Showing a neighborhood is a surface Suppose that the function $h:\mathbb{R}^3 \rightarrow \mathbb{R}$ is contiuously differentiable. Let $p$ be a point in $\mathbb{R}^3$ at which $\nabla h(p) \neq 0$ and define $c=h(p)$. Show that there is a neighborhood, $N$, of $p$ such that $\mathcal{S}$ $=\{ (x,y,z) \in N | h(x,y,z)=c \} $ is a surface.
 A: Since $\nabla h(p) \neq 0$, there is some index $i$ such that $\langle \nabla h(p), e_i \rangle = \frac{\partial h(p)}{\partial x_i}\neq 0$. Let $j,k$ be the two other indices. Reparameterize $\mathbb{R}^3$ as follows: Let $\pi:\mathbb{R} \times \mathbb{R}^2 \to \mathbb{R}^3$ with $\pi(x,y) = x e_i + y_1 e_j + y_2 e_k$. Let $\pi(\bar{x}, \bar{y}) = p$. Note that $\pi$ is an invertible (in fact, orthogonal) linear map.
Let $\phi:\mathbb{R} \times \mathbb{R}^2 \to \mathbb{R}$ be given by $\phi(x,y) = h(\pi(x,y)) -c$. Then $\phi(\bar{x}, \bar{y}) = 0$ and $\frac{\partial \phi(\bar{x}, \bar{y}) }{\partial x} \neq 0$, hence by the implicit function theorem, there exists a neighborhood $U$ of $\bar{y}$, a neighborhood $V$ of $\bar{x}$, and a $C^1$ function $\eta:U \to V$ such that $\eta(\bar{y}) = \bar{x}$ and $\phi(\eta(y), y) = 0$ for $y \in U$. Furthermore, if $(x,y) \in U \times V$ and $\phi(x,y) = 0$, then $x=\eta(y)$. 
Let $N = \pi(U\times V)$. Since $\pi$ is invertible, $N$ is an open neighborhood of $p$.  It follows that $S \cap N = \{ \pi(\eta(y), y) \, | \, y \in U \}$. If we let $\gamma(y) =
\pi(\eta(y), y)$, we can see that $\gamma$ has an inverse on $S \cap N$ given by $\gamma^{-1} (q) = (q_j,q_k)$. Hence $U \subset \mathbb{R}^2$ is diffeomorphic to $S \cap N$, and so $S\cap N$ is a surface.
