Lemma 101.14.7. Let f : \mathcal{X} \to \mathcal{Y} be a morphism of algebraic stacks. The following are equivalent:
f is universally injective,
for every affine scheme Z and any morphism Z \to \mathcal{Y} the morphism Z \times _\mathcal {Y} \mathcal{X} \to Z is universally injective, and
add more here.
Proof. The implication (1) \Rightarrow (2) is immediate. Assume (2) holds. We will show that \Delta _ f : \mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X} is surjective, which implies (1) by Lemma 101.14.5. Consider an affine scheme V and a smooth morphism V \to \mathcal{Y}. Since g : V \times _\mathcal {Y} \mathcal{X} \to V is universally injective by (2), we see that \Delta _ g is surjective. However, \Delta _ g is the base change of \Delta _ f by the smooth morphism V \to \mathcal{Y}. Since the collection of these morphisms V \to \mathcal{Y} are jointly surjective, we conclude \Delta _ f is surjective. \square
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