I am writing the proof that, $M:=\{(x,|x|):x\in \Bbb R\}$ is not a submanifold of $\Bbb R^2$. Can anyone help me by finding the fault in the proof.
If possible let $M$ be a submanifold of $\Bbb R^2$. Then we have homeomorphism $\phi:U\to \Bbb R$ where $(0,0)\in U\subseteq_{\text{open}}M$ such that following holds,
$\phi(0,0)=0$,
There is an set $\tilde U\subseteq_{\text{open}}\Bbb R^2$ and a smooth function $\tilde\phi :\tilde U\to \Bbb R$ with $U=\tilde U\cap M$ and $\phi=\tilde \phi\big|_{U}$.
$(f,g)=\phi^{-1}:\Bbb R\to U\subseteq R^2$ is a smooth map.
Now, $\tilde\phi\circ\phi^{-1}:\Bbb R\to \Bbb R$ is identity map implies $(f'(0),g'(0))=(d\phi^{-1})_0$ is injective. But, $\phi^{-1}:\Bbb R\to U\subseteq M$ implies $|f|=g$. Also $f(0)=0$. So that, $|f'(0)|=g'(0)$. Hence, $f'(0)\not=0$. Next notice that, $f:\Bbb R\to (a,b)$ is a smooth bijective function for $(a,b)$ is an open interval containing $0$. Since continouos image of a connected set is connected and $f$ is bijective $f(0,\infty)$ and $f(-\infty,0)$ are intevels not containing $0$. Without loss of generality let $f(0,\infty) =(0,b)$ and $f(-\infty,0)=(a,0)$. Now $$\lim_{x\to 0+}\frac{g(x+0)-g(0)}{x}=\bigg|\lim_{x\to 0+}\frac{f(x)}{x}\bigg|=|f'(0)|>0,$$ $$\lim_{x\to 0-}\frac{g(x+0)-g(0)}{x}=-\bigg|\lim_{x\to 0-}\frac{f(x)}{x}\bigg|=-|f'(0)|<0.$$ This is a contradiction. So we are done.
Now my next question is the following, every topological manifold of dimension $1,2,3$ has a unique compatible smooth structure. So $M$ has also so and let $\psi:V\to \Bbb R$ be a chart at $(0,0)$ in this structure. Now $M$ is not a submanifolds of $\Bbb R^2$ either $\psi$ doesn't have any smooth extension as in fact 2. or the component functions of $\psi^{-1}$ are not smooth as a map from $\Bbb R$ to $\Bbb R$. Am I right?