Lemma 74.11.30. 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 73.9.3 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 74.11.15 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 73.9.5.
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 66.17.1. \square
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