The Stacks project

Proof. Let $a, b : Z \to X$ be two morphisms of schemes such that $j \circ a = j \circ b$. Then (1) implies $a = b$ as underlying maps of topological spaces. For any $z \in Z$ we have $a^\sharp _ z \circ j^\sharp _{a(z)} = b^\sharp _ z \circ j^\sharp _{b(z)}$ as maps $\mathcal{O}_{Y, j(a(z))} \to \mathcal{O}_{Z, z}$. The surjectivity of the maps $j^\sharp _ x$ forces $a^\sharp _ z = b^\sharp _ z$, $\forall z \in Z$. This implies that $a^\sharp = b^\sharp$. Hence we conclude $a = b$ as morphisms of schemes as desired. $\square$