Lemma 66.25.6. Let $S$ be a scheme. Let $f : Y \to X$ be a surjective finite morphism of decent and locally Noetherian algebraic spaces. Let $\delta : |X| \to \mathbf{Z}$ be a function. If $\delta \circ |f|$ is a dimension function, then $\delta$ is a dimension function.

Proof. Let $x \mapsto x'$, $x \not= x'$ be a specialization in $|X|$. Choose $y \in |Y|$ with $|f|(y) = x$. Since $|f|$ is closed (Morphisms of Spaces, Lemma 65.45.9) we find a specialization $y \leadsto y'$ with $|f|(y') = x'$. Thus we conclude that $\delta (x) = \delta (|f|(y)) > \delta (|f|(y')) = \delta (x')$ (see Topology, Definition 5.20.1). If $x \leadsto x'$ is an immediate specialization, then $y \leadsto y'$ is an immediate specialization too: namely if $y \leadsto y'' \leadsto y'$, then $|f|(y'')$ must be either $x$ or $x'$ and there are no nontrivial specializations between points of fibres of $|f|$ by Lemma 66.18.10. $\square$

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