Let $f\colon[0,1]\to \mathbb{R^2}$ be continuous such that $f(0)=f(1)$.

If want to find a 1-Lipschitz function $g : [0,a]\to f([0,1])$ such that $g(0)=g(a)$ and $g$ is surjective ($a>0$).

I had the following idea using the total variation of $f$: Denote $V_T(f) = \sup \left\lbrace\sum_{i=1}^n \lVert f(x_i) - f(x_{i-1})\rVert_2 \ \biggm| \ n\in\mathbb{N}, \ 0=x_0<x_1<\dots<x_n=1\right\rbrace$. Suppose morevover that $V_T(f)<+\infty$.

Define $g:[0,V_T(f)]\to f([0,1])$ such that $g(x) = f\left(\frac{x}{V_T(f)}\right)$.

Then $g(0)=g(V_T(f))$ and $g$ is surjective. But I cannot prove that $\Vert g(x_1)-g(x_2)\Vert_2\leq |x_1-x_2|$. Maybe it is not true, in that case is there another definition of $g$ that would be suitable?