Lemma 101.14.7. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The following are equivalent:

1. $f$ is universally injective,

2. 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

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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