Lemma 5.8.14 (004Z)—The Stacks project

Lemma 5.8.14. Let $f : X \to Y$ be a continuous map of topological spaces. Assume that (a) $Y$ is irreducible, (b) $f$ is open, and (c) there exists a dense collection of points $y \in Y$ such that $f^{-1}(y)$ is irreducible. Then $X$ is irreducible.

Proof. Suppose $X = Z_1 \cup Z_2$ with $Z_ i$ closed. Consider the open sets $U_1 = Z_1 \setminus Z_2 = X \setminus Z_2$ and $U_2 = Z_2 \setminus Z_1 = X \setminus Z_1$. To get a contradiction assume that $U_1$ and $U_2$ are both nonempty. By (b) we see that $f(U_ i)$ is open. By (a) we have $Y$ irreducible and hence $f(U_1) \cap f(U_2) \not= \emptyset $. By (c) there is a point $y$ which corresponds to a point of this intersection such that the fibre $X_ y = f^{-1}(y)$ is irreducible. Then $X_ y \cap U_1$ and $X_ y \cap U_2$ are nonempty disjoint open subsets of $X_ y$ which is a contradiction. $\square$

Comments (5)

Comment #631 by Wei Xu on May 30, 2014 at 09:14

Typo, $Y$ in "Suppose $Y = Z_1 \cup Z_2$ with $Z_i$ closed." should be $X$ .

Comment #1147 by sid on November 07, 2014 at 07:41

Typo, Y in "U1=Z1∖Z2=Y∖Z" should be X.

Comment #1168 by Johan on December 02, 2014 at 13:37

Thanks! See here.

Comment #5951 by Fabian Korthauer on March 01, 2021 at 16:50

The assumption that $f$ is continuous seems to be superfluous. Furthermore one could maybe add a comment (resp. add an analogous lemma somewhere else) that the statement remains true with essentially the same prove if one exchanges the word "irreducible" everywhere by the word "connected".

Comment #6137 by Johan on April 30, 2021 at 18:41

@#5951. OK, yes, I agree. There are probably many variants of this lemma. For now I will leave it as is, unless you want a reference for a publication. The variants for connectedness are Lemmas 5.7.5 and 5.7.6.

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9 comment(s) on Section 5.8: Irreducible components

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