Multivariable Inverse Function Theorem¶

Theorem¶

Let $A\subseteq\mathbb{R}^n$ be open, let $\boldsymbol{f}\colon A\to\mathbb{R}^n$ be $C^1$ , and let $\boldsymbol{x}_0\in A$ with $\boldsymbol{y}_0=\boldsymbol{f}(\boldsymbol{x}_0)$ $f (x_{0})$ . If

\det(\boldsymbol{D}\boldsymbol{f}(\boldsymbol{x}_0))\neq0,

then there exists an open neighbourhood $V\subseteq A$ of $\boldsymbol{x}_0$ such that $\boldsymbol{f}|_V\colon V\to \boldsymbol{f}(V)$ is bijective. The image $\boldsymbol{f}(V)$ is an open neighbourhood of $\boldsymbol{y}_0$ . Its inverse is $C^1$ and

\boldsymbol{D}(\boldsymbol{f}|_V^{-1})(\boldsymbol{y}_0)=\left(\boldsymbol{D}\boldsymbol{f}(\boldsymbol{x}_0)\right)^{-1}.

Interpretation¶

An invertible Jacobian makes the function locally reversible: if the linear approximation of $\boldsymbol{f}$ at $\boldsymbol{x}_0$ can be undone, so can $\boldsymbol{f}$ itself near $\boldsymbol{x}_0$ , and the inverse is again $C^1$ .

A square neighbourhood V of x0 mapped by f onto a curved region f(V) around y0 = f(x0), with an arrow back labelled as the inverse map.

The square $V$ around $\boldsymbol{x}_0$ is sent onto a curved region $\boldsymbol{f}(V)$ around $\boldsymbol{y}_0$ ; every point of that region comes from exactly one point of the square. The arrow back is the local inverse, defined only on the curved region.

The determinant condition is the square full-rank condition: $n$ equations in $n$ unknowns. The $C^1$ hypothesis keeps the Jacobian invertible near $\boldsymbol{x}_0$ ; differentiability alone is not enough. For $f(x)=x+2x^2\sin(1/x)$ with $f(0)=0$ $0$ , $f'(0)=1$ $1$ , yet $f'$ takes the values $-1$ and $3$ arbitrarily close to $0$ , so $f$ is not injective on any interval around $0$ . If $\det\boldsymbol{D}\boldsymbol{f}(\boldsymbol{x}_0)=0$ $0$ , the theorem is silent: $(x^2,y)$ has no local inverse at the origin, but $(x^3,y)$ has one.

Differentiating $\boldsymbol{f}|_V^{-1}(\boldsymbol{f}(\boldsymbol{x}))=\boldsymbol{x}$ with the multivariable chain rule gives $\boldsymbol{D}(\boldsymbol{f}|_V^{-1})(\boldsymbol{y}_0)\,\boldsymbol{D}\boldsymbol{f}(\boldsymbol{x}_0)=\boldsymbol{I}$ $I$ , the derivative formula of the Theorem. For $n=1$ it is the rule $(f^{-1})'(y)=1/f'(x)$ of the one-variable inverse function theorem.

The conclusion is local bijectivity of a restricted map, not a global inverse on the original domain. For $(\mathrm{e}^x\cos y,\mathrm{e}^x\sin y)$ from the Examples the determinant $\mathrm{e}^{2x}$ is never $0$ , yet $(x,y)$ and $(x,y+2\pi)$ have the same image.

Key Results¶

The determinant condition is the square full-rank condition.
The derivative of the local inverse is the inverse matrix of the original derivative.
The theorem can be derived from the implicit function theorem and vice versa.

Examples¶

For $\boldsymbol{f}(x,y)=(\mathrm{e}^x\cos y,\mathrm{e}^x\sin y)$ $(e^{x} cos y, e^{x} sin y)$ , the determinant is $\mathrm{e}^{2x}$ , so the map is locally invertible everywhere.
For $\boldsymbol{f}(x,y)=(x^2,y)$ at $(0,0)$ , the determinant is $0$ , so the theorem does not apply there.

Prerequisites¶

Sources¶

Math IV: Multivariate Calculus · Week 3, Lecture 1

Aliases: inverse function theorem in R^n, inverse function theorem in Rn Keywords: local inverse theorem, Jacobian inverse theorem Symbols: element of, pi, real numbers, subset

Last updated: 20 October 2026 03:02