Lemma 5.23.11. Let $f : X \to Y$ be a continuous map of topological spaces. If
$X$ and $Y$ are spectral,
$f$ is spectral and bijective, and
generalizations (resp. specializations) lift along $f$.
Then $f$ is a homeomorphism.
Lemma 5.23.11. Let $f : X \to Y$ be a continuous map of topological spaces. If
$X$ and $Y$ are spectral,
$f$ is spectral and bijective, and
generalizations (resp. specializations) lift along $f$.
Then $f$ is a homeomorphism.
Proof. Since $f$ is spectral it defines a continuous map between $X$ and $Y$ in the constructible topology. By Lemmas 5.23.2 and 5.17.8 it follows that $X \to Y$ is a homeomorphism in the constructible topology. Let $U \subset X$ be quasi-compact open. Then $f(U)$ is constructible in $Y$. Let $y \in Y$ specialize to a point in $f(U)$. By the last assumption we see that $f^{-1}(y)$ specializes to a point of $U$. Hence $f^{-1}(y) \in U$. Thus $y \in f(U)$. It follows that $f(U)$ is open, see Lemma 5.23.6. Whence $f$ is a homeomorphism. To prove the lemma in case specializations lift along $f$ one shows instead that $f(Z)$ is closed if $X \setminus Z$ is a quasi-compact open of $X$. $\square$
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