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. $\square$

Comment #632 by Wei Xu on

Typo, $p$ should be $f$.

There are also:

• 8 comment(s) on Section 5.8: Irreducible components

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).