MAT257Y1/Unit 6
Unit 6 · Analysis II

The Inverse Function Theorem

If the linear approximation to $f$ at $a$ is invertible, then $f$ itself is invertible near $a$. That sentence is the theorem, and it is the first place in the course where a statement about the derivative — an infinitesimal object — buys you a statement about $f$ on an actual neighbourhood.

By the end you can
Why it matters & where it comes from
Why it exists

Solving $f(x)=y$ for $x$ is impossible in closed form for almost every $f$. The theorem says you do not need a formula: an invertible derivative is already a guarantee that a solution exists, is unique nearby, and depends smoothly on $y$.

In the real world

Newton's method is the proof turned into an algorithm, and its quadratic convergence comes from the same estimate. Coordinate changes in mechanics, the change of variables formula in Unit 10, and the chart maps of Unit 13 all invoke it.

Story & open questions

Banach's fixed point theorem (1922) is what made the modern proof short, and the same argument runs unchanged in Banach spaces, giving the infinite-dimensional version used for differential equations. Whether a global version can hold under a determinant condition alone is the still-open Jacobian conjecture in the polynomial setting.

The contraction mapping principle

Theorem Banach fixed point

Let $X$ be a nonempty closed subset of $\mathbb{R}^n$ and $T:X\to X$ satisfy $\|T(x)-T(y)\|\le c\|x-y\|$ for a constant $c<1$. Then $T$ has exactly one fixed point in $X$.

Pick any $x_0$ and iterate $x_{k+1}=T(x_k)$. Then $\|x_{k+1}-x_k\|\le c^k\|x_1-x_0\|$, so the tail sums are dominated by a geometric series and $(x_k)$ is Cauchy.
$\mathbb{R}^n$ is complete and $X$ is closed, so $x_k\to x\in X$. Contractions are continuous, so $T(x)=\lim T(x_k)=\lim x_{k+1}=x$.
If $T(x)=x$ and $T(y)=y$ then $\|x-y\|\le c\|x-y\|$, and $c<1$ forces $\|x-y\|=0$. ∎

Completeness is doing the real work. This is why the theorem lives in analysis and not in algebra.

Rewriting an equation as a fixed point is the move. Solving $f(x)=y$ becomes finding a fixed point of $T(x)=x+Df(a)^{-1}(y-f(x))$: the correction term vanishes exactly when $f(x)=y$. Everything after this is showing $T$ contracts.

Statement and proof

Theorem Inverse function theorem

Let $f:A\to\mathbb{R}^n$ be $C^1$ on an open $A\subseteq\mathbb{R}^n$, and suppose $Df(a)$ is invertible. Then there are open sets $U$ containing $a$ and $V$ containing $f(a)$ such that $f:U\to V$ is a bijection, the inverse $f^{-1}:V\to U$ is $C^1$, and

$D(f^{-1})(y)=\big(Df(f^{-1}(y))\big)^{-1}$ for $y\in V$.

Note the domain and codomain have the same dimension. Nothing here applies to $f:\mathbb{R}^2\to\mathbb{R}^3$.

Proof sketch Four steps
Normalise. Replacing $f$ by $Df(a)^{-1}\circ(f-f(a))$ and translating, assume $a=0$, $f(0)=0$, and $Df(0)=I$. Neither operation changes invertibility.
Contract. Let $g(x)=x-f(x)$, so $Dg(0)=0$. Since $f$ is $C^1$, $\|Dg\|\le\tfrac12$ on some closed ball $\bar B_r$, and the mean value inequality gives $\|g(x)-g(x')\|\le\tfrac12\|x-x'\|$ there.
Solve. For $\|y\|<r/2$ the map $T_y(x)=y+g(x)$ sends $\bar B_r$ into itself and contracts, so it has a unique fixed point — and $T_y(x)=x$ says exactly $f(x)=y$. This gives a well-defined inverse on $V=B_{r/2}$, with $U=f^{-1}(V)\cap B_r$ open by continuity.
Differentiate. The contraction estimate also gives $\|f(x)-f(x')\|\ge\tfrac12\|x-x'\|$, so $f^{-1}$ is Lipschitz. Feeding that into the definition of the derivative shows $f^{-1}$ is differentiable with the stated derivative, and the formula shows the derivative is continuous because matrix inversion is. ∎
Example Polar coordinates

$g(r,\theta)=(r\cos\theta,r\sin\theta)$ has $\det Jg=r$, nonzero whenever $r>0$. So $g$ is locally invertible away from the origin. It is not globally injective — $\theta$ and $\theta+2\pi$ collide — which is precisely the gap between local and global that the theorem does not close.

