Lemma 5.23.3. Let $f : X \to Y$ be a spectral map of spectral spaces. Then

1. $f$ is continuous in the constructible topology,

2. the fibres of $f$ are quasi-compact, and

3. the image is closed in the constructible topology.

Proof. Let $X'$ and $Y'$ denote $X$ and $Y$ endowed with the constructible topology which are quasi-compact Hausdorff spaces by Lemma 5.23.2. Part (1) says $X' \to Y'$ is continuous and follows immediately from the definitions. Part (3) follows as $f(X')$ is a quasi-compact subset of the Hausdorff space $Y'$, see Lemma 5.12.4. We have a commutative diagram

$\xymatrix{ X' \ar[r] \ar[d] & X \ar[d] \\ Y' \ar[r] & Y }$

of continuous maps of topological spaces. Since $Y'$ is Hausdorff we see that the fibres $X'_ y$ are closed in $X'$. As $X'$ is quasi-compact we see that $X'_ y$ is quasi-compact (Lemma 5.12.3). As $X'_ y \to X_ y$ is a surjective continuous map we conclude that $X_ y$ is quasi-compact (Lemma 5.12.7). $\square$

Comment #5058 by Laurent Moret-Bailly on

In fact, (1) is an equivalence in the sense that if $f$ is continuous, then it is spectral iff it is continuous for the constructible topology. Proof of "if": if $U\subset Y$ is quasicompact open, then it is constructibly clopen and so is $f^{-1}(U)$, which is therefore constructibly compact, hence quasicompact in $X$.

Comment #9049 by andy on

Doesn't (2) just follow b/c the point is compact and the fibre is inverse image of a point and the map is compact? Also very minor thing--might be helpful to say that the last part is proving (2)

Comment #9050 by andy on

Ah oops I guess compact map only asks for inverse of compact open is compact, sorry!

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