't achieved much. Here is my attempt. By symmetry, it suffices to show that $p_1$ is surjective. If we can construct a scheme $Z$ with morphisms $f_i: Z \to X_i$ such that $f_1$ is surjective, then by the universal property we get a morphism $\psi: Z \to X_1 \times_S X_2$ such that $f_1 = p_1 \circ \psi$ (from which the surjectivity of $p_1$ follows). If $X_1 = X_2$, we can just take $\psi$ to be the diagonal embedding and then we're done. In all other cases, I do not see how I can proceed.
Let $X\subset Y$ be proper closed subscheme and let $S=Y$, with the natural maps $X\to S, Y\to S$. Then $X\times_S Y\to Y$ has image precisely $X\neq Y$.