Showing a set is/isn't an embedded submanifold using inclusion maps I am using the definition in "Introduction to Smooth Manifolds" by Lee
Defn: Suppose $M$ is a smooth manifold.  An embedded smooth manifold of $M$ is a subset $S\subset M$  which is a manifold in the subspace topology, endowed with a smooth structure with respect to which the inclusion map $i:S\to M$ is a smooth embedding.
My question has to do with showing that {$(x,y):x=|y|=\sqrt{y^2} $} is not a smooth submanifold of $R^2$, but more specifically to do with the inclusion map.  I suppose a second part to this question is if I am thinking about this the correct way, or if am I completely wrong.
I want to show that the inclusion map isn't a smooth embedding (hopefully this is the correct approach).  
I am wondering if I can define $i:S\to M$ by $$i(x,y)=(x,y)=(|y|,y)$$
so that my derivative is $$(1,\frac{y}{\sqrt{y^2}});(0,1)$$
My conclusion is that this doesn't exist at $(0,0)$ so the inclusion map isn't a smooth embedding. 
(I don't know how to make a matrix on here, so first set of parenthesis is first row of Jacobian, second set is second row )
I understand that this question is similar to some from the past, by I haven't found any regarding how to define the inclusion map to break the definition.
Thanks!
 A: So one thing to start with here regarding the 2x2 Jacobian you have, if $S$ is a submanifold, it has to be a
1-manifold. You can see that by noting that the sets $S^+ = \{(y,y) | y > 0\}$ and
$S^- = \{(-y,y) | y < 0\}$ (the two rays that make up the V-shape of S) individually are
embedded submanifolds of $\mathbb{R}^2$ and they're each embedded 1-submanifolds of $\mathbb{R}^2$.
If $\iota$ was a smooth embedding, its restriction to $S^+$ would be a smooth embedding. We can show an example of $S^+$ being a smooth embedded 1-submanifold, and the smooth structure that makes $\left.\iota\right|_{S^+}$ a smooth embedding is unique, so that means any smooth structure we hook up on $S$ has to be that of a 1-manifold when restricted to $S^+$, so $S$ itself has to be a 1-manifold.
That means we can pick a chart around the point $(0,0)$,
$\phi : U \rightarrow \mathbb{R}$. Shrinking $U$ if necessary, and translating, we can say $\phi(U) = (-\epsilon,\epsilon)$ for
some $\epsilon > 0$ and $\phi(0) = (0,0)$ (shrinking to a coordinate ball centered at $(0,0)$). $\phi$ is a
diffeomorphism and $\iota$ is a smooth immersion, so $g = \iota \circ \phi^{-1} : (-\epsilon,\epsilon) \rightarrow \mathbb{R}^2$ is a
smooth immersion of $(-\epsilon,\epsilon)$ into $\mathbb{R}^2$. We can write it in coordinates as $g(t) =
(g_x(t),g_y(t))$, swapping $g(t)$ with $g(-t)$ if necessary, and noting the
shape of $S$, we really have:
$$
g(t) = \left\{\begin{aligned}
&(g_y(t),g_y(t)) && t < 0\\
&(-g_y(t),g_y(t)) && t > 0 \\
& (0,0) && t = 0
\end{aligned}
\right.$$
We can also evaluate the tangent vector of this away from $t=0$:
$$
g'(t) = \left\{\begin{aligned}
&g_y'(t)(1,1) && t < 0\\
&g_y'(t)(-1,1) && t > 0 \\
\end{aligned}
\right.$$
Now the only way that can be smooth is if $g_y'(t)$ smoothly goes to zero at $t = 0$, and takes the value $g_y'(0) = 0$.
If it does that, since $|g_x'(t)| = |g_y'(t)|$ near $t=0$, the whole tangent vector to the
curve, $g'(t)$ must smoothly go to zero. That's not possible though, because $g$ is a smooth
immersion, so it has an injective pushforward. $g'(t)$, the tangent vector,
is related to the pushforward of $g$ by $g_*(\left.\frac{d}{dt}\right|_0) = g'(0) = 0$, but $\frac{d}{dt}$ isn't zero so
that's not injective. That means $\iota$ can't be a smooth embedding.
