Suppose that a subset $S$ of $R^2$ has the property that locally on $S$ one of the coordinates is a $C^∞$ function of the other coordinate. Show that $S$ is a regular submanifold of $R^2$
My prof gave me a hint saying we must take charts $(U,x,y-f(x))$ and $(V,x,y-g(x))$ and show they are compatible and so are their adapted charts. What does he mean?
A subset $S$ of a manifold N of dimension $n$ is a regular sub- manifold of dimension k if for every p ∈ S there is a coordinate neighborhood $(U, φ ) = (U, x_1 , . . . , x_n )$ of p in the maximal atlas of N such that U ∩ S is defined by the vanishing of n − k of the coordinate functions. By renumbering the coordinates, we may assume that these n − k coordinate functions are $x_{k+1} , . . . , x_n$ .
This is one of the many uses of the inverse function theorem. The following is an outline of the proof.
Let $p\in S$, and without loss of generality, the $y$ coordinate is a smooth function of the $x$ coordinate around $p$. In other words, a piece of the curve $S$ around $p$ is parametrized by$$x\mapsto(x,y(x)).$$
Without loss of generality, this parametrization is regular on the interval $(-\epsilon,\epsilon)$. Define a function$$\varphi:(-\epsilon,\epsilon)\times\mathbb{R}\to\mathbb{R}^2$$by$$(x_1,x_2)\mapsto(x_1,y(x_1)+x_2).$$Use the inverse function theorem to show that $\varphi$ is a diffeomorphism on some neighborhood of $(0,0)$. In other words, $\varphi$ is a parametrization of a neighborhood in $\mathbb{R}^2$. In the image of $\varphi$, the curve $S$ is given by $X_2=0$.