Proof that cone not diffeomorphic to plane What is the simplest way to show that the cone $\{(x,y,z)\in\mathbb R^3\,|\,z=\sqrt{x^2 + y^2}, z\geq 0\}$ is not diffeomorphic to $\mathbb R^2$?
After some comments, I realize that this question The cone is not immersed in $\mathbb{R}^3$ is more in the the spirit of what I wanted to know.
What I'm really trying to ask in a physical sense is in what sense is the vertex of the cone a point at which the cone is not smooth, and how can one prove this?
 A: Your original question makes perfect sense.  A diffeomorphism between a set $X \subset \mathbb R^n$ and a set $M \subset \mathbb R^k$ means a 1-1 continuous function $f : X \to M$ which is `smooth' in the sense that for every $p \in X$ there is an open neighbourhood $U$ of $p$ in $\mathbb R^n$ and a smooth map 
$$ \tilde f : U \to \mathbb R^k$$
such that $\tilde f(q) = f(q)$ for all $q \in U \cap X$.  You also demand that $f^{-1} : M \to X$ is smooth. 
The reason why a cone isn't diffeomorphic to a euclidean space basically just boils down to an implicit function theorem argument.  You ask yourself, what is the tangent space at the cone point?  The implicit function theorem tells you that near any point in the manifold, the manifold is the graph of a smooth function from the tangent space to its orthogonal complement.  So you just have to argue that regardless of what the tangent space is, there's no smooth function that it's the graph of.  There's also a coordinate-axis version of the implicit function theorem which is more convenient in this case.  If it was a smooth manifold, the function $z = \sqrt{x^2+y^2}$ would have to be a smooth function. 
