# $\text{graph}(f)$ from a pre-image

Let $V$ be an open subset of $\mathbb{R}^{n-1}$, $f\in C^1(V,\mathbb{R})$, and $S:=\{(x_1,\ldots,x_{n-1},f(x_1,\ldots,x_{n-1})) : (x_1,\ldots,x_{n-1}) \in V\}$ the graph of the function $f$. Now, is there a function $g\in C^1(\mathbb{R}^n)$ such that $S=g^{-1}(0)$. I think $0$ must be a regular value of $g$ but I cannot imagine what $g$ looks like.

No, not in general. Since $g$ is continuous, $g^{-1}(0)$ is closed, while $S$ need not be closed. For example, take $n=2$, $V=(0,1)$, and $f(x)=x$.
Put $g(x_1,\ldots, x_n):=f(x_1,\ldots,x_{n-1})-x_n\$.
That defines $g$ only on $V \times \mathbb{R}$. –  Chris Eagle Apr 22 '12 at 10:25