Implicit function theorem for a surjective derivative

Statement

Let $U \subseteq \mathbb{R}^n$ be open, let $m \le n$, and let $F \colon U \to \mathbb{R}^m$ be of class $C^1$. Let $p \in U$ with $F(p) = 0$, and suppose that $DF(p) \colon \mathbb{R}^n \to \mathbb{R}^m$ is surjective. Then there are $m$ coordinates $x_{i_1}, \dots, x_{i_m}$ and an open neighbourhood $W$ of $p$ with this property. The set $\{x \in W : F(x) = 0\}$ is the graph of a $C^1$ function of the other $n - m$ coordinates.

Remarks

The derivative is surjective if and only if the Jacobian matrix $J_F(p)$ has rank $m$. Then some $m$ columns of $J_F(p)$ are linearly independent, and these columns choose the coordinates. After a renumbering of coordinates, the statement is the implicit function theorem with $y = (x_{i_1}, \dots, x_{i_m})$. If $DF(x)$ is surjective at each point of the level set $F^{-1}(0)$, this set is locally a graph near each of its points. So it is a $C^1$ submanifold of dimension $n - m$. Example: for $F(x, y, z) = x^2 + y^2 + z^2 - 1$, $DF$ is surjective on the unit sphere. So the sphere is locally the graph of a function of two coordinates.

Explanations

L1 human

The condition that the derivative is onto means the $m$ equations are independent at the point. Then $m$ of the variables are determined by the others, and the solution set has $n - m$ free directions.