Timeline for Do proper polynomial mappings have a path-lifting property?
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 23, 2010 at 19:40 | vote | accept | Kevin M Pilgrim | ||
| Nov 23, 2010 at 8:06 | answer | added | Sergei Ivanov | timeline score: 9 | |
| Nov 23, 2010 at 7:47 | comment | added | André Henriques | If $n=1$, the answer is "yes", which makes me think that the answer is always "yes". Note: already for $n=1$, the question is quite non-trivial. But in that case, it is enough to look at the map $z\mapsto z^n$ (that's the local model). Given a continuous path $\gamma:[0,1]\to \mathbb C^n$, one considers the closed subset $\gamma^{-1}(0)$, and its complement, which is a countable disjoint union of intervals. The lifting problem can then be solved independently on each component of that complement. | |
| Nov 23, 2010 at 4:35 | history | edited | David Roberts♦ | CC BY-SA 2.5 |
fixed TeX
|
| Nov 23, 2010 at 3:35 | history | asked | Kevin M Pilgrim | CC BY-SA 2.5 |