---
tag: 2MCN92CX58
kind: proof
name: Proof of the implicit function theorem for a surjective derivative
msc: ["26B10"]
proves: 0NE2W2XPQJ
uses: [0BXBADR94C, 0K9HSJFCMJ, 0YF0VXKXGB, 0EWQEYRVA8, 0EVHJFTG9Z, 0EAAS2EHDR]
sources:
  - url: https://en.wikipedia.org/wiki/Implicit_function_theorem
    locator: Statement of the theorem
---

## Proof

We choose $m$ coordinates in which the derivative is invertible, and apply the implicit function theorem.

1. Let $U$, $m \le n$, $F$ and $p$ be as in the statement. Let $J = J_F(p)$ be the $m \times n$ [[Jacobian matrix|0YF0VXKXGB]], so $DF(p) v = J v$.
2. The image of $DF(p)$ is the column space of $J$. Since $DF(p)$ is surjective, the column space is $\R^m$, so the [[rank|0EWQEYRVA8]] of $J$ is $m$.
3. A spanning set of $\R^m$ contains a basis. So some $m$ columns of $J$, with indices $i_1 < \dots < i_m$, are [[linearly independent|0EVHJFTG9Z]]. The $m \times m$ matrix $C$ of these columns is invertible.
4. Case $m = n$. Then $C = J$, and $DF(p)$ is invertible. By the [[inverse function theorem|0K9HSJFCMJ]], $F$ is injective on an open neighbourhood $W$ of $p$. So $\set{x \in W : F(x) = 0} = \set{p}$. This is the graph of the constant function of the empty list of remaining coordinates.
5. Case $m < n$. For $x \in \R^n$, let $x'' = (x_{i_1}, \dots, x_{i_m}) \in \R^m$, and let $x' \in \R^{n-m}$ list the other coordinates in their order. Let $P(x) = (x', x'')$.
6. The map $P$ is a linear bijection that only reorders coordinates, so it preserves norms. Hence $\Omega = P(U)$ is open, and $G = F \circ P^{-1} \colon \Omega \to \R^m$ is of class $C^1$.
7. Write $P(p) = (p', p'')$. By the [[chain rule|0EAAS2EHDR]], the partial derivatives of $G$ with respect to the coordinates of $x''$ at $(p', p'')$ are the columns $i_1, \dots, i_m$ of $J$. So $D_{x''} G(p', p'') = C$, which is invertible.
8. Also $G(p', p'') = F(p) = 0$. By the [[implicit function theorem|0BXBADR94C]], there are open sets $U' \ni p'$ in $\R^{n-m}$ and $V' \ni p''$ in $\R^m$ and a $C^1$ function $\varphi \colon U' \to V'$ with this property. For $(x', x'') \in U' \times V'$, $G(x', x'') = 0$ if and only if $x'' = \varphi(x')$.
9. The function $G$ is evaluated on $U' \times V'$ in this statement, so $U' \times V' \subseteq \Omega$. Let $W = P^{-1}(U' \times V')$. It is open, since $P$ is continuous, and $p \in W \subseteq U$.
10. Let $x \in W$. Then $(x', x'') = P(x) \in U' \times V'$ and $F(x) = G(x', x'')$. So $F(x) = 0$ if and only if $x'' = \varphi(x')$.
11. Hence $\set{x \in W : F(x) = 0}$ is the set of $x \in \R^n$ with $x' \in U'$ and $x'' = \varphi(x')$. This is the graph of the $C^1$ function $\varphi$ of the $n - m$ coordinates in $x'$.
12. So the coordinates $x_{i_1}, \dots, x_{i_m}$ and the neighbourhood $W$ have the property in the statement.

## Idea

Surjectivity of $DF(p)$ means that the Jacobian matrix has $m$ independent columns. The coordinates of these columns play the role of $y$ in the implicit function theorem, and the remaining coordinates play the role of $x$. After this renumbering, the statement is the implicit function theorem.