# Proof that any unit-speed-reparametrization of a curve preserves orientation and is an inverse of an arc length function based at some $t_0$.

I am not able to prove the following two facts about a unit speed reparametrization of a curve.

1. Let $\alpha$ be defined on some interval $I$ and define for $t_0\in I$ $$\lambda_{t_{0}}(t)=\int_{t_0}^t\|\alpha'(x)\|\,dx.$$ Any unit-speed-reparametrization (usp) $\phi_{t_{0}}$ is then given by $$\phi_{t_{0}}=\lambda_{t_{0}}^{-1}.$$

2. Any unit speed reparametrization preserves orientation.

I think the facts assume that any usp $h$ is a diffeomorphism, but I'm not sure where this comes from. I would appreciate any explanations on the above facts.

My question arose from the same problem as the one from the possible duplicate question, however, the answer in that did not explain the above two, which I don't have a justification for.

• possible duplicate of Unit speed reparametrization of curve – Michael Hoppe Apr 15 '15 at 20:58
• @MichaelHoppe I looked at the answer and that's where I got the above questions. You said they were basic facts, but I can't prove them. – takecare Apr 16 '15 at 3:19
• Well, obviously (1) defines all usp parametrizations of the curve; there's nothing to prove. And hence, by construction, all usp preserve orientation. – Michael Hoppe Apr 16 '15 at 8:23
• @MichaelHoppe Sorry but I don't see how it is obvious. It's saying that any usp $h$ must be an inverse of some arc length parametrization. How is this clear? – takecare Apr 16 '15 at 15:59
• What else should a usp be?! – Michael Hoppe Apr 16 '15 at 18:41