Take the 2-minute tour ×

Hello. I have a question.

Does there exist a continuous mapping


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.

share|improve this question
Please don't use LaTeX in the title. It makes the main page load more slowly. –  Harry Gindi Mar 10 '10 at 10:26
Also, what is $F$? –  Harry Gindi Mar 10 '10 at 10:31
Most likely $f=F$. –  Konrad Swanepoel Mar 10 '10 at 10:31
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. –  Loop Space Mar 10 '10 at 10:35
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 –  Gerry Myerson Mar 11 '10 at 3:40

3 Answers 3

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.

share|improve this answer

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 $\mathbb{R}^2$.

share|improve this answer
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 $Y\subset E^n$. –  Petya Mar 10 '10 at 13:43
@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. –  Pete L. Clark Mar 10 '10 at 13:52

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 $F(z_{1})=F(z_{2})=c$.

share|improve this answer

Your Answer


