The Stacks project

Lemma 41.20.2. Assume $f : X \to S$ is submersive and any étale base change of $f$ is submersive. Then the functor ( is fully faithful.

Proof. By Lemma 41.20.1 the functor is faithful. Let $U_1 \to S$ and $U_2 \to S$ be étale morphisms and let $a : X \times _ S U_1 \to X \times _ S U_2$ be a morphism compatible with canonical descent data. We will prove that $a$ is the base change of a morphism $U_1 \to U_2$.

Let $U'_2 \subset U_2$ be an open subscheme. Consider $W = a^{-1}(X \times _ S U'_2)$. This is an open subscheme of $X \times _ S U_1$ which is compatible with the canonical descent datum on $V_1 = X \times _ S U_1$. This means that the two inverse images of $W$ by the projections $V_1 \times _{U_1} V_1 \to V_1$ agree. Since $V_1 \to U_1$ is surjective (as the base change of $X \to S$) we conclude that $W$ is the inverse image of some subset $U'_1 \subset U_1$. Since $W$ is open, our assumption on $f$ implies that $U'_1 \subset U_1$ is open.

Let $U_2 = \bigcup U_{2, i}$ be an affine open covering. By the result of the preceding paragraph we obtain an open covering $U_1 = \bigcup U_{1, i}$ such that $X \times _ S U_{1, i} = a^{-1}(X \times _ S U_{2, i})$. If we can prove there exists a morphism $U_{1, i} \to U_{2, i}$ whose base change is the morphism $a_ i : X \times _ S U_{1, i} \to X \times _ S U_{2, i}$ then we can glue these morphisms to a morphism $U_1 \to U_2$ (using faithfulness). In this way we reduce to the case that $U_2$ is affine. In particular $U_2 \to S$ is separated (Schemes, Lemma 26.21.13).

Assume $U_2 \to S$ is separated. Then the graph $\Gamma _ a$ of $a$ is a closed subscheme of

\[ V = (X \times _ S U_1) \times _ X (X \times _ S U_2) = X \times _ S U_1 \times _ S U_2 \]

by Schemes, Lemma 26.21.10. On the other hand the graph is open for example because it is a section of an étale morphism (Proposition 41.6.1). Since $a$ is a morphism of descent data, the two inverse images of $\Gamma _ a \subset V$ under the projections $V \times _{U_1 \times _ S U_2} V \to V$ are the same. Hence arguing as in the second paragraph of the proof we find an open and closed subscheme $\Gamma \subset U_1 \times _ S U_2$ whose base change to $X$ gives $\Gamma _ a$. Then $\Gamma \to U_1$ is an étale morphism whose base change to $X$ is an isomorphism. This means that $\Gamma \to U_1$ is universally bijective, hence an isomorphism by Theorem 41.14.1. Thus $\Gamma $ is the graph of a morphism $U_1 \to U_2$ and the base change of this morphism is $a$ as desired. $\square$

Comments (2)

Comment #2872 by Ko Aoki on

Typo in the proof: "In particular is separated" should be replaced by "In particular is separated".

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