Show that the set $M$ is not an Embedded submanifold

Question

How can I prove that $M=\{(x,y)\in \mathbb{R}^2\ ; y=|x|\}$ is not an embedded smooth submanifold of $\mathbb{R}^2$?

Answer 1

Being an embedded smooth submanifold of codimension $k$ implies (either as the definition or a consequence of it) that every point $p\in M$ has a neighborhood $U$ such that $$M\cap U = \{x\in U: g_1(x)=\dots = g_k(x)=0\}$$ where the functions $g_1,\dots,g_k$ are smooth and their gradients $\nabla g_j(p)$ are linearly independent.

If your set $M$ was a smooth embedded submanifold, $\nabla g_1(0,0)$ would be orthogonal to both vectors $(1,1)$ and $(1,-1)$. Indeed, the directional derivatives $\lim_{h\to 0}\frac{1}{h}(g_1(\pm h,h)-g_1(0,0))$ are zero because $g_1$ vanishes on $M$. A nonzero vector in two dimensions can't be orthogonal to two linearly independent vectors.

By the way, $M$ is a $1$-dimensional embedded topological submanifold.