Lemma 5.19.9. Let $f : X \to Y$ be a morphism of topological spaces. Suppose that $Y$ is a sober topological space, and $f$ is surjective. If either specializations or generalizations lift along $f$, then $\dim (X) \geq \dim (Y)$.
Proof. Assume specializations lift along $f$. Let $Z_0 \subset Z_1 \subset \ldots Z_ e \subset Y$ be a chain of irreducible closed subsets of $X$. Let $\xi _ e \in X$ be a point mapping to the generic point of $Z_ e$. By assumption there exists a specialization $\xi _ e \leadsto \xi _{e - 1}$ in $X$ such that $\xi _{e - 1}$ maps to the generic point of $Z_{e - 1}$. Continuing in this manner we find a sequence of specializations
\[ \xi _ e \leadsto \xi _{e - 1} \leadsto \ldots \leadsto \xi _0 \]
with $\xi _ i$ mapping to the generic point of $Z_ i$. This clearly implies the sequence of irreducible closed subsets
\[ \overline{\{ \xi _0\} } \subset \overline{\{ \xi _1\} } \subset \ldots \overline{\{ \xi _ e\} } \]
is a chain of length $e$ in $X$. The case when generalizations lift along $f$ is similar. $\square$
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