# Prove that a compact cone is not diffeomorphic to the 2-sphere

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.

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Your cone $C$ is not a smooth submanifold of $\mathbb R^3$. What exactly does «$f:C\to S^2$ is smooth at the origin» mean? –  Mariano Suárez-Alvarez Jul 15 '13 at 6:30
I checked the relevant passage using Google books, and the explanation is really bad. As Mariano notes, what he draws is not a manifold. Just ignore this passage, and perhaps find a better book. –  Potato Jul 15 '13 at 6:34
@Avitus: You can define the partial derivatives once you extend the notion of ''smooth'' as in the definition given in my second-to-last comment. –  Dan Douglas Jul 15 '13 at 6:51
We can't get off the ground here: there is no homeomorphism from $C$ to $S^2$ since the former is a homeomorph of the disc, so contractible, while the $2$-sphere is not contractible (non-trivial $\pi_2$). –  Fran Burstall Jul 15 '13 at 7:06
@DanDouglas If you're feeling more adventurous, Lee's An Introduction to Smooth Manifolds is the standard text for introductory graduate courses. It might be useful as an additional reference. –  Potato Jul 15 '13 at 7:16

there does not exist a smooth local extension of $f$ about the origin.
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?).
Thank you. There has been a lot of confusion on this thread, so I want to tell people exactly the question you answered which is what I really wanted to be answered except that I failed to ask the question successfully: Let C be defined as above. It is clear that a small neighborhood, U, of 0 in C is homeomorphic to some small open, V, in $S^2$. Call this homeomorhism $f$. Prove that there does not exist a smooth bijection locally extending $f$ whose inverse is also smooth (which, a fortiori, implies that $f$ has an inverible derivative). –  Dan Douglas Jul 16 '13 at 6:17
*where above the extension is to some open ball in $\mathbb{R}^3$. –  Dan Douglas Jul 16 '13 at 6:20