How to prove this assertion? Do I need to use Inverse function Theorem?

Question

Let $f:R^2\to R$ be a continuously differentiable function. Show that there exist a continuous one-one function $g:[0,1]\to R^2$ such that $f\circ g:[0,1]\to R $ is constant.

Answer 1

If $f$ is constant, any $g$ will do. Otherwise there is a point $(p,q)\in{\mathbb R}^2$ with $\nabla f(p,q)\ne0$. Let $f(p,q)=:c$, and assume that $f_y(p,q)\ne0$. By the implicit function theorem there is a $C^1$-function $$\psi:\quad x\mapsto y:=\psi(x)\ ,$$ defined in some neighborhood $U=\ ]p-h, p+h[\ $ of $p$, with $\psi(p)=q$, and $$f\bigl(x,\psi(x)\bigr)=c\quad(x\in U)\ .$$ (This is expressing the fact that we can solve the equation $f(x,y)=c$ for $y$ in the neighborhood of $(p,q)$.) Now put $$g(x):=\bigl(x,\psi(x)\bigr)\qquad(p-h<x<p+h)\ .$$ Note that $g$ is injective since the first component of $g$ is. – It is an easy matter to rescale the interval $U$ (or a slightly smaller compact interval) to $[0,1]$, if so desired.