Lemma 70.4.8. With notation and assumptions as in Lemma 70.4.7. Assume $g$ is locally quasi-finite, or more generally that for every $y' \in |Y'|$ the transcendence degree of $y'/g(y')$ is $0$. Then $\text{Ass}_{X'/Y'}(\mathcal{F}')$ is the inverse image of $\text{Ass}_{X/Y}(\mathcal{F})$.

**Proof.**
The transcendence degree of a point over its image is defined in Morphisms of Spaces, Definition 66.33.1. Let $x' \in |X'|$ with image $x \in |X|$. Choose a scheme $V$ and a surjective étale morphism $V \to Y$. Choose a scheme $U$ and a surjective étale morphism $U \to V \times _ Y X$. Choose a scheme $V'$ and a surjective étale morphism $V' \to V \times _ Y Y'$. Then $U' = V' \times _ V U$ is a scheme and the morphism $U' \to X'$ is surjective and étale. Choose $u \in U$ mapping to $x$. Denote $v \in V$ the image of $u$. Then $x \in \text{Ass}_{X/Y}(\mathcal{F})$ is equivalent to $u \in \text{Ass}(\mathcal{F}|_{U_ v})$ by definition. Choose a point $u' \in U'$ mapping to $x' \in |X'|$ and to $u \in U$ (possible by Properties of Spaces, Lemma 65.4.3). Let $v' \in V'$ be the image of $u'$. Then $x' \in \text{Ass}_{X'/Y'}(\mathcal{F}')$ is equivalent to $u' \in \text{Ass}(\mathcal{F}'|_{U'_{v'}})$ by definition. Now the lemma follows from the discussion in Divisors, Remark 31.7.4 applied to $u' \in \mathop{\mathrm{Spec}}(\kappa (v') \otimes _{\kappa (v)} \kappa (u))$.
$\square$

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