Given the curve $C$ simple, smooth and $\gamma:[a,b]\rightarrow {\mathbb{R}^{3}}$ regular parametrization of the given curve. For each $t\in[a,b]$ let $g(t)$ the arclength curve between two points $\gamma(a)$ and $\gamma(t)$ we have that:
$g(t)= \int_{a}^{t} \left\lVert\gamma'(\tau) \right\rVert\, d\tau$
Where the function $g(t)$ is $C^{\infty}$ with $g'(t)\neq 0$ $\forall t\in[a,b]$ and $g'(t) = \left\lVert\gamma'(t) \right\rVert$ is the instantaneous variation of the arclength in the point (i.e. position) $\gamma(t)$ $\Longrightarrow$ $ \left\lVert\gamma'(t) \right\rVert\ $ is the speed.
Let $\varphi(s):\gamma (g^{-1}(s))$, where $g^{-1}(s)$ is $C^{\infty}$ . Prove that $\varphi(s):[0,l]\rightarrow {\mathbb{R}^{3}}$ where $l$ is the length of the curve $C$ is a regular parametrization of the curve $C$ such that the arclength that goes from $\varphi(0)$ to $\varphi(s)$ is $s$.
Honestly, I'm not sure where to start and how. I will be very thankful if someone helps me with this.
Note that $\varphi'(s)=(\gamma\circ g^{-1})'(s)=\gamma'\bigl(g^{-1}(s)\bigr)(g^{-1})'(s)\neq0$. Therefore, $\varphi$ is a regular curve.
On the other hand\begin{align}\int_0^l\bigl\|\varphi'(s)\bigr\|\,\mathrm ds&=\int_0^l\bigl\|\gamma'\bigl(g^{-1}(s)\bigr)(g^{-1})'(s)\bigr\|\,\mathrm ds\\&=\int_0^l\left\|\frac{\gamma'\bigl(g^{-1}(s)\bigr)}{g'\bigl(g^{-1}(s)\bigr)}\right\|\,\mathrm ds\\&=\int_0^l\left\|\frac{\gamma'\bigl(g^{-1}(s)\bigr)}{\bigl\|\gamma'\bigl(g^{-1}(s)\bigr)\bigr\|}\right\|\,\mathrm ds\\&=\int_0^l1\,\mathrm ds\\&=l.\end{align}