The Stacks project

Proof. Suppose we have $h : X' \to X$ surjective and flat and $a, b : X \to Y$ morphisms such that $a \circ h = b \circ h$. As $h$ is surjective we see that $a$ and $b$ agree on underlying topological spaces. Pick $x' \in X'$ and set $x = h(x')$ and $y = a(x) = b(x)$. Consider the local ring maps

These become equal when composed with the flat local homomorphism $h^\sharp _{x'} : \mathcal{O}_{X, x} \to \mathcal{O}_{X', x'}$. Since a flat local homomorphism is faithfully flat (Algebra, Lemma 10.39.17) we conclude that $h^\sharp _{x'}$ is injective. Hence $a^\sharp _ x = b^\sharp _ x$ which implies $a = b$ as desired. $\square$