I assume that this must be true because the parametrization describes the same object, but I cannot recall a theorem that would state this explicitly.
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What definition of smooth are we using this time? – jimjim Sep 11 '19 at 00:01
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@Arjang infinitely differentiable. – Daphne Sep 11 '19 at 00:03
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in that case if use a parametrisation from no where differentiable curve like Weistraus's example then I guess the answer is no. – jimjim Sep 11 '19 at 00:07
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@Arjang did you mean parametrization of e.g. Weierstrass function? I assume the curve being parametrized to be smooth. – Daphne Sep 11 '19 at 00:11
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1Define a parametrisation from Weistaus function to your smooth curve. – jimjim Sep 11 '19 at 00:13
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No. Take, for instance, the curve in $\mathbb R^2$ defined by $\gamma(t)=(t,t)$ ($t\in[-1,1]$). And now take the reparametrization $\eta(t)=\left(\sqrt[3]t,\sqrt[3]t\right)$ (again, with $t\in[-1,1]$).
José Carlos Santos
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Would a parametrisation from Weistaus function to the smooth curve also be used as counter example? – jimjim Sep 11 '19 at 01:15
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1Yes. And a more radical counterexample than mine, since it would be differentiable nowhere. – José Carlos Santos Sep 11 '19 at 01:28