Lemma 76.3.6. In Situation 76.2.1.

1. If the support of $\mathcal{F}$ is proper over $Y$, then $\mathcal{F}$ is universally pure relative to $Y$.

2. If $f$ is proper, then $\mathcal{F}$ is universally pure relative to $Y$.

3. If $f$ is proper, then $X$ is universally pure relative to $Y$.

Proof. First we reduce (1) to (2). Namely, let $Z \subset X$ be the scheme theoretic support of $\mathcal{F}$ (Morphisms of Spaces, Definition 66.15.4). Let $i : Z \to X$ be the corresponding closed immersion and write $\mathcal{F} = i_*\mathcal{G}$ for some finite type quasi-coherent $\mathcal{O}_ Z$-module $\mathcal{G}$. In case (1) $Z \to Y$ is proper by assumption. Thus by Lemma 76.3.5 case (1) reduces to case (2).

Assume $f$ is proper. Let $(g : T \to Y, t' \leadsto t, \xi )$ be an impurity of $\mathcal{F}$ above $y$. Since $f$ is proper, it is universally closed. Hence $f_ T : X_ T \to T$ is closed. Since $f_ T(\xi ) = t'$ this implies that $t \in f(\overline{\{ \xi \} })$ which is a contradiction. $\square$

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