The Inverse Function Theorem

Core Idea

Let f map an open set in ℝⁿ to ℝⁿ, be continuously differentiable, and have an invertible derivative matrix Df(a) at a point a. The inverse function theorem says that f is then invertible near a: there are open sets V containing a and W containing f(a) such that f maps V one-to-one onto W, and the inverse is also continuously differentiable. Its derivative is the inverse matrix, D(f⁻¹)(f(x)) = [Df(x)]⁻¹. The theorem turns a linear-algebra condition at one point (det Df(a) ≠ 0) into a nonlinear conclusion on a neighborhood.

How It's Best Learned

Start in one variable: a C¹ function with f'(a) ≠ 0 is strictly monotone near a, so it has a local inverse with derivative 1/f'. Then look at maps of the plane, such as polar coordinates, and compute where the Jacobian determinant vanishes. Use the complex exponential map (x, y) ↦ (eˣ cos y, eˣ sin y) to show the result is local. Present the proof as solving f(x) = y by a contraction mapping.

Common Misconceptions

• If det Df(x) ≠ 0 everywhere, f is one-to-one on its whole domain. (The map (x, y) ↦ (eˣ cos y, eˣ sin y) has Jacobian determinant e²ˣ > 0 everywhere but repeats its values when y increases by 2π.)
• If Df(a) is singular, f cannot be one-to-one near a. (f(x) = x³ is one-to-one on ℝ although f'(0) = 0; what fails is differentiability of the inverse at 0.)
• Differentiability of f at a is enough. (The theorem needs Df to be continuous near a; there are differentiable functions with f'(0) ≠ 0 that are not one-to-one on any interval around 0.)

Explainer

In one variable, if f is continuously differentiable and f'(a) ≠ 0, then f' keeps the same sign near a. So f is strictly increasing or strictly decreasing there and has an inverse function nearby, with (f⁻¹)'(f(a)) = 1/f'(a). The inverse function theorem extends this to maps from ℝⁿ to ℝⁿ, where "f'(a) ≠ 0" becomes "the matrix Df(a) is invertible."

Theorem. Let U ⊆ ℝⁿ be open, f: U → ℝⁿ continuously differentiable, and a ∈ U with det Df(a) ≠ 0. Then there are open sets V ∋ a and W ∋ f(a) such that f: V → W is a bijection, its inverse g: W → V is continuously differentiable, and Dg(f(x)) = [Df(x)]⁻¹ for x in V.

The idea is linear approximation. Near a, f(x) ≈ f(a) + Df(a)(x - a). If Df(a) is invertible, the linear part is a bijection, and the theorem says the small error in the approximation cannot destroy that near a. To make this precise, fix y near f(a) and try to solve f(x) = y. Rewrite the equation as a fixed-point problem: x = x + Df(a)⁻¹(y - f(x)). The map on the right has derivative I - Df(a)⁻¹Df(x), which is small for x near a because Df is continuous. So the map is a contraction on a small closed ball, and the contraction mapping theorem gives exactly one solution x in that ball. This produces the local inverse. A further estimate shows it is differentiable, and the chain rule applied to g(f(x)) = x gives the formula for Dg.

The conclusion is local, and that cannot be improved. The map f(x, y) = (eˣ cos y, eˣ sin y) has Jacobian determinant e²ˣ, which is never zero, so it is locally invertible everywhere. Yet it is not one-to-one, because adding 2π to y gives the same output. This is the complex exponential function in real coordinates.

The hypothesis is sufficient but not necessary for a one-to-one map. The function x³ is a bijection of ℝ with derivative 0 at the origin, but its inverse fails to be differentiable there. Invertibility of Df(a) is what makes the inverse differentiable.

The implicit function theorem, which says when F(x, y) = 0 defines y as a function of x, follows by applying this theorem to the map (x, y) ↦ (x, F(x, y)). Together they are the basic tools for working with smooth manifolds.