In Tapp's "Matrix Groups for Undergraduates" he briefly states (p.103) that a compact cone (he just shows a picture of a manifold with a ''cone point'') is not diffeomorphic to a 2-sphere. I would love for someone to give me a simple proof, using only elementary analysis/topology methods, why this is true. To be on the same page:
Let $C = \left\{x \in \mathbb{R}^3 \mid 0 \leq z = \sqrt{x^2 + y^2} \leq 1\right\}$ be the compact cone. Let $f : C \to S^2 \subset \mathbb{R}^3$ be a homeomorphism. Prove that $f$ is not a diffeomorphism by proving that $f$ is not smooth at the origin; that is, there does not exist a smooth local extension of $f$ about the origin.
Should be added: "with invertible derivative" (i.e., nonvanishing Jacobian). It is possible to map a 3D neighborhood of the cone point homeomorphically onto a 3D neighborhood of a point on the sphere, with cone surface going to sphere surface, and the map differentiable everywhere, with zero derivative at the cone point.
Assuming nonzero derivative, the proof can go like this: assume $f$ is such a map of 3D neighborhoods. The function $|f|^2=f_1^2+f_2^2+f_3^2$ is smooth, has nonzero gradient (why?), and is constant on the surface of the cone. Let $v$ be the cone vertex. The directional derivative of $f$ at $v$ in the direction of any vector tangent to the cone is zero. Therefore, all such vectors are orthogonal to $\nabla f(v)$. But this is impossible (why?).