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I'm learning differential geometry and I found this definition about the equivalence of two parametrized curves. enter image description here

I want to understand the intuition behind those conditions. What I have been understanding so far is the condition $\gamma_2(\varphi(t))=\gamma_1(t), \forall t \in I_1$ means the two parametrized curves have the same trace. What about the first condition $\varphi'(t) \neq 0, \forall t \in I_1$? What is the intuition behind it?

UPDATE: Consider the two parametrized curves $(C_1): r_3(t)=(t^3, 0), t \in \mathbb{R}$ and $(C_1): r_4(t)=(t, 0), t \in \mathbb{R}$. They are not equivalent even they have the same direction and the same trace. The only thing here is the kind of velocity they travel is different. Like the third one has the constant acceleration but the fourth is not. How the kind of velocity affect the equivalence?

Thank you for reading my question. Any help would be appreciated.

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If $\gamma_2\bigl(\varphi(t)\bigr)=\gamma_1(t)$, then $\gamma_1\bigl(\varphi^{-1}(t)\bigr)=\gamma_2(t)$. However, without the assumption that you always have $\varphi'(t)\ne0$, $\varphi^{-1}$ would not be differentiable, and therefore two parametrizations of the same curve being equivalent would not be an equivalence relation. But it is natural to expect that it is.

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  • $\begingroup$ In the aspect of well-defining, I think the first condition is to make the inverse map differentiable. I have updated my questions, would you mind have a look again? Thank you, sir. $\endgroup$
    – RopuToran
    Jun 27, 2021 at 14:21
  • $\begingroup$ Concerning your assertion that “the first condition is to make the inverse map differentiable”, that is exactly what I wrote in my answer. And this has nothing to do with direction. The curves $\gamma_1,\gamma_2\colon[0,2\pi]\longrightarrow\Bbb R^2$ defined by $\gamma_1(t)=(\cos(t),\sin(t))$ and by $\gamma_2(t)=(\cos(t),\sin(-t))$ are equivalent; just take $\varphi(t)=2\pi-t$. $\endgroup$
    – José Carlos Santos
    Jun 27, 2021 at 14:26
  • $\begingroup$ Thanks for pointing out my errors. I deleted that example. How about the other one? $\endgroup$
    – RopuToran
    Jun 27, 2021 at 15:34
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    $\begingroup$ What do you expect to say other than “Yes, $r_1$ and $r_2$ are indeed not equivalent”? The only thing that I can add is that the speed of $r_3$ is $0$ at some point, whereas with $r_4$ that does not occur. $\endgroup$
    – José Carlos Santos
    Jun 27, 2021 at 16:18
  • $\begingroup$ Two equivalent curves will have the same trace but the converse is not correct. So I just want to know what makes the converse statement false and I'm just thinking of the speed and direction. Therefore I carelessly claimed the equivalence in the first example. $\endgroup$
    – RopuToran
    Jun 28, 2021 at 2:42

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