The Inverse Function Theorem

The Inverse Function Theorem (IFT) is the single most useful theorem in multivariable analysis. It answers a fundamental question: given a smooth map $f:\\mathbb{R}^n\\to\\mathbb{R}^n$, can we solve $f(\\mathbf{x})=\\mathbf{y}$ for $\\mathbf{x}$ as a smooth function of $\\mathbf{y}$? The answer hinges on one condition: is the Jacobian $Df$ invertible at the point? If yes, the function is locally invertible too. The IFT formalizes a deep intuition: a smooth map behaves locally like its linear approximation. If the linear approximation is invertible, the full nonlinear map is invertible in a neighbourhood.

The statement

Let $f:\\mathbb{R}^n\\to\\mathbb{R}^n$ be continuously differentiable ($C^1$) on an open set containing $\\mathbf{a}$. If $\\det Df(\\mathbf{a})\\neq 0$, then:

• There exist open sets $U$ containing $\\mathbf{a}$ and $V$ containing $f(\\mathbf{a})$ such that $f:U\\to V$ is a bijection.
• The inverse $f^{-1}:V\\to U$ is continuously differentiable.
• Its derivative is the matrix inverse of $Df$: D(f^{-1})(f(\\mathbf{a}))=\[Df(\\mathbf{a})\]^{-1}.

The condition $\\det Df(\\mathbf{a})\\neq 0$ is exactly the condition that the linear approximation $Df(\\mathbf{a})$ has a matrix inverse. The theorem says: if the linear approximation is invertible, the nonlinear function is locally invertible too.

Why "local" matters

The IFT only gives a local inverse. A standard example: $f(x)=x^2$ has $f'(1)=2\\neq 0$, so by the 1D IFT, $f$ is locally invertible near $x=1$. But $f(-1)=f(1)=1$, so $f$ is not globally injective. Locally near $x=1$ the inverse is $y\\mapsto\\sqrt{y}$; you cannot extend this to a global inverse on all of $\\mathbb{R}$. The IFT promises a neighbourhood, not the whole space.

The mapping-arrows widget above shows how each input gets sent to its output. The IFT says: if you zoom in enough around $\\mathbf{a}$, the arrows form a tidy bijection between a small input region and a small output region, even if globally the map sends multiple inputs to the same output. Local invertibility is a microscope-level property.

• Robotics, inverse kinematics: The forward-kinematics map $\\mathbf{p}=f(\\boldsymbol{\\theta})$ sends joint angles to end-effector positions. To plan a motion you need the inverse: given a desired $\\mathbf{p}$, find joint angles $\\boldsymbol{\\theta}$. The IFT guarantees the local inverse exists wherever the Jacobian $J(\\boldsymbol{\\theta})$ is non-singular. Singular configurations ("gimbal lock") are exactly where $\\det J=0$, control algorithms must avoid these.
• Continuous optimization, Newton's method: Each Newton step solves a linearized inverse problem: given the residual $f(\\mathbf{x}\_k)$, find a small step $\\Delta\\mathbf{x}$ such that the linearization $f(\\mathbf{x}\_k)+Df(\\mathbf{x}\_k)\\Delta\\mathbf{x}=0$. This is exactly inverting $Df$. The IFT guarantees Newton converges quadratically when $Df$ is non-singular at the root.
• Differential geometry, smooth manifolds: The IFT is the proof engine behind the manifold concept itself. A smooth manifold is one where every point has a neighbourhood that is the image of a Jacobian-invertible map from $\\mathbb{R}^n$, the IFT translates "non-singular Jacobian" into "looks like Euclidean space locally."

Pause and think: The map $f(x,y)=(e^x\\cos y,\\ e^x\\sin y)$ has $\\det Df=e^{2x}>0$ everywhere. So the IFT applies at every point. Yet $f$ is NOT globally injective: $f(0,0)=f(0,2\\pi)=(1,0)$. How do you reconcile this? (Answer: local invertibility is everywhere; global invertibility fails because the map wraps around in $y$.)

Try it

• Predict first: for which points $(x,y)$ does $f(x,y)=(x+y^2,\\ y+x^2)$ satisfy the IFT? Compute $\\det Df=1-4xy$; the IFT applies wherever $4xy\\neq 1$.
• The map $f(x)=x+\\sin(x)/2$ has $f'(x)=1+\\cos(x)/2>0$ everywhere. Argue (using the IFT) that $f$ is globally invertible.
• Find all points where $f(x,y)=(\\sin x\\cos y,\\ \\sin x\\sin y)$ fails the IFT condition. (Hint: compute the Jacobian determinant and look for zeros.)
• If Df(\\mathbf{a})=\\begin{pmatrix}2&1\\\\1&3\\end{pmatrix}, compute $D(f^{-1})(f(\\mathbf{a}))$.
• Trap: write down a function where the IFT condition holds everywhere but the function is not globally injective. (The map $f(x,y)=(e^x\\cos y,e^x\\sin y)$ above works.)

A trap to watch for

The IFT requires $f$ to be continuously differentiable, not just differentiable. A function whose partials exist but are discontinuous can fail to be locally invertible even with non-zero Jacobian determinant at a point. In practice you almost always work with smooth (C^\\infty) functions, so this technicality rarely bites, but it is the reason the standard statement of the theorem includes the $C^1$ hypothesis explicitly.

What you now know

You can apply the IFT in 1D, 2D, and higher: compute the Jacobian, evaluate its determinant at the point of interest, and conclude local invertibility from non-singularity. The next section presents the Implicit Function Theorem, the closely related answer to "when does $F(\\mathbf{x},\\mathbf{y})=0$ define $\\mathbf{y}$ as a function of $\\mathbf{x}$?"

Mark section complete →

References

• Garrity, T. (2002). All the Mathematics You Missed. Cambridge UP, ch. 3.
• Spivak, M. (1965). Calculus on Manifolds. W. A. Benjamin, ch. 2.
• Munkres, J. R. (1991). Analysis on Manifolds. Westview Press, ch. 2 and 3.
• Rudin, W. (1976). Principles of Mathematical Analysis (3rd ed.). McGraw-Hill, ch. 9 (Theorem 9.24).
• Apostol, T. M. (1974). Mathematical Analysis (2nd ed.). Addison-Wesley, ch. 13.