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I'm given two curves:

$\gamma_1: t \to (\cos t, \sin t)$ and $\gamma_2: t \to (\cos 2t, \sin 2t)$, both defined in $I=[0,2\pi[$.

Clearly these two have the same image, the unit circle, but $\gamma_2$ has twice the velocity.

Question 1 : How do I show that they actually do have the same image?

Question 2 : How do I prove that there is no smooth diffeomorphism $\phi: I \to I$ such that $\gamma_2=\gamma_1 \circ \phi$

Thank you in advance for your help: ideas, etc

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  1. As you wrote, the image of both is the unit circle. Or you could note that $\gamma_2(t) = \gamma_1(2t)$ for $t \in [0,\pi]$ and $\gamma_1(2t-2\pi)$ for $t \in [\pi,2\pi]$.

  2. $\gamma_1$ is one-to-one, and so would be $\gamma_1 \circ \phi$. $\gamma_2$ is not.

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  • $\begingroup$ Thank you. Can I state that both curves are equivalent in every subinterval of $I$ of amplitude $\pi$ ? $\endgroup$
    – hugh_maths
    Apr 16, 2021 at 16:38

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