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I have read " Let $\gamma : [a,b]→\mathbb{C}$ be a smooth path. Then the directed smooth curve with parameterization $\gamma$ is defined as an equivalence class of smooth path. A smooth path $\sigma : [c,d]→\mathbb{C}$ belongs to the equivalence class of $\gamma$ iff there exists a bijective differentiable strictly increasing function $\phi :[c,d]→[a,b]$ such that $\sigma=\gamma \circ \phi$." Here, smooth path $\gamma$ means that $\gamma \in C^1[a,b]$ and $\gamma'(t)\neq 0,~\forall t \in [a,b].$

Are there any smooth curves with same direction which represents the image of $\gamma$, but is not equivalent to $\gamma$?

Intuitively, I thought that a smooth path $\sigma: [c,d] \to \mathbb{C}$ with same direction which represents the image of $\gamma$ is equivalent to $\gamma,$ i.e., there exists a bijective differentiable strictly increasing function $\phi :[c,d]→[a,b]$ such that $\sigma=\gamma \circ \phi$." Is it wrong?

I would be thankful for any comments about my question.

04170706
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    Is there any difference between the descriptions of equivalence in both paragraphs? – Amitai Yuval Jun 05 '17 at 03:02
  • I am sorry for ambiguous question. My question is "are there any smooth curves with same direction which represents the image of $\gamma$ but is not equivalent to $\gamma$?". I have edited my question. – 04170706 Jun 05 '17 at 03:09
  • What do you mean by "with same direction"? – Eric Wofsey Jun 05 '17 at 03:18
  • For the image of a given smooth curve in complex plane, there are two directions. For example, a circle has two directions such as counter-clockwise or clockwise. – 04170706 Jun 05 '17 at 03:26
  • If the curve does not intersect itself, then the answer is yes: two regular parametrizations of the same curve admit a change of variables function. If the curve does intersect itself (in particular a tangential self intersection) there can be two (topologically) distinct parametrizations. – Willie Wong Jun 05 '17 at 03:29
  • In the non-self-intersecting case, the proof is straight forward but annoying. I have some old lecture notes in which I wrote out the proof in detail, and it is about 2 pages long . (The bijection is easy to construct since $\sigma$ and $\gamma$ are assumed to be bijective to their images. Monotonicity follows immediately once you prove differentiability. Differentiability is an involved game of epsilons and deltas.) – Willie Wong Jun 05 '17 at 03:36
  • Thank you for the answer. You mean a change of variable function $\phi$ is differentiable. I will try to prove it. I have one more question. If $\phi$ is a change of variables function, is $\phi'$ integrable? – 04170706 Jun 05 '17 at 03:43
  • Looking at my proof, if you assume that $\gamma$ and $\sigma$ are both $C^1$, then the condition that $\gamma' \neq 0 \neq \sigma'$ tells you that $\phi$ has Lipschitz constant $< C \sup |\gamma'| / \inf |\sigma'|$ (or maybe the other way around). And we know that Lipschitz implies integrable derivative. – Willie Wong Jun 06 '17 at 02:58

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