3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
6
$\begingroup$

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 $S^1\times \{0,1\}$:

enter image description here

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.