Does anyone know how to prove the existence of a smooth curve $c$ from the unit interval of the real numbers to a connected smooth manifold $M$ such that $c(0)=p$ and $c(1)=q$ with $p$ and $q$ points on $M$ ? This is a question from Spivak´s Differential Geometry Vol 1. Thanks for the help.
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Possible duplicate of https://math.stackexchange.com/questions/1415147/perturb-a-piecewise-linear-path-to-make-it-c-infty?noredirect=1&lq=1 and https://math.stackexchange.com/questions/243295/smooth-curves-on-a-path-connected-smooth-manifold/2713251#2713251 – Kelvin Lois Mar 29 '18 at 14:11
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1Possible duplicate of Smooth curves on a path connected smooth manifold – Dap Mar 29 '18 at 18:24
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I don't have enough reputation to comment, so I just put it as an answer:
Your question is answered e.g. here: Link
user12390
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Yeah but my question is about finding a smooth function with this property, the answer you showed me just find a continuous, even piece-wise smooth but not smooth that is what I want. Thanks anyway. – kvicente Jul 07 '17 at 19:26
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1@kvicente Once you have the continuous path, cover it with finitely many charts (since it is compact) and smooth out on each chart since each chart is just an $\mathbb{R}^n$.. – Bettybel Jul 07 '17 at 19:28
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How do I precisely smooth out, that´s the main issue, is there any procedure to do this? – kvicente Jul 07 '17 at 19:29
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1@kvicente I can assume the charts are just balls from $\mathbb{R}^n$. Deep inside the balls (like radius minus epsilon) you can replace the path with straight lines if you like by joining the first entry point to the last entry point. In the intersection of charts use flat functions (the $e^{-1/x^2}$ business) to join together the line segments from each chart. – Bettybel Jul 07 '17 at 19:32
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1I see, thanks for pointing that out. I still think that one should be able to find the answer to this on the web with not much effort. For example, take a look at the accepted answer here: Link – user12390 Jul 07 '17 at 19:33
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Yeah thanks both of you for the ideas, I had the same idea but I had troubles trying to make it rigorously. Thanks for everything. – kvicente Jul 07 '17 at 19:38
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Is it clear to you now? I think the answer of Nils Matthes in the link I put in the comments explains it in great detail. – user12390 Jul 07 '17 at 19:41
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The heuristic behind Nils Matthes answer is often described as follows: if you have a line with some kind of kink, it can still be the image of a smooth curve. The idea is to parametrize the line in such a way that you slow down to zero (i.e., the derivative of the curve becomes zero) at the kink when you run through the curve. – user12390 Jul 07 '17 at 20:05