Lemma 67.19.6 (03MX)—The Stacks project

Lemma 67.19.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:

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

Proof. We will use that being universally injective is preserved under base change (Lemma 67.19.5) without further mention in this proof. It is clear that (1) $\Rightarrow $ (2) $\Rightarrow $ (3) $\Rightarrow $ (4).

Assume $g : Z \to Y$ as in (4). Let $y : \mathop{\mathrm{Spec}}(K) \to Y$ be a morphism from the spectrum of a field into $Y$. By assumption we can find an extension field $\alpha : K \subset K'$ and a morphism $z : \mathop{\mathrm{Spec}}(K') \to Z$ such that $y \circ \alpha = g \circ z$ (with obvious abuse of notation). By assumption the morphism $Z \times _ Y X \to Z$ is universally injective, hence there is at most one lift of $g \circ z : \mathop{\mathrm{Spec}}(K') \to Y$ to a morphism into $X$. Since $\{ \alpha : \mathop{\mathrm{Spec}}(K') \to \mathop{\mathrm{Spec}}(K)\} $ is a fpqc covering this implies there is at most one lift of $y : \mathop{\mathrm{Spec}}(K) \to Y$ to a morphism into $X$, see Properties of Spaces, Proposition 66.17.1. Thus we see that (1) holds.

We omit the verification that (5) is equivalent to (1). $\square$

The Stacks project

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