---
tag: 0K9HSJFCMJ
kind: statement
role: theorem
name: Inverse function theorem
aliases: [inverse-mapping-theorem]
msc: [26B10]
uses: [0TCBZMEYGN, 03W6Q1MFR5, 08EVNYZF42, 0G65751VSW]
sources:
  - url: https://en.wikipedia.org/wiki/Inverse_function_theorem
    locator: Several variables
  - url: https://en.wikipedia.org/wiki/Inverse_function_theorem
    locator: A proof using the contraction mapping principle
same_as:
  wikidata: Q931001
  wikipedia: Inverse function theorem
---

## Statement

Let $U \subseteq \mathbb{R}^n$ be open, let $k \ge 1$, and let $f \colon U \to \mathbb{R}^n$ be of [[class C^k|0TCBZMEYGN]] on $U$.
Let $a \in U$, and suppose that the derivative $Df(a)$ is invertible, that is, the [[Jacobian determinant|03W6Q1MFR5]] of $f$ at $a$ is not zero.
Then there are open sets $V \subseteq U$ with $a \in V$ and $W \subseteq \mathbb{R}^n$ with $f(a) \in W$ such that $f$ restricted to $V$ is a bijection from $V$ onto $W$.
Its inverse $f^{-1} \colon W \to V$ is of class $C^k$, and $D(f^{-1})(f(x)) = Df(x)^{-1}$ for each $x \in V$.

## Remarks

The theorem is local. The map $f(x, y) = (e^x \cos y, e^x \sin y)$ has invertible derivative at each point of $\mathbb{R}^2$, but it is not injective, since $f(x, y + 2\pi) = f(x, y)$.
Continuity of the derivative matters. The function $g(x) = x + 2 x^2 \sin(1/x)$, $g(0) = 0$, has $g'(0) = 1$, but $g$ is not injective on any interval around $0$.
A standard proof applies the contraction mapping principle to $x \mapsto x + Df(a)^{-1}(y - f(x))$ near $a$.
The formula for the derivative of the inverse is the [[derivative of the inverse function|0G65751VSW]].

## Explanations

### L1 human

If the linear approximation of $f$ at $a$ is invertible, then $f$ itself is invertible near $a$.
So one can solve the system $f(x) = y$ for $x$ when $y$ is close to $f(a)$, and the solution depends smoothly on $y$.