Given $\mathbf{r}:(t\_0-\varepsilon, t\_0+\varepsilon)\to\mathbb{R}^2$ be a $C^{1}$ application, with $\mathbf{r}'(t)=(0,0)\Longleftrightarrow\ t=t\_0$, and


$$\mathbf{r}(t)=\begin{cases} \mathbf{r}\_1 (t), \ t\in (t\_0-\varepsilon, t\_0] \\ \mathbf{r}\_2 (t), \ t\in [t\_0, t\_0+\varepsilon)\end{cases},\ \mathbf{r}\_1\in C^{\infty}((t\_0-\varepsilon, t\_0]),\ \mathbf{r}\_2\in C^{\infty}([t\_0, t\_0+\varepsilon))$$


Is it true that we can always find $\mathbf{s}:(t\_0-\varepsilon, t\_0+\varepsilon)\to\mathbb{R}^2$, $\mathbf{s}\in C^{\infty}((t\_0-\varepsilon, t\_0+\varepsilon))$, such that $\mathbf{s}((t\_0-\varepsilon, t\_0+\varepsilon))=\mathbf{r}((t\_0-\varepsilon, t\_0+\varepsilon))$?



Glueing smooth functions give a smooth function if reparametrized
Yes, even if $\mathbf r$ is only $C^0$. Just reparameterize it in such a way that it stops infinitely flat at 0 (use $e^{-1/t^2}$).


