The two sheeted cone is $\{(x,y,z) \in \mathbb R^3 : x^2+y^2-z^2=0\}$.
I would like to use this proposition:
to show that the two sheeted cone is not a regular surface. I know that the point of failure has to be the $(0,0,0)$, but I am not sure how to use this proposition to show this.
Best Answer
Suppose the two sheeted cone $S$ is a regular surface. As you note $p=(0,0,0)\in S$. Let $U\subset S$ be an open neighbourhood of $p$ in $S$ such that $U$ is the graph of a differentiable function of one of the forms $$z=f(x,y),\qquad y=g(x,z),\qquad x=h(y,z).$$ By definition of the topologies on $S$ and $\Bbb{R}^3$ there exists an open ball $B_{\varepsilon}(p)\subset\Bbb{R}^3$, centered at $p$ and with radius $\varepsilon$, such that $V:=B_{\varepsilon}(p)\cap S$ and $V\subset U$. Then $V$ is also the graph of a function as above. Note that $V$ contains a point $(x,y,z)$ with $x\neq0$, $y\neq0$ and $z\neq0$. Then it also contains the points $$(x,y,-z),\qquad (x,-y,z)\qquad\text{ and }\qquad (-x,y,z).$$ But this contradicts the fact that $$z=f(x,y),\qquad y=g(x,z)\qquad\text{ or }\qquad x=h(y,z),$$ respectively. Hence the two sheeted cone is not a regular surface.