If we are given a $C^1$ map $F:\mathbb{R}^{n+m}\rightarrow\mathbb{R}^{m}$ and a point $(x_{0},y_{0})\in\mathbb{R}^{n}\times\mathbb{R}^{m}$ such that $F(x_{0},y_{0})=0$, the implicit function theorem says that we can uniquely solve $F(x,y)=0$ for $y$ as a function of $x$ in a neighborhood of $(x_{0},y_{0})$ provided $\det{D_{y}}F(x_{0},y_{0})\neq 0$.
My question is: if we give up uniqueness, can we show the existence of some local implicit function $y(x)$ with the same $C^{1}$ regularity under a weaker hypothesis on $D_{y}F(x_{0},y_{0})$? Something like "$D_{y}F(x_{0},y_{0})$ has constant rank in a neighborhood of $(x_{0},y_{0})$", for example.
Best Answer
Your example of a weaker condition will not work. Consider $F(x,y)=x-1$. Then $F_y(x,y)=0$ for all $(x,y)$ and so the Jacobian has constant rank (zero) for all $(x,y)$. Clearly, $F(x,y)=0$ does not define $y$ as a function of $x$ around any point.
A condition like $F_y(x,y)\neq 0$ for all $(x,y)\neq (x_0,y_0)$ in some neighbourhood of $(x_0,y_0)$ would also not be enough.
Examples of cases satisfying this condition but not having the desired properties are as follows:
$F(x,y)=x^2+y^2-1$ with $(x_0,y_0)=(1,0)$: The equation $F(x,y)=0$ does not give at least one value for $y$ for each $x$ in the neighbourhood of $(1,0)$, uniquely or otherwise. This is because the equation is not satisfied at $x>1$.
$F(x,y)=x^2-y^3$ with $(x_0,y_0)=(0,0)$: In this case, the equation does define $y$ as a function of $x$ around $(0,0)$. That function is given by $y(x)=x^{2/3}$. However, this function is not differentiable at $x=0$ (there is a cusp).