Let $f\colon X \to Y$ be a surjective open map between topological spaces. Show that if each set $f^{-1}({y})$ is irreducible , and if $Y$ is irreducible, then $X$ is irreducible.

Def: A topological space $X$ is irreducible if $X$ is nonempty and $X$ is not the union of two closed subsets different from $X$.(ofcourse there are many equivalent definition of irreducible topological space)

Because The fibres $f^{-1}(y)$ are just the equivalence classes under the relation $x\sim x'\leftrightarrow f(x)=f(x')$.Hence $X$ is union of disjoint equivalance classes,By using this I've proved this result by assuming $f$ to be a Closed map.But I'm unable in proving if $f$ is open map.Any ideas?

  • $\begingroup$ there is an alternate criterion for irreducibility i.e any two non empty open sets must intersect. You can try to solve this using this criterion $\endgroup$ – happymath Sep 7 '15 at 5:36
  • $\begingroup$ @happymath yeah I know this definition also.Can you say a bit more how to prove using this definition? $\endgroup$ – Arpit Kansal Sep 7 '15 at 5:38
  • $\begingroup$ I have written an answer assuming you know that definition $\endgroup$ – happymath Sep 7 '15 at 5:41

HINT: Suppose that $X$ is not irreducible, and let $U$ and $V$ be disjoint non-empty open subsets of $X$; since $f$ is open, $f[U]$ and $f[V]$ are non-empty open subsets of $Y$. Suppose that $y\in f[U]\cap f[V]$, and consider the fibre $f^{-1}[\{y\}]$.

  • $\begingroup$ Its not a hint,its "almost" a solution :) Thank you,Best Regards, $\endgroup$ – Arpit Kansal Sep 7 '15 at 5:45
  • $\begingroup$ @Arpit: It’s a very broad hint. :-) You’re welcome. $\endgroup$ – Brian M. Scott Sep 7 '15 at 5:46
  • $\begingroup$ May I know why do you write $f[U]$ instead of $f(U)$? $\endgroup$ – Arpit Kansal Sep 7 '15 at 5:47
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    $\begingroup$ @Arpit: The notation $f(U)$ makes it look as if $U$ is an element of the domain of $f$. Here, however, it’s a subset of that domain, and what we want is really $\{f(x):x\in U\}$; one common convention for distinguishing the two notions is to use square brackets for the image of a subset of the domain, so that $f[U]=\{f(x):x\in U\}$. $\endgroup$ – Brian M. Scott Sep 7 '15 at 5:49
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    $\begingroup$ @Arpit: Your questions look like more than basic stuff to me! Anyway, you’re welcome, and thank you. $\endgroup$ – Brian M. Scott Sep 7 '15 at 6:01

Say that there exists open sets $U_1$ and $U_2$ which are non empty and do not intersect. Then take image $f(U_1)$ and $f(U_2)$ they are open and have an intersection. Say $y$ belongs to the intersection. Then $f^{-1}(y) \cap U_1$ and $f^{-1}(y) \cap U_2$ are two open sets in $f^{-1}(y)$ which do not intersect contradicting the irreducibility of $f^{-1}(y)$


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