Lemma 5.19.5. Suppose $f : X \to Y$ and $g : Y \to Z$ are continuous maps of topological spaces. If specializations lift along both $f$ and $g$ then specializations lift along $g \circ f$. Similarly for “generalizations lift along”.
Proof. Let $z'\leadsto z$ be a specialization in $Z$ and let $x' \in X$ such as $g\circ f (x') = z'$. Then because specializations lift along $g$, there exists a specialization $f(x') \leadsto y$ of $f(x')$ in $Y$ such that $g(y) = z$. Likewise, because specializations lift along $f$, there exists a specialization $x' \leadsto x$ of $x'$ in $X$ such that $f(x) = y$. It provides a specialization $x' \leadsto x$ of $x'$ in $X$ such that $g\circ f(x) = z$. In other words, specialization lift along $g\circ f$. $\square$
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