Lemma 67.10.6 (042Q)—The Stacks project

Lemma 67.10.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent

$f$ is a monomorphism,
for every scheme $Z$ and morphism $Z \to Y$ the base change $Z \times _ Y X \to Z$ of $f$ is a monomorphism,
for every affine scheme $Z$ and every morphism $Z \to Y$ the base change $Z \times _ Y X \to Z$ of $f$ is a monomorphism,
there exists a scheme $V$ and a surjective étale morphism $V \to Y$ such that the base change $V \times _ Y X \to V$ is a monomorphism, and
there exists a Zariski covering $Y = \bigcup Y_ i$ such that each of the morphisms $f^{-1}(Y_ i) \to Y_ i$ is a monomorphism.

Proof. We will use without further mention that a base change of a monomorphism is a monomorphism, see Lemma 67.10.5. In particular it is clear that (1) $\Rightarrow $ (2) $\Rightarrow $ (3) $\Rightarrow $ (4) (by taking $V$ to be a disjoint union of affine schemes étale over $Y$, see Properties of Spaces, Lemma 66.6.1). Let $V$ be a scheme, and let $V \to Y$ be a surjective étale morphism. If $V \times _ Y X \to V$ is a monomorphism, then it follows that $X \to Y$ is a monomorphism. Namely, given any cartesian diagram of sheaves

\[ \vcenter { \xymatrix{ \mathcal{F} \ar[r]_ a \ar[d]_ b & \mathcal{G} \ar[d]^ c \\ \mathcal{H} \ar[r]^ d & \mathcal{I} } } \quad \quad \mathcal{F} = \mathcal{H} \times _\mathcal {I} \mathcal{G} \]

if $c$ is a surjection of sheaves, and $a$ is injective, then also $d$ is injective. Thus (4) implies (1). Proof of the equivalence of (5) and (1) is omitted. $\square$

The Stacks project

Comments (0)

Post a comment