Lemma 35.23.31. The property $\mathcal{P}(f) =$“$f$ is a monomorphism” is fpqc local on the base.

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 $a = b$. $\square$

There are also:

• 2 comment(s) on Section 35.23: Properties of morphisms local in the fpqc topology on the target

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).