Lemma 76.6.4. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Let $y \in |Y|$. Assume

$f$ is decent and of finite type,

$\mathcal{F}$ is of finite type,

$\mathcal{F}$ is flat over $Y$ at all points lying over $y$, and

$\mathcal{F}$ is pure above $y$.

Then $\mathcal{F}$ is universally pure above $y$.

**Proof.**
Consider the morphism $\mathop{\mathrm{Spec}}(\mathcal{O}_{Y, \overline{y}}) \to Y$. This is a flat morphism from the spectrum of a stricly henselian local ring which maps the closed point to $y$. By Lemma 76.3.4 we reduce to the case described in the next paragraph.

Assume $Y$ is the spectrum of a strictly henselian local ring $R$ with closed point $y$. By Lemma 76.4.6 there exists an étale morphism $g : X' \to X$ with $g(|X'|) \supset |X_ y|$, with $X'$ affine, and with $\Gamma (X', g^*\mathcal{F})$ a free $R$-module. Then $g^*\mathcal{F}$ is universally pure relative to $Y$, see More on Flatness, Lemma 38.17.4. Hence it suffices to prove that $g(|X'|)$ contains $\text{Ass}_{X/Y}(\mathcal{F})$ by Lemma 76.6.3 part (1). This in turn follows from Lemma 76.6.3 part (2).
$\square$

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