▲ | Koshkin 8 months ago | ||||||||||||||||||||||||||||||||||||||||
> Every knot is “homeomorphic” to the circle Here's an explanation: https://math.stackexchange.com/questions/3791238/introductio... | |||||||||||||||||||||||||||||||||||||||||
▲ | noqc 8 months ago | parent | next [-] | ||||||||||||||||||||||||||||||||||||||||
This isn't true. Knots are not topological spaces, they are imbeddings, K:S^1 -> S^3 (more generally, S^n-> S^n+2). Therefore there isn't an obvious notion of homeomorphism. What the poster points out is that restriction to the image is a homeomorphism from the circle (because it is an imbedding of the circle) As maps between topological spaces (and almost always we pick these to be from the smooth or piecewise linear categories, which further restricts them), the closest "natural" interpretation of isomorphism is pairs of homeomorphisms, f:S^1 -> S^1, g:S^3 -> S^3 satisfying Kf=gK. Ie natural transformations or commuting squares. This gives us almost what we want, except that I can flip the orientation of space, or the orientation of the knot with homeomorphisms, neither of which correspond to the physical phenomenon of knots, so we give our spaces an orientation, which requires us to move to the PL or smooth category, or use homotopy/isotopy. | |||||||||||||||||||||||||||||||||||||||||
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▲ | bmitc 8 months ago | parent | prev | next [-] | ||||||||||||||||||||||||||||||||||||||||
Intuitively, just imagine picking a starting point on each of the circle and the knot. Now walk at different speeds such that you get back to the starting point at the same time. In fact, that's what the knot is: a continuous, bijective mapping from the circle to the image of the mapping, i.e., the knot. (As the linked answer says.) Edit: I see now that the article already has this intuitive explanation but with ants. | |||||||||||||||||||||||||||||||||||||||||
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▲ | Nihilartikel 8 months ago | parent | prev [-] | ||||||||||||||||||||||||||||||||||||||||
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