Why does a surface F(x,y,z) that never fold back on itself have $\nabla F$ $\cdot \vec{p} \neq 0$?

Question

Why does an implicit surface F(x,y,z) that never fold back on itself have $\nabla F$ $\cdot \vec{p} \neq 0$?

Answer 1

The assumption $\nabla F({\bf x})\ne{\bf 0}$ for all ${\bf x}\in S$ guarantees that each point ${\bf x}\in S$ has a neighborhood $U$ such that within $U$ the surface behaves as desired: no self-intersections, cusps, etc. One then can select a variable, say $z$, such that within $U$ the surface $S$ appears as graph of a $C^1$-function $z=\psi(x,y)$. The assumption $\nabla F({\bf x})\cdot{\bf p}\ne0$, whereby ${\bf p}=(0,0,1)$, enforces that we can actually choose $z$ as this "special" variable.

In points ${\bf x}\in S$ where $\nabla F({\bf x})\cdot{\bf p}=0$ the gradient of $F$, while $\ne{\bf 0}$, is parallel to the $(x,y)$-plane. Therefore the tangent plane of $S$ is vertical in such points. This has the consequence that in the neighborhood of such points the surface $S$ is still smooth, but it does not project properly onto the $(x,y)$-plane.