Let $F: \mathbb{R^{3}}\to \mathbb{ R}$ a $C^{1}$-image and assume that $(dF) (x, y, z)\neq 0$ as soon as $F (x, y, z) = 0$.
Define $O= \{(x, y, z) ∈ \mathbb{R^{3}}| F (x, y, z) = 0\}$.
Prove that every point of $O$ has an environment that looks like a piece of the xy plane in $\mathbb{R^{3}}$.
I think this means that I have to prove that for every point $(a, b, c) \in O$ there open parts $U, V \subset \mathbb{R^{3}}$ exist and a bijective $C^{1}$- image $\phi: U → V$ with $C^{1}$-inverse such that $(0, 0, 0) \in U, (a, b, c) \in V$ and $O \cap V = \phi (U \cap (\mathbb{R^{2}} \times {0})$
My attempt:
Let's take $(a,b,c) \in O$ random. Let's assume that (1x3)- matric $(dF)(x,y,z)$ is of rang $1$. Because of the implicit function theorem there exist
- an open part $V\subset \mathbb{R^{3}}$ so $(a,b,c)\in V$
- an open part $U \subset \mathbb{R^{2}}$ with $(0,0) \in U$
- a $C^{1}$-image $\phi:U\to \mathbb{R}$ with $\phi((0,0))=(a,b,c)$
and $ V \cap O=\phi(U)$ and the image of $(d\phi)(y)$ equals de kern of $(dF)(\phi(y))$ for all $y\in U$.
Because the condition that $(dF)(x,y,z) \neq 0$, $U$ needs to be a part of $\mathbb{R^{3}}$. This means that $(0,0,0) \in U$ and $O \cap V=\phi(U\cap (\mathbb{R^{2} \times {0} ))}$
I'm not that sure that the conclusions I made about the condition $(dF)(x,y,z) \neq 0$ are correct. Or maybe I started the question wrong. Can someone check my answer and see what I do wrong or right.
Here is an alternative, more straightforward solution: By the regular level set theorem, $O=F^{-1}(0)$ is an embedded submanifold of codimension 1. Hence it has a smooth structure which makes it into a manifold of dimension 2.