4 ms·
No it's just permutations. Just watch the video and you'll see.
by almostgotcaught 2y ago
No it's just permutations. Just watch the video and you'll see.
- lupire 2y agoNot true. The proof depends on continuous loops and non-invertible functions. It's significantly related to Riemann surfaces. https://math.stackexchange.com/questions/3349111/where-exactly-riemann-surfaces-and-monodromy-groups-appear-in-this-video https://math.stackexchange.com/questions/3349111/where-exact... https://web.williams.edu/Mathematics/lg5/394/ArnoldQuintic.pdf https://web.williams.edu/Mathematics/lg5/394/ArnoldQuintic.p... https://projecteuclid.org/journals/topological-methods-in-nonlinear-analysis/volume-16/issue-2/The-topological-proof-of-Abel-Ruffini-theorem/tmna/1471875703.pdf https://projecteuclid.org/journals/topological-methods-in-no...
- almostgotcaught 2y agoYou're responding to wrong comment. In the comment you meant to respond to I admitted I wasn't sure whether continuity was required to make the proof go through and you still haven't demonstrated that it is. Riemann surface here is just a fancy word for the loop itself and indeed the key part of the definition is connectedness not C. You linked a math SE question with no responses (yes I read the comments), Arnold's book, and the same proof I've already read. So I'm not convinced.