How can I prove that $M=\{(x, |x|), x\in\mathbb{R}\}$ is not an embedded smooth($C^\infty$) submanifold of $\mathbb{R}^2$.
I tried to say there is any ($C^{\infty}$) immersion from $\mathbb{R}$ into $\mathbb{R}^2$ such that its image is $M$, but I couldn't.
I attend all of you to the following point( My friends!! please do not beguile for the graph of non-differentiable function $f:\mathbb{R}\longrightarrow\mathbb{R} $ defined by $f(x)=|x|$ at $x=0$). Since, I can suppose that
$g(x)=\begin{cases}\mathrm{e}^{-1/x^2} & x>0\\ 0 & x=0 \\ -\mathrm{e}^{-1/x^2} & x<0\end{cases}$ and $h:\mathbb{R}\longrightarrow\mathbb{R}^2$ defined by $h(x)=(g(x), |g(x)|)$. I could prove that $h$ is differentiable at $x=0$ and $C^{\infty}$ on $\mathbb{R}$ and its image is $M$ and $h$ is not an immersion at $x=0$.
Indeed, for answering to my main problem, we should show that $M$ is not the image of any ($C^{\infty}$) immersion of $\mathbb{R}$ into $\mathbb{R}^2$. How can I do this work, exactly ????