Showing that a level set is not a submanifold Is there a criterion to show that a level set of some map is not an (embedded) submanifold? In particular, an exercise in Lee's smooth manifolds book asks to show that the sets defined by $x^3 - y^2 = 0$ and $x^2 - y^2 = 0$ are not embedded submanifolds.
In general, is it possible that a level set of a map which does not has constant rank on the set still defines a embedded submanifold? 
 A: It is certainly possible for a level set of a map which does not have constant rank on the set to still be an embedded submanifold.  For example, the set defined by $x^3 - y^3 = 0$ is an embedded curve (it is the same as the line $y=x$), despite the fact that $F(x,y) = x^3 - y^3$ has a critical point at $(0,0)$.
The set defined by $x^2 - y^2 = 0$ is not an embedded submanifold, because it is the union of the lines $y=x$ and $y=-x$, and is therefore not locally Euclidean at the origin.  To prove that no neighborhood of the origin is homeomorphic to an open interval, observe that any open interval splits into exactly two connected components when a point is removed, but any neighborhood of the origin in the set $x^2 - y^2$ has at least four components after the point $(0,0)$ is removed.
The set $x^3-y^2 = 0$ is an embedded topological submanifold, but it is not a smooth submanifold, since the embedding is not an immersion.  There are many ways to prove that this set is not a smooth embedded submanifold, but one possibility is to observe that any smooth embedded curve in $\mathbb{R}^2$ must locally be of the form $y = f(x)$ or $x = f(y)$, where $f$ is some differentiable function.  (This follows from the local characterization of smooth embedded submanifolds as level sets of submersions, together with the Implicit Function Theorem.)  The given curve does not have this form, so it cannot be a smooth embedded submanifold.
A: The set given by $(x^2+y^2)(x^2+y^2-1)=0$ is an embedded submanifold in $\mathbb R^2$
 but it has components of different dimension and so I guess the map does not have constant rank on the set, but I haven't checked.
