The Stacks project

Proof. Let $f : X \to S$ be a morphism of schemes. Let $\{ S_ i \to S\} $ be an fpqc covering, and assume each of the base changes $f_ i : X_ i \to S_ i$ of $f$ is a monomorphism. Let $a, b : T \to X$ be two morphisms such that $f \circ a = f \circ b$. We have to show that $a = b$. Since $f_ i$ is a monomorphism we see that $a_ i = b_ i$, where $a_ i, b_ i : S_ i \times _ S T \to X_ i$ are the base changes. In particular the compositions $S_ i \times _ S T \to T \to X$ are equal. Since $\coprod S_ i \times _ S T \to T$ is an epimorphism (see e.g. Lemma 35.13.7) we conclude $a = b$. $\square$