Arc length parametrisation is reparametrisation I was trying to show that if $\gamma : [a,b] \to \mathbb R^3$ is a curve and $$ p(t) = \int_{t_0}^t |\gamma'(\tau)|d\tau$$ where $t_0 \in [a,b]$ then $p^{-1}: [c,d] \to [a,b]$ is a reparametrisation of $\gamma$. This is exercise 2 here. 
But I am not even sure what I need to show: 


*

*$p^{-1}$ is a bijection

*and $(p^{-1})'(s) \neq 0$ for all $s$

*and $\gamma ( p^{-1}(s)) = \gamma (s)$ for all $s$?
I tried to show 3. but the problem is that it does not even have same domain on both side. How to show that $p^{-1}$ is a reparametrisation of $\gamma$? Thank you for help.

You must show that $p^{-1}$ is continuous and injective. To achieve this it is enough to show that $p$ is strictly increasing and continuous.
some details
You have a $C^1$ function $\gamma\colon [a,b]\to \mathbb R^3$ with $|\gamma'(t)|\neq 0$ for all $t$ and define
$$
  p(t) = \int_a^t |\gamma'(\tau)|\, d\tau.
$$
Let $L=p(b)$. L is the length of the curve $\gamma$. The function
$$
  p\colon [a,b] \to [0,L]
$$
is differentiable and increasing because by the fundamental theorem of calculus:
$$
  p'(t) = |\gamma'(t)| > 0.
$$
Hence $p$ is a bijection. It is injective because it is strictly increasing. It is surjective in view of the intermediate values theorem since $p(0)=0$, $p(b)=L$. So $p$ is invertible. Since $p'(t)\neq 0$ the inverse is also differentiable and 
$$
  (p^{-1})'(s) = \frac{1}{p'(p^{-1}(s))}> 0.
$$
