# The double cone is not a surface.

My question is that

A double cone ( also named as "circular cone") is not a surface.

I know its reason. But I cannot show this mathematically.

Suppose $\sigma : U \to S\cap W$ Is a surface patch.

Because the vertex $(0,0,0,)$ is a problem, S is not a surface.

I can see it. But I cannot express it in the mathematical way.

When I remove the vertex point $(0,0,0)$, the double cone is a surface.

Please can someone show/write these mathematically?

If you remove a point from an open subset in $\mathbb{R}^2$, it remains connected. A surface patch around the vertex has to give a homeomorphism to an open subset of $\mathbb{R}^2$. However, any open neighborhood of the vertex has the property that if you remove the vertex it becomes disconnected. Thus it cannot be homeomorphic to a subset of $\mathbb{R}^2$.

• Then, I define an open set $U=\Bbb R^2 /\{(0,0)\}$. And $\sigma : U \to \Bbb R^3$ Such that $\sigma(u,v)=(u,v,- or + \sqrt{u^2+v^2})$
– 1190
Commented Nov 23, 2013 at 14:37
• I do not really know how to express my ideas for this proof.
– 1190
Commented Nov 23, 2013 at 14:40
• Can you help me more explicitly?
– 1190
Commented Nov 23, 2013 at 14:41
• Hmm, is this answer enough to get successful point for a homework? I know all these. But I cannot express properly. So i have asked Dear I.Solomon.
– 1190
Commented Nov 23, 2013 at 14:51

Note that if S where to be a regular surface, then it would be, in an open nhood of $(0,0,0)$ in $S$, the graph of differentiable function of the form: $x=f(y,z)$, $y=g(x,z)$ or $z=h(x,y)$. But that obviously cannot be the case, once the projections of S onto the coordinate planes aren't injective.

• for future readers, this is proposition 2.23 in Ros-Montiel Curves and Surfaces (the explanation comes from lemma 2.21) Commented Apr 22, 2018 at 20:53