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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. – Doug 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
up vote 1 down vote accepted

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?).

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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). – Doug Jul 16 '13 at 6:17
*where above the extension is to some open ball in $\mathbb{R}^3$. – Doug Jul 16 '13 at 6:20

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