The Stacks project

Lemma 5.8.2. Let $f : X \to Y$ be a continuous map of topological spaces. If $E \subset X$ is an irreducible subset, then $f(E) \subset Y$ is irreducible as well.

Proof. Clearly we may assume $E = X$ (i.e., $X$ irreducible) and $f(E) = Y$ (i.e., $f$ surjective). First, $Y \not= \emptyset $ since $X \not= \emptyset $. Next, assume $Y = Y_1 \cup Y_2$ with $Y_1$, $Y_2$ closed. Then $X = X_1 \cup X_2$ with $X_ i = f^{-1}(Y_ i)$ closed in $X$. By assumption on $X$, we must have $X = X_1$ or $X = X_2$, hence $Y = Y_1$ or $Y = Y_2$ since $f$ is surjective. $\square$


Comments (4)

Comment #118 by Adeel Ahmad Khan on

Here is a proof:

Suppose is the union of and , for two distinct closed subsets and of ; this is equal to the intersection , so is then contained in the union . For the irreducibility of it suffices to show that it is contained in either or . The relation shows that ; as the right-hand side is clearly equal to and since , it follows that , from which one concludes by the irreducibility of that or . Hence one sees that either or .

Comment #3565 by Laurent Moret-Bailly on

Things get simpler if you first replace by . The statement becomes: "if is irreducible and surjective then is irreducible". Now of course is nonempty if is, and then if with , closed, then from the irreducibility assumption you get that for some , hence since is surjective.

Comment #3689 by on

OK, yes what you say makes sense, but since the proof is correct and since this is so early in the theory, it isn't worth it to me to make the change. If you want to make the edits to the latex file and send it to me, then I'll reconsider.

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

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