Manifold in $\mathbb{R}^2$ Is the set $\left\lbrace  (x,y) \in \mathbb{R}^2 \mid x^2 = y^2 \right\rbrace $ a  manifold in $\mathbb{R}^2 $?
I know I could use the level set theorem if I had $\left\lbrace  x \in \mathbb{R}^2 \mid x_1^2 = x_2^2=c  \right\rbrace $, provided  $ c \neq 0$;
but the set I have is in fact the union of all such sets along with the case where $c =0$.
Thanks!
 A: Your set—let's call it $M$—is not a manifold. Since $x^2 = y^2$ if and only if $|x| = |y|$ if and only if $x = \pm y$, 
$$
 M := \{(x,y) \in \mathbb{R}^2 \mid x^2 = y^2\} = \{(x,y) \in \mathbb{R}^2 \mid y = x\} \cup \{(x,y) \in \mathbb{R}^2 \mid y = -x\},
$$
i.e., your set is the union of the lines $y=x$ and $y=-x$. Away from the origin, $M$ is readily seen to be locally homeomorphic to $\mathbb{R}$, as you would expect for a $1$-manifold, e.g., for $x \neq 0$,
$$
 \phi_{x,\pm} : \left(-\tfrac{|x|}{2},\tfrac{|x|}{2} \right) \to M, \quad t \mapsto (x+t,\pm(x+t))
$$
defines a would-be coordinate chart at $(x,\pm x)$. However, for any neighbourhood $U$ of $(0,0)$ in $M$, $U$ cannot possibly be homeomorphic to some interval $(-\epsilon,\epsilon)$, for if $\phi : (-\epsilon,\epsilon) \to U$ were such a homemorphism, it would restrict to a homeomorphism
$$
 (-\epsilon,\epsilon) \setminus \{\phi^{-1}(0,0)\} \to U \setminus \{(0,0)\},
$$
where $(-\epsilon,\epsilon) \setminus \{\phi^{-1}(0,0)\}$ has two connected components, whilst $U \setminus \{(0,0)\}$ has at least four.
In terms of level sets, the key observation is that $M = f^{-1}(0)$ for $f(x,y) = x^2-y^2$, where $$\nabla f(x,y) = (2x,-2y)^T$$ and hence $\nabla f(0,0) = 0$, ruling out the use of the level set theorem. Indeed, this is an excellent illustration of why regularity ($\nabla f(x,y) \neq 0$ everywhere) is essential to get a manifold; things can go horribly wrong precisely where $\nabla f(x,y) = 0$, e.g., at $(x,y) = (0,0)$ in this example.
