Lemma 42.11.2 (02R2)—The Stacks project

Lemma 42.11.2. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ , $Y$ be locally of finite type over $S$ . Let $f : X \to Y$ be a morphism. Assume $f$ is quasi-compact, and $\{ Z_ i\} _{i \in I}$ is a locally finite collection of closed subsets of $X$ . Then $\{ \overline{f(Z_ i)}\} _{i \in I}$ is a locally finite collection of closed subsets of $Y$ .

Proof. Let $V \subset Y$ be a quasi-compact open subset. Since $f$ is quasi-compact the open $f^{-1}(V)$ is quasi-compact. Hence the set $\{ i \in I \mid Z_ i \cap f^{-1}(V) \not= \emptyset \}$ is finite by a simple topological argument which we omit. Since this is the same as the set

$\{ i \in I \mid f(Z_ i) \cap V \not= \emptyset \} = \{ i \in I \mid \overline{f(Z_ i)} \cap V \not= \emptyset \}$

the lemma is proved. $\square$

Comments (2)

Comment #123 by Adeel Ahmad Khan on February 14, 2013 at 16:26

I think that one should note that (1) the quasi-compact open subsets form a basis for the topology, and (2) therefore a collection of subspaces $Z_i$ of $X$ is locally finite iff for all quasi-compact open subspaces $U \subset X$ , the number of $Z_i$ intersecting $U$ is finite; one uses (1) for the reverse implication.

Comment #126 by Johan on February 14, 2013 at 18:37

OK, I tried to improve the exposition. The confusion I think arose because we are not using the most general definition of locally finite collection of subspaces, i.e., we have incorporated the thing with the quasi-compact opens into Definition 42.8.1. We really should define locally finite collections of subspaces in complete generality for subsets of topological spaces and then have some lemmas about this notion...

Comments (2)

Post a comment