Image of any curve can be parametrized without zero derivative

Question

Let $\gamma :[a,b]\to\mathbb{R}^2$ be a $C^{1} ([a,b])$ injective application. Prove that there is another continuous parametrization $\rho:[c,d]\to\mathbb{R}^2$ such that the following two sets are equal:

$\gamma([a,b])=\rho([c,d])$,

and $\rho$ admits lateral derivatives at each point on $[c,d]$ with the property that $\rho'_+ (t)\neq (0,0), \ \rho'_{-}(t)\neq (0,0),\ \forall\ t\in [c,d]$.

P.S. I have use the following notations: $\rho'_{+}(t_0)=\lim\limits_{t\searrow t_0}\dfrac{\rho(t)-\rho(t_0)}{t-t_0}$ and $\rho'_{-}(t_0)=\lim\limits_{t\nearrow t_0}\dfrac{\rho(t)-\rho(t_0)}{t-t_0}$.

Answer 1

It's not true (even if you disregard constant functions). It is possible to define a $C^\infty$ function $\gamma$ from $[0,1]$ into $\mathbb R^2$ such that $\gamma(t) = [0,0]$ for infinitely many $t$ but $\gamma$ is not constant on any interval, and $\gamma([0,1])$ is the union of infinitely many loops each containing $[0,0]$. But for your $\rho$, this can't happen: it's easy to show that there can be at most finitely many $t$ with $\rho(t) = [0,0]$.

Answer 2

Even if $\gamma$ is injective, it's not true. Here's a counterexample: $$ \gamma(t) = \begin{cases} (-e^{1/t},0), & t<0,\\ (0,0), &t=0,\\ \left( e^{-1/t},\ e^{-1/t}\sin \frac{1}{t}\right), & t>0. \end{cases} $$ This is $C^1$ (in fact $C^\infty$) and injective, but there's no $C^1$ reparametrization with nonvanishing right-hand derivative at the origin. To see why, note that the slope of the curve at $\gamma(t)$ for $t>0$ is $y'(t)/x'(t)$, which reduces to $\sin\frac 1 t - \cos \frac 1 t$, which has no limit as $t\searrow 0$.