28
$\begingroup$

Hello. I have a question.

Does there exist a continuous mapping

$F:\mathbb{R}^2\rightarrow\mathbb{R}^2$

such that for every $c\in F(\mathbb{R}^2)$

there are two and only two points $z_{1}$, $z_{2}$

such that $F(z_{1})=F(z_{2})=c$ ?

Thank you very much for attention.

$\endgroup$
8
  • 1
    $\begingroup$ Please don't use LaTeX in the title. It makes the main page load more slowly. $\endgroup$ Mar 10, 2010 at 10:26
  • 1
    $\begingroup$ Most likely $f=F$. $\endgroup$ Mar 10, 2010 at 10:31
  • 2
    $\begingroup$ Presumably $F = f$. My instinct is that the answer is "no". If there were such a map, you could define a $\mathbb{Z}_2$-action on the plane by swapping the points. Subject to some minor technicalities, the quotient would then be $B\mathbb{Z}_2$ so you're asking for an injection from $B\mathbb{Z}_2 \to \mathbb{R}^2$ which seems highly unlikely. But there may be some super-snazzy-technicalities that I've overlooked. $\endgroup$ Mar 10, 2010 at 10:35
  • 1
    $\begingroup$ We attempted to discuss this problem on the Usenet newsgroup sci.math in October - we turned up a couple of references, but I don't know that we made too much progress. You can find it under Subject: an exotic continuous complex function $\endgroup$ Mar 11, 2010 at 3:40
  • 2
    $\begingroup$ If it exists, I want to see it! :) $\endgroup$
    – Qfwfq
    Feb 25, 2018 at 2:40

5 Answers 5

14
$\begingroup$

Look at the paper "Two-to-one mappings of manifolds" by Paul Civin Duke Math. J. Volume 10, Number 1 (1943), 49-57. He proved that there is no such a closed continuous mapping on ${\mathbb R}^2$ (i.e. transforming closed sets into closed sets).

Update: accordingly to the paper http://www.dml.cz/bitstream/handle/10338.dmlcz/700959/Toposym_01-1961-1_63.pdf there exists 2-to-1 map on ${\mathbb R}^2$ but I do not understand what is the image.

$\endgroup$
0
9
$\begingroup$

Amazingly, it seems that the answer is yes:

Mioduszewski, J. On two-to-one continuous functions. (Russian summary) Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 9 1961 129--132.

The author announces results concerning two-to-one functions $f$ on a locally compact separable space $X$, proofs of which appear in Rozprawy Mat. 24 (1962), 1--41. Let $\phi$ be the (discontinuous) involution defined by $\varphi(x)=f^{-1}f(x)-x$. A result of the reviewer [Duke Math. J. 10 (1943), 49--57; MR0008697 (5,47e)] asserts that if $X$ is a compact manifold or $f$ is closed and $X$ is a locally compact manifold, then the investigation of $\phi$ is equivalent to the investigation of a continuous involution. The author calls a point $x\in X$ pseudo-Euclidean if it has a neighborhood $H$ such that the closure of the component of $x$ in $H$ is a Euclidean solid sphere. The principal theorem asserts that if $x$ is a pseudo-Euclidean point with $K$ as the solid sphere of the definition, and if $\psi=\varphi|K$, that $\lim\text{}\sup_{y\rightarrow x}\psi(y)=x\bigcup\varphi(x)$ is impossible. This yields an extension of the result of the reviewer quoted above. The author indicates the existence of a plane simply connected domain $G$ whose boundary is an irreducible cut of the plane into two domains and such that there exists a two-to-one mapping defined on $\overline G$. This is in contrast to the result of Roberts [ibid. 6 (1940), 256--262; MR0001923 (1,319d)], which asserts the non-existence of two-to-one mappings defined on two-cells. The existence of two-to-one mappings defined on Euclidean spaces $E^n$, $n\geq 2$, is shown. However, the question of the existence of two-to-one mappings defined on $n$-cells, $n>3$, remains open. [MathSciNet review by P. Civin.]

I can't access this paper, so I can't say anything about the construction. It would be nice to see some corroboration for this result and/or a more (physically) accessible contemporary treatment.

Addendum: Petya's response gives a link to the paper, from which one can see that the function is essentially defined in terms of the involution $\iota$, so it is not immediately clear what the codomain is or whether it can be embedded in $\mathbb{R}^2$.

$\endgroup$
2
  • 2
    $\begingroup$ Pete, the theorem claims that there is a topological space $Y$ and 2-to-1 mapping $E^n$ to $Y$. I am not sure that one can take $Y\subset E^n$. $\endgroup$
    – Petya
    Mar 10, 2010 at 13:43
  • $\begingroup$ @Petya: yes, you're right. The review I quoted above doesn't give a careful definition of what a 2:1 function is, so I think that what I said above was actually correct: it does seem, from reading the review, that the answer is yes. But I agree, the question is not fully answered yet. $\endgroup$ Mar 10, 2010 at 13:52
3
$\begingroup$

As I see, from the article by Mioduszewski it follows that if we take his two-to-one continuous function $F:\mathbb{R}^{2}\to Y$ then $Y$ can be embedded in $\mathbb{R}^{4}$.

So, there exists a continuous two-to-one function $F:\mathbb{R}^{2}\to\mathbb{R}^{4}$ (but it is not surjective).

That is, for every $c\in F(\mathbb{R}^{2})$ there are two and two points $z_{1}$, $z_{2}$ such that $F(z_{1})=F(z_{2})=c$.

$\endgroup$
1
$\begingroup$

I am sorry for the intervention. I read the above mentioned article by Mioduszewski (it is available at the link http://pldml.icm.edu.pl/pldml/element/bwmeta1.element.zamlynska-2e6943ca-7faf-46aa-ab2d-bbc5d5f1bcff ) but it does not contain a clear proof of the above statement: 'there exists a continuous two-to-one function $F:\mathbb{R}^{2}\to\mathbb{R}^{4}$ (but it is not surjective), that is, for every $c\in F(\mathbb{R}^{2})$ there are two and only two points $z_{1}$, $z_{2}$ such that $F(z_{1})=F(z_{2})=c$.' It is not obviously that this 'statement' is really true. Why $\mathbb{R}^{4}$ for $\mathbb{R}^{2}$ ?

$\endgroup$
1
$\begingroup$

My $5$ cents. The simplest connected space with a continuous $2$-to-$1$ map that I can imagine is an infinite binary tree (with no root nor leaves). Say all branches have unit length. Then, going down along branches by a unit length, produces a continuous $2$-to-$1$ map $f:T\to T$ (also mapping nodes to nodes). We can embed $T$ in $\mathbb{R}^2$. Can we extend $f$ to a continuous $2$-to-$1$ map on $\mathbb{R}^2$, maybe a covering projection on $\mathbb{R}^2\setminus T$?

$\endgroup$
1
  • $\begingroup$ Or more likely, one could try to extend a 2-to-1 map $f:S\to T$, with $S\subset\mathbb{R}^2$ $\endgroup$ Mar 3, 2018 at 11:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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