# Why is this proof that a circular cone is not a surface not rigorous?

In example $4.1.5$, page $73$ of Pressley's Elementary Differential Geometry, a "heuristic" argument is given to prove that the circular cone with vertex the origin and angle $\pi/4$, is not a surface. Here are the exact words:

To see that it is not a surface, suppose that $\sigma: U \to S \cap W$ is a surface patch containing the vertex $(0,0,0)$ of the cone, and let $a\in U$ correspond to the vertex. We can assume that $U$ is an open ball with center $a$, since any open set $U$ containing $a$ must contain such an open ball. The open set $W$ must obviously contain a point $p$ in the lower half $S_{-}$ of $S$ where $z < 0$ and a point $q$ in the upper half $S^{+}$ where $z>0$; let $b$ and $c$ be the corresponding points in $U$. It is clear that there's a curve $\pi$ in $U$ passing through $b$ and $c$, but not passing through $a$. This is mapped by $\sigma$ into a curve $\gamma = \sigma \circ \pi$ lying entirely in $S$, passing through $p$ and $q$, and not passing through the vertex. (It is true that $\gamma$ will, in general, only be continuous, and not smooth, but this does not affect the argument.) This is clearly impossible. (Readers familiar with point set topology will be able to make this heuristic argument rigorous).

Why is this argument not considered rigorous? Can someone give an outline of how a rigorous argument should be?

• It's not rigorous because "clearly impossible" is itself not a rigorous claim. But note that virtually all proofs that you see are variously informal/unrigorous, i.e., in practice, rigor is a matter of degree, not of absolutes. – Travis Willse Apr 22 '16 at 12:59
• To make this more rogorous (without really changing the argument): Any surface remains (pathwise) connected after removing a single point - argue as above with the open neighbourhood of the gap. Bit the cone is not connected (positive vs negative part) after removing its vertex – Hagen von Eitzen Apr 22 '16 at 13:03
• The argument does seem a bit roundabout -- it would be clearer to observe that $S\cap W$ has the property that removing a single point can make it disconnected, whereas $U$ doesn't -- so they cannot be homeomorphic. – hmakholm left over Monica Apr 22 '16 at 13:03

The problem is at the center of coordinates, that is at the point $$(0,0,0)$$. The usual parametrization for a cone of one sheet is the following, $$z=\sqrt{x^2 +y^2}$$ which is not differentiable at $$(0,0,0)$$. But could there be another parametrization such that the cone is a regular surface? Here comes to rescue the following result which is a converse to the usual result that the graph of a differentiable function is a regular surface. In this case the converse is local.
Now if the cone where to be a regular surface, there should be a neighborhood $$V$$ of $$(0,0,0)$$ such that $$V$$ is the graph of a differentiable function which has one of those forms. If you haven't seen the proof of this proposition, what you actually use to construct the function is (not surprisingly) the inverse function theorem applied to the composition of the parametrization of the surface $$X$$ and the projection $$\pi$$ to one of the planes $$\{z=0\} \hspace{0.2cm} \text {or} \hspace{0.2cm} \{y=0\} \hspace{0.2cm} \text {or} \hspace{0.2cm} \{x=0\}$$.
In this case in particular of the cone, it can be seen that a neighborhood of $$(0,0,0)$$ whenever is projected to any other plane different than $$\{z=0\}$$, the projection can never be a bijection much less a diffeomorphism, so that you wouldn't be able to use the inverse function theorem. This tells us that the graph of the differentiable function ought to be of the type $$z=f(x,y)$$. This finishes the proof because now you know from the usual parametrization of the cone that $$f(x,y)= \sqrt{x^2 +y^2}$$ in a neighborhood of $$(0,0,0)$$. This function is not differentiable at $$(0,0)$$ so that you have arrived at a contradiction. This contradiction arises from supposing that the cone was a regular surface.