Lemma 73.11.30. The property $\mathcal{P}(f) =$“$f$ is a monomorphism” is fpqc local on the base.

**Proof.**
Let $f : X \to Y$ be a morphism of algebraic spaces. Let $\{ Y_ i \to Y\} $ be an fpqc covering, and assume each of the base changes $f_ i : X_ i \to Y_ i$ of $f$ is a monomorphism. We have to show that $f$ is a monomorphism.

First proof. Note that $f$ is a monomorphism if and only if $\Delta : X \to X \times _ Y X$ is an isomorphism. By applying this to $f_ i$ we see that each of the morphisms

is an isomorphism. The base change of an fpqc covering is an fpqc covering, see Topologies on Spaces, Lemma 72.9.3 hence $\{ Y_ i \times _ Y (X \times _ Y X) \to X \times _ Y X\} $ is an fpqc covering of algebraic spaces. Moreover, each $\Delta _ i$ is the base change of the morphism $\Delta : X \to X \times _ Y X$. Hence it follows from Lemma 73.11.15 that $\Delta $ is an isomorphism, i.e., $f$ is a monomorphism.

Second proof. Let $V$ be a scheme, and let $V \to Y$ be a surjective étale morphism. If we can show that $V \times _ Y X \to V$ is a monomorphism, then it follows that $X \to Y$ is a monomorphism. Namely, given any cartesian diagram of sheaves

if $c$ is a surjection of sheaves, and $a$ is injective, then also $d$ is injective. This reduces the problem to the case where $Y$ is a scheme. Moreover, in this case we may assume that the algebraic spaces $Y_ i$ are schemes also, since we can always refine the covering to place ourselves in this situation, see Topologies on Spaces, Lemma 72.9.5.

Assume $\{ Y_ i \to Y\} $ is an fpqc covering of schemes. 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 $a_ i, b_ i : Y_ i \times _ Y T \to X_ i$ are the base changes. In particular the compositions $Y_ i \times _ Y T \to T \to X$ are equal. Since $\{ Y_ i \times _ Y T \to T\} $ is an fpqc covering we deduce that $a = b$ from Properties of Spaces, Proposition 65.17.1. $\square$

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