Two-to-one continuous mapping from R² to R² - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T07:37:27Z http://mathoverflow.net/feeds/question/17707 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/17707/two-to-one-continuous-mapping-from-r-to-r Two-to-one continuous mapping from R² to R² unknown (google) 2010-03-10T09:59:31Z 2012-09-16T13:02:38Z <p>Hello. I have a question.</p> <p>Does there exist a continuous mapping</p> <p>$F:\mathbb{R}^2\rightarrow\mathbb{R}^2$</p> <p>such that for every $c\in F(\mathbb{R}^2)$</p> <p>there are two and only two points $z_{1}$, $z_{2}$</p> <p>such that $F(z_{1})=F(z_{2})=c$ ?</p> <p>Thank you very much for attention.</p> http://mathoverflow.net/questions/17707/two-to-one-continuous-mapping-from-r-to-r/17710#17710 Answer by unknown (google) for Two-to-one continuous mapping from R² to R² unknown (google) 2010-03-10T11:33:24Z 2010-03-10T11:33:24Z <p>I'm sorry, f=F.</p> <p>Does there exist a continuous mapping</p> <p>$F:\mathbb{R}^2\rightarrow\mathbb{R}^2$</p> <p>such that for every $c\in F(\mathbb{R}^2)$</p> <p>there are two and only two points $z_{1}$, $z_{2}$</p> <p>such that $F(z_{1})=F(z_{2})=c$ ?</p> http://mathoverflow.net/questions/17707/two-to-one-continuous-mapping-from-r-to-r/17718#17718 Answer by Petya for Two-to-one continuous mapping from R² to R² Petya 2010-03-10T13:21:50Z 2010-03-10T13:34:47Z <p>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).</p> <p>Update: accordingly to the paper <a href="http://www.dml.cz/bitstream/handle/10338.dmlcz/700959/Toposym_01-1961-1_63.pdf" rel="nofollow">http://www.dml.cz/bitstream/handle/10338.dmlcz/700959/Toposym_01-1961-1_63.pdf</a> there exists 2-to-1 map on ${\mathbb R}^2$ but I do not understand what is the image. </p> http://mathoverflow.net/questions/17707/two-to-one-continuous-mapping-from-r-to-r/17723#17723 Answer by Pete L. Clark for Two-to-one continuous mapping from R² to R² Pete L. Clark 2010-03-10T13:37:31Z 2010-03-10T13:54:53Z <p>Amazingly, it seems that the answer is <b>yes</b>:</p> <blockquote> <p>Mioduszewski, J. On two-to-one continuous functions. (Russian summary) Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 9 1961 129--132.</p> </blockquote> <p>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.]</p> <p>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.</p> <p><b>Addendum</b>: 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 $\mathbb{R}^2$. </p> http://mathoverflow.net/questions/17707/two-to-one-continuous-mapping-from-r-to-r/102897#102897 Answer by unknown (google) for Two-to-one continuous mapping from R² to R² unknown (google) 2012-07-22T21:00:56Z 2012-07-22T21:00:56Z <p>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 $\mathbb{R}^{4}$.</p> <p>So, there exists a continuous two-to-one function $F:\mathbb{R}^{2}\to\mathbb{R}^{4}$ (but it is not surjective).</p> <p>That is, for every $c\in F(\mathbb{R}^{2})$ there are two and two points $z_{1}$, $z_{2}$ such that $F(z_{1})=F(z_{2})=c$.</p> http://mathoverflow.net/questions/17707/two-to-one-continuous-mapping-from-r-to-r/107316#107316 Answer by Shuba Kumar for Two-to-one continuous mapping from R² to R² Shuba Kumar 2012-09-16T13:02:38Z 2012-09-16T13:02:38Z <p>Please check this reply (answered Jul 22 at 21:00). Is the information in it true?</p>