Let $X, Y$ be topological spaces, we say that a continuous $f:X\rightarrow Y$ is irreducible if $f$ is onto $Y$ and whenever $F \subset X$ is closed, if $f[F]=Y$ then $F=X$.

I'm looking for an example of $X, Y$ compact Hausdorff spaces and $f:X\rightarrow Y$ irreducible such that for every $y \in F$, $|f^{-1}[\{y\}]|>1$. I have tried some compact subsets of $\mathbb R$ and some one point compactifications and compact ordinals, but nothing worked.

  • $\begingroup$ This usage of the word irreducible is unfamiliar to me. Is it standard terminology? What's a reference? (It's not included on this list.) $\endgroup$ – Alex Kruckman Apr 14 '16 at 1:31
  • $\begingroup$ Ryszard Engelking, General Topology, exercise 3.1.C. This exercise tells me to prove some stuff related to irreducible functions, and asks for this example, I did everything else but I couldn't think of such an example :p $\endgroup$ – Vinicius Rodrigues Apr 14 '16 at 2:58

Let $X$ be the double arrow space; $$X=[0,1]\times \{0,1\}$$ with the lexicographic order topology. $X$ is compact Hausdorff (though non-metrizable). Let $$Y=[0,1]$$and take $f:X\to Y$ to be the first coordinate projection, i.e., $$f(r,n)=r$$ for $r\in[0,1]$ and $n\in \{0,1\}$.

Then (i) $f$ is continuous, and (ii) every set that maps onto $Y$ is dense in $X$.

Therefore $f$ is irreducible. Clearly $|f^{-1}\{y\}|=2$ for each $y\in Y$.

If you can picture the basic open subsets of $X$, then (i) and (ii) are easy to prove. The middle section in Double Arrow Space will help you think of these open sets.

EDIT: It has occurred to me that the points $(0,1)$ and $(1,0)$ are isolated, and therefore (ii) may not hold. We can fix this problem by replacing all instances of $[0,1]$ with $S^1$ - it is not linearly ordered, but there is a simple double-arrow-ification of $S^1\times \{0,1\}$:

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  • $\begingroup$ I have not seen the double arrow space before, is the topology on $X$ the order topology induced by the lexicographic order? $\endgroup$ – Vinicius Rodrigues Apr 14 '16 at 2:23
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    $\begingroup$ @ViniciusRodrigues please see my edit. there was a minor problem. $\endgroup$ – Forever Mozart Apr 14 '16 at 4:08
  • $\begingroup$ @ViniciusRodrigues I was bored so I made a picture for you :) $\endgroup$ – Forever Mozart Apr 14 '16 at 6:43
  • $\begingroup$ It may be mildly helpful to note that the space in the edit can be obtained from the original $X$ by identifying $\langle 0,0\rangle$ with $\langle 1,0\rangle$ and $\langle 0,1\rangle$ with $\langle 1,1\rangle$. This avoids any question about how to define a ‘circular’ lexicographic order. $\endgroup$ – Brian M. Scott Apr 14 '16 at 7:38
  • $\begingroup$ In the classic space-filling curve, a continuous surjection $f: [0,1] \to [0,1]^2,$ are there necessarily any $ p$ such that $|f^{-1}\{p\}|=1?$ If not, then it is an example, because if $C\subset [0,1]$ and $\phi =C\cap ((m-1)4^{-n},m 4^{-n})$ for some $m.n\in N$ with $m\leq 4^n,$ then for some $m',m'' \in N $ with $m'\leq 2^n\geq m''$, we have $\phi=$ $f(C)\cap (\;(m'-1)2^{-n},m' 2^{-n})\times$ $ ((m''-1)2^{-n},m'' 2^{-n})\;)$. $\endgroup$ – DanielWainfleet Apr 14 '16 at 8:22

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