Let $\sum$ be an oriented Riemannian 2-manifold. Prove that for any piecewise smooth curve $\alpha:[a,b] \rightarrow \sum$, where $\sum$ whose derivative is never zero, there exists piecewise smooth $\alpha _1:[0,l(\alpha)] \to \sum$ such that $\alpha _1$ is parametrized by arc-length, $\alpha$ and $\alpha _1$ have the same image and they also have same total arc-length. Moreover, prove that $\alpha _1(t)=t$.
So here's my take on the above problem. If $\alpha [a,b]\rightarrow M$ be a smooth curve. The length $l(\alpha)$ of $\alpha$ is the real number $l(\alpha)=\int_{a}^b \vert\dot\alpha (t)\vert dt$ and arc-length along $\alpha$ is the function $s:[a,b]\rightarrow R$ given by $s(t)=\int_{a}^b \vert \dot \alpha (t)\vert dt$.
Given a curve $\alpha :[a,b]\rightarrow M$ is parametrized by arc-length if $\vert \dot \alpha (t)\vert=1$ for all $t \in [a,b]$. Then $s(t)=t-a$ for all $t \in [a,b]$. and for any curve $\alpha :[a,b]\rightarrow M$ with $\vert \dot \alpha (t)\vert \ne 0$ for all $t \in [a,b]$, a new curve $\alpha _1:[0,l(\alpha)]\rightarrow M$, parametrized by arc length is obtained by setting $\alpha _1=\alpha \circ s^{-1}$ then Im $\alpha _1$ = Im $\alpha$ and $l(\alpha _1)=l(\alpha)$.
But I don't know for $\sum$, a Riemannian 2-manifold how that would be satisfied?