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 on $U$. Let $a \in U$, and suppose that the derivative $Df(a)$ is invertible, that is, the Jacobian determinant 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.

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 $y$.