Inverse Function Theorem

The Inverse Function Theorem states that a continuously differentiable function with nonsingular Jacobian has a local inverse that is also continuously differentiable. This fundamental result ensures that smooth functions are locally invertible near non-degenerate points and provides a formula for the derivative of the inverse. Applications include coordinate transformations, differential geometry, and optimization.

Statement

Theorem10.1Inverse Function Theorem

Let $f : U \subseteq \mathbb{R}^n \to \mathbb{R}^n$ be continuously differentiable on an open set $U$ , and let $\mathbf{a} \in U$ . If the Jacobian matrix $Jf(\mathbf{a})$ is invertible (i.e., $\det Jf(\mathbf{a}) \neq 0$ ), then:

There exist open neighborhoods $V$ of $\mathbf{a}$ and $W$ of $f(\mathbf{a})$ such that $f : V \to W$ is a bijection.
The inverse $g = f^{-1} : W \to V$ is continuously differentiable.
For $\mathbf{y} \in W$ , $Jg(\mathbf{y}) = [Jf(g(\mathbf{y}))]^{-1}$ .

RemarkLocal invertibility

The theorem is local: it guarantees invertibility only in neighborhoods, not globally. For example, $f(x) = x^3$ is globally invertible, but $f(x) = x^2$ is only locally invertible away from $x = 0$ (where $f'(0) = 0$ ).

Examples

ExamplePolar to Cartesian coordinates

The map $f(r, \theta) = (r\cos\theta, r\sin\theta)$ has Jacobian

$Jf = \begin{pmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{pmatrix}.$

$\det Jf = r$ , so $Jf$ is invertible when $r \neq 0$ . Thus $f$ is locally invertible away from the origin, giving local polar coordinates.

ExampleOne-dimensional case

For $f : \mathbb{R} \to \mathbb{R}$ , the condition $Jf(a) \neq 0$ is just $f'(a) \neq 0$ . The theorem says: if $f'(a) \neq 0$ , then $f$ is locally invertible near $a$ with $(f^{-1})'(f(a)) = 1/f'(a)$ (the usual inverse derivative formula).

ExampleFailure when Jacobian is singular

Let $f(x, y) = (e^x \cos y, e^x \sin y)$ (complex exponential in real coordinates). At any point,

$\det Jf = e^{2x} > 0,$

so $f$ is locally invertible everywhere. However, $f$ is not globally injective: $f(x, y) = f(x, y + 2\pi)$ (periodicity in $y$ ).

Proof idea

RemarkProof via contraction mapping

The proof uses the Banach Fixed-Point Theorem. Given $\mathbf{y}$ near $f(\mathbf{a})$ , we seek $\mathbf{x}$ with $f(\mathbf{x}) = \mathbf{y}$ . Define $T(\mathbf{x}) = \mathbf{x} + [Jf(\mathbf{a})]^{-1}(\mathbf{y} - f(\mathbf{x}))$ . Then $T$ is a contraction near $\mathbf{a}$ (using that $Jf$ is continuous and invertible), so $T$ has a unique fixed point $\mathbf{x}$ with $f(\mathbf{x}) = \mathbf{y}$ . This gives the local inverse.

Applications

ExampleConstrained optimization

In optimization with constraints, the Inverse Function Theorem guarantees that constraint manifolds are locally parametrizable, enabling Lagrange multipliers and constraint qualification.

ExampleCoordinate charts in differential geometry

Smooth manifolds are defined by "gluing" open sets of $\mathbb{R}^n$ via coordinate charts. The Inverse Function Theorem ensures transition maps between charts are smooth.

Summary

The Inverse Function Theorem:

Nonsingular Jacobian $\Rightarrow$ local invertibility.
Inverse is smooth with derivative $(f^{-1})' = (f')^{-1}$ .
Proof uses Banach Fixed-Point Theorem.
Applications: coordinate transformations, manifolds, optimization.

See Inverse Function Theorem proof and Implicit Function Theorem.