TITLE: Elementary question in differential geometry
QUESTION [7 upvotes]: I am trying to learn differential geometry (i.e., teach myself!)
So here is a question that came up.
For some $h > 0$, consider the cone 
$C_h = \{ (x,y,z) \; : \;  0 \le z = \sqrt{x^2 + y^2} < h \} \subset \mathbb{R}^3$
endowed with subspace topology. It seems that we can cover this with a single chart $(U,\phi)$ where $U = C_h$ and $\phi$ is the projection $\phi(x,y,z) = (x,y)$. So it seems that this defines a differentiable structure and we get a smooth ($C^\infty$) 2-dimensional manifold. (Is it correct?)
Now consider the inclusion map $i : C_h \to \mathbb{R}^3$, is this maps smooth? It doesn't seem to me that it is. The expression of $i$ in the chart above is not smooth at $(0,0)$ and I don't seem to be able to find any other compatible chart around zero which has a smooth representation. (Haven't given it  much thought though). If this is true how one shows that this map is not smooth. (Also, if this is true, a vague question is whether removing the origin is the only way to fix this problem)

REPLY [3 votes]: You've endowed $C_h$ with the structure of an abstract manifold but $C_h$ is not a submanifold of $\mathbb R^3$.  The fact that your set isn't a submanifold boils down to two observations: 
(1) The fact your map $i$ is not differentiable at the origin
and
(2) An application of the implicit function theorem gives the proof by contradiction. The implicit function theorem says that if your set was a submanifold, $i$ would have to be smooth -- technically you have to consider the two other coordinate projections $(x,y,z) \to (y,z)$, $(x,y,z) \to (x,z)$ but $C_h$ does not satisfy the "vertical line rule" so it can't be a graph of a function of $(y,z)$ or $(x,z)$.