Lemma 5.8.15 (037A)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 5: Topology
Section 5.8: Irreducible components
Lemma 5.8.15 (cite)

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Lemma 5.8.15. Let $f : X \to Y$ be a continuous map of topological spaces. Assume that (a) $f$ is open, and (b) for every $y \in Y$ the fibre $f^{-1}(y)$ is irreducible. Then $f$ induces a bijection between irreducible components.

Proof. We point out that assumption (b) implies that $f$ is surjective (see Definition 5.8.1). Let $T \subset Y$ be an irreducible component. Note that $T$ is closed, see Lemma 5.8.3. The lemma follows if we show that $f^{-1}(T)$ is irreducible because any irreducible subset of $X$ maps into an irreducible component of $Y$ by Lemma 5.8.2. Note that $f^{-1}(T) \to T$ satisfies the assumptions of Lemma 5.8.14. Hence we win. $\square$

Comments (2)

Comment #632 by Wei Xu on May 30, 2014 at 09:15

Typo, $p$ should be $f$ .

Comment #641 by Johan on June 01, 2014 at 12:03

Thanks for this and comments #624 -- #632. Fixed here.

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