Lemma 71.4.7. Let $S$ be a scheme. Let

\[ \xymatrix{ X' \ar[d]_{f'} \ar[r]_{g'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y } \]

be a cartesian diagram of algebraic spaces over $S$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module and set $\mathcal{F}' = (g')^*\mathcal{F}$. If $f$ is locally of finite type, then

$x' \in \text{Ass}_{X'/Y'}(\mathcal{F}') \Rightarrow g'(x') \in \text{Ass}_{X/Y}(\mathcal{F})$

if $x \in \text{Ass}_{X/Y}(\mathcal{F})$, then given $y' \in |Y'|$ with $f(x) = g(y')$, there exists an $x' \in \text{Ass}_{X'/Y'}(\mathcal{F}')$ with $g'(x') = x$ and $f'(x') = y'$.

**Proof.**
This follows from the case of schemes by étale localization. We write out the details completely. 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.

Proof of (1). 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 (writing $\text{Ass}$ instead of $\text{WeakAss}$ makes sense as $U'_{v'}$ is locally Noetherian). Applying Divisors, Lemma 31.7.3 we see that the image $u \in U$ of $u'$ is in $\text{Ass}(\mathcal{F}|_{U_ v})$ where $v \in V$ is the image of $u$. This in turn means $g'(x') \in \text{Ass}_{X/Y}(\mathcal{F})$.

Proof of (2). 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 $v' \in V'$ mapping to $y' \in |Y'|$ and to $v \in V$ (possible by Properties of Spaces, Lemma 66.4.3). Let $t \in \mathop{\mathrm{Spec}}(\kappa (v') \otimes _{\kappa (v)} \kappa (u))$ be a generic point of an irreducible component. Let $u' \in U'$ be the image of $t$. Applying Divisors, Lemma 31.7.3 we see that $u' \in \text{Ass}(\mathcal{F}'|_{U'_{v'}})$. This in turn means $x' \in \text{Ass}_{X'/Y'}(\mathcal{F}')$ where $x' \in |X'|$ is the image of $u'$.
$\square$

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