How can one prove that the set $\{(x,|x|)\in \mathbb{R}^2 \mid x\in \mathbb{R}\}$ cannot be the image of an immersion of a smooth manifold?
This was my homework exercise in a course about differentiable manifold a few years ago.
However, I couldn't give a rigorous proof so far.
suppose there is a differentiable map $$ f:U \rightarrow \mathbb{R}^2 $$ where $U$ is a one dimensional manifold. You can assume that $U$ such that $f(p)=(0,0)$ and $f(U)$ is your set. If $df|_p$ is injective, take any non-zero tangent vector $X|_p$ at $p$ and think about the non-zero tangent-vector $$ df|_p (X|_p) $$ of $T_{(0,0)}\mathbb{R}^2$. If $\phi$ is any differentiable function near $(0,0)$, then $$ df|_p (X|_p)\phi = X|_p(\phi \circ f) $$ By http://en.wikipedia.org/wiki/Hadamard's_lemma you can write $$ \phi = \phi(0) + xg_1(x,y) + yg_2(x,y) $$ so if you want to reach the contradiction $df_p(X_p)=0$ it is enough to show that $$ df|_p (X|_p)x = X|_p(x \circ f) = 0\qquad \text{and} \qquad df|_p (X|_p)y = X|_p(y \circ f) = 0 $$ This is easy: $y\circ f$ has a minimum at $p$ and $(x\circ f)^2=(y\circ f)^2$
