# prove that arc length map is diffeomorphism

Given a smooth curve $$\alpha :I\to \mathbb{R}^n$$,define arc length map $$s:I\to \mathbb{R}$$ as: $$s(t) = \int_{t_0}^t |\alpha'(u)|du$$

Show that if $$|a'(u)| \ne 0$$ ,map s on $$I \to s(I)$$ is a diffeomorphism.

I can prove that $$s \in C^{\infty}$$ by calculation.and injective since if$$s(t_1) = s(t_2)$$ then $$\int_{t_1}^{t_2}|\alpha'|du = 0$$ with $$|a'(u)|>0$$ so $$t_1 = t_2$$.

What I don't know is how to consider the inverse of $$s$$

My idea is using inverse function theorem,so inverse function is $$C^1$$ with inverse derivative being explicitly expressed,so $$C^n$$ is given by calculation?

• Using the IFT sounds good, what is the problem? – Jan Bohr Jun 30 at 8:56
• @Jan Bohr can you provide the hint? – yi li Jun 30 at 9:32
• From the question it is not clear, what you are struggling with. You mention IFT, so you should check whether it is applicable and what it tells you in this context. Then you proceed by arguing why $s$ is a diffeomorphism. If you have trouble executing either of these steps, please tell us precisely where and why and we'll try to help. – Jan Bohr Jun 30 at 12:39

What I believe you're missing is an induction argument. Let $$g=s^{-1}$$. By the Inverse Function Theorem, $$g'(u) = \frac 1{s'(g(u))}.$$ Since $$s\in C^1$$, we see that $$g'$$ is continuous and so $$g\in C^1$$. Now suppose you've shown that $$s\in C^k$$ implies that $$g\in C^k$$. Use this formula to deduce that if $$s\in C^{k+1}$$, then $$g\in C^{k+1}$$.
• Of course. Since $\alpha$ is smooth and $\alpha'\ne 0$, the function $|\alpha'|$ is smooth and so is its antiderivative. – Ted Shifrin Jul 2 at 15:23