Timeline for Does there exist a continuous 2-to-1 function from the sphere to itself?
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| Dec 12, 2022 at 11:47 | comment | added | cnikbesku | @YCor The answer to your follow-up question (a') is that it is possible. For every pair of antipodal points $(a,b)$ define its twist to be the mapping $z\mapsto z^2$ after we identify the sphere with the complex projective space $\mathbb{C}P^1$ by projecting it from the point $b$ to the complex plane that is tangent to the sphere at point $a$ so that $a$ is paired with $0$ and $b$ is paired with $\infty$. Then the composition of twists for two different pairs of antipodal points is a mapping that is 2-to-1 thru 4-to-1. There is a similar but more complicated construction for 2-to-1 and 3-to-1. | |
| Dec 23, 2018 at 9:52 | comment | added | YCor | Follow-up questions: (a) same question with 2 replaced by $n\ge 2$ (a') same with the assumption that for every $x$, $f^{-1}(\{x\})$ is finite of cardinal $\ge 2$ (b) original question on higher-dimensional spheres (d) which closed 3-manifolds admit a continuous self-map in which every preimage has cardinal 2? | |
| Dec 23, 2018 at 5:25 | vote | accept | Nathaniel Butler | ||
| Dec 23, 2018 at 2:08 | answer | added | GH from MO | timeline score: 32 | |
| Dec 22, 2018 at 9:25 | comment | added | Tyler Lawson | If a 2-to-1 map $X \to Y$ is sufficiently regular (eg if there are compatible triangulations or CW decompositions of the source and target), then you get an identity of Euler characteristics $2 \chi(Y) = \chi(X)$. This would give a contradiction in your case because both the source and target have Euler characteristic 2. I haven't been able to think of a sufficiently clever argument that the Euler characteristic formula does or does not hold for a general 2-to-1 map of $S^2$. | |
| Dec 22, 2018 at 7:55 | review | First posts | |||
| Dec 22, 2018 at 10:05 | |||||
| Dec 22, 2018 at 7:50 | history | asked | Nathaniel Butler | CC BY-SA 4.0 |