Regular and simple curves with the same support are equivalent I would like to know if anyone knows a simple proof of the following statement:

Let $\gamma: I \to \Bbb R^n$ and $\phi: J \to \Bbb R^n$ be curves such that:

*

*$\gamma$ and $\phi$ are simple (i.e.: they are injective in the interior of $I$ and $J$ respectively);

*$\gamma$ and $\phi$ are regular (i.e.: class $C^1$ with non-null derivative)

*$\gamma$ and $\phi$ have the same support (i.e.: the same image)
then there exists a diffeomorphism relating the two curves (i.e.: they are equivalent).


I have built the candidate for the diffeomorphism by using the injectivity but I couldn't prove that it is continuous, let alone differentiable. I think it may have something to do with the theorem on the limit of the composition of functions but I have not managed to go further than that.

As always, any comment or answer is welcome and let me know if I can explain myself clearer!
 A: Hint: First consider the two curves
$$\alpha:[0,1]\to\mathbb{R}^2, t\mapsto (\cos(2\pi t),\sin(2\pi t)),$$
$$\beta:[0,1]\to\mathbb{R}^2, t\mapsto (\cos(2\pi (t-1/2)),\sin(2\pi (t-1/2))).$$
Then $\alpha$ and $\beta$ are both simple and regular curves with the unit circle on the plane the common image. If $[0,1]$ is given the standard orientation and the unit circle is given the counterclockwise orientation, $\alpha$ and $\beta$ are also orientation preserving. Any diffeomorphism $\Phi: [0,1]\to [0,1]$ such that $\beta\circ \Phi=\alpha$ would have to preserve the boundary, but $\alpha(0)=\alpha(1)=(1,0)$ and $\beta(0)=\beta(1)=(-1,0)$; thus $\alpha$ and $\beta$ are not reparameterizations of each other.
For a positive proof let us assume that both $\gamma$ and $\phi$ are injective ($\dagger$) and that both $I$ and $J$ are compact intervals. Call $C$ the common image with positive length $\ell(C)$. Replacing $\phi$ by $\phi\circ \iota$ if necessary, where $\iota:J\to J$ is the affine diffeomorphism changing the orientation of $J$, we may assume that $\gamma(\min(I))=\phi(\min(J))$. By the hypotheses on $\gamma$ and $\phi$ they are of positive and finite length. Define the arclength function $\Lambda_{\gamma}: I\to [0,\ell(C)]$, by
$$\Lambda_\gamma: t\mapsto  \int_{\min(I)}^{t} \left|\dfrac{d\gamma}{dt}(\tau)\right|\, d\tau.$$
Note that $\ell(C)=\Lambda_\gamma(\max(I))=\Lambda_\phi(\max(J))$ by the definition of the length and the change of variables formula for integrals. Both $\gamma$ and $\phi$ are regular, thus the integrants involved are both continuous, and consequently both $\Lambda_\gamma$ and $\Lambda_\phi$ are $C^1$; by the regularity of $\gamma$ and $\phi$ the functions $\Lambda_\gamma$ and $\Lambda_\phi$ are also strictly increasing, hence are diffeomorphisms (in general $\Lambda_\gamma$ is a $C^k$ diffeomorphism if $\gamma$ is $C^k$). Then we have two $C^1$ parameterizations $\gamma\circ \Lambda_\gamma^{-1}, \phi\circ\Lambda_\phi^{-1}: [0,\ell(C)]\to C$ of unit speed with $\gamma\circ \Lambda_\gamma^{-1}(0)=\phi\circ\Lambda_\phi^{-1}(0)$. Thus they have to coincide, so that $\gamma$ and $\phi$ are reparameterizations of each other.
($\dagger$) A similar argument works if $\gamma$ and $\phi$ are simple but not injective, provided that the endpoints are the same.
