Smooth immersion $\gamma$ from $\mathbb R \to \mathbb R^2$ such that $\operatorname{Im}\gamma=\{(x,\lvert x\rvert)\mid x\in \mathbb R\}$

Question

Can I find a smooth immersion $\gamma$ from $\mathbb R \to \mathbb R^2$ such that $\operatorname{Im}\gamma=\{(x,\lvert x\rvert)\mid x\in \mathbb R\}$, I cannot take $\gamma(t)=(t,\lvert t\rvert)$ as this is not smooth, any hint? Can i take $\gamma(t)=(t^2,t^2)$? which is not an immersion :-(

Answer 1

You can't have an immersion like that. It would be monotone in the first argument, so at some point $p$ you would have $\gamma(p)=(0,0)$. But then you would also have $\lim_{x\to p^-} D\gamma(x)(1)=\lim_{x\to p^+}D\gamma(x)(1)$, which is only possible if both are zero (because $D\gamma(x)(1)\perp D\gamma(y)(1)$ whenever $x<p<y$), but then $D\gamma(0)=0$.

It's not hard to find a smooth homeomorphism which is an immersion everywhere but $0$, though. Just compose the standard $\gamma(t)=(t,\lvert t\rvert)$ with a suitable function ${\bf R}\to{\bf R}$ with all derivatives zero at zero.