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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 ????

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3  
Thanks for improving the question. If you want to show that $M$ is not an embedded submanifold, shouldn't you try to show that there can't be any immersion from $\mathbb R$ into $\mathbb R^2$ whose image is $M$? (What would its derivative be at $(0,0)$?) – Rahul Aug 20 '12 at 7:00
1  
It depends what kinds of manifolds you're talking about. In the category of PL (piecewise linear) manifolds, the embedding is OK. So probably your question is in the category of $C^\infty$ or smooth manifolds, and it would help if you would specify that. Also, please note that an embedding is not the same as an immersion (any embedding is an immersion, but many immersions are not embeddings). – Hew Wolff Aug 21 '12 at 22:44

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