Example Local but never global

$f(x,y)=(e^x\cos y,\,e^x\sin y)$ has $\det Jf=e^{2x}>0$ everywhere, so it is locally invertible at every point of $\mathbb{R}^2$, yet it is $2\pi$-periodic in $y$ and therefore infinitely-to-one. This is the complex exponential, and it is the standard demonstration that "invertible derivative everywhere" and "invertible" are different claims.

What each hypothesis is buying

Theorem $C^1$ cannot be dropped

$f(x)=x+2x^2\sin(1/x)$ with $f(0)=0$ has $f'(0)=1\ne0$, but $f'$ is discontinuous at $0$ and takes negative values in every neighbourhood. So $f$ is not monotone near $0$ and has no local inverse there.

Differentiability at the single point $a$ is not enough. The proof used control of $Df$ on a whole ball, and this example shows that use was not laziness.

Theorem Open mapping

If $f$ is $C^1$ on $A$ and $Df(x)$ is invertible for every $x\in A$, then $f(A)$ is open, and more generally $f$ maps open subsets of $A$ to open sets. Apply the theorem at each point: every image point has an open $V$ around it inside the image.

Local means local. The theorem produces a neighbourhood and says nothing about its size. When you need a quantitative radius you must reread the proof and track the constants; the statement alone will not give you one.
Common pitfalls

• Concluding $f$ is injective on all of $A$ from an everywhere-nonzero Jacobian determinant. The complex exponential above refutes it.
• Using the theorem when domain and codomain have different dimensions. That case is the rank theorem, not this one.
• Checking only that $Df(a)\ne0$. The condition is that $Df(a)$ is invertible, that is $\det Jf(a)\ne0$.
• Assuming differentiability at $a$ suffices. The hypothesis is $C^1$ on a neighbourhood.
• Forgetting the inverse inherits smoothness: if $f$ is $C^k$ so is $f^{-1}$, which follows by induction on the formula for $D(f^{-1})$.

Practice

Original problems written for this guide in the style of Spivak and Munkres exercises. Write a full solution before opening the answer.

Practice 1 · Where it applies

For $f(x,y)=(x^2-y^2,\,2xy)$, at which points does the theorem apply?

Solution
$\det Jf=4(x^2+y^2)$, nonzero except at the origin. So $f$ is locally invertible everywhere but the origin. Globally it is two-to-one, since $f(-x,-y)=f(x,y)$: this is $z\mapsto z^2$.
Practice 2 · Derivative of an inverse

With the same $f$, compute $D(f^{-1})$ at $f(1,1)=(0,2)$.

Solution
$Jf(1,1)=\begin{smallmatrix}2&-2\\2&2\end{smallmatrix}$ with determinant $8$, so the inverse is $\tfrac18\begin{smallmatrix}2&2\\-2&2\end{smallmatrix}$. No formula for $f^{-1}$ was needed.
Practice 3 · A contraction

Show $T(x)=\tfrac12\cos x$ has a unique fixed point on $\mathbb{R}$.

Solution
$|T'(x)|=\tfrac12|\sin x|\le\tfrac12$, so $T$ is a contraction with $c=\tfrac12$ by the mean value theorem, and $\mathbb{R}$ is closed and complete. One fixed point, and iterating from any start converges to it.
Practice 4 · Newton's method

Explain why $x_{k+1}=x_k-Df(x_k)^{-1}f(x_k)$ is the proof made computational.

Solution
It is fixed-point iteration for $T(x)=x-Df(x)^{-1}f(x)$, whose fixed points are the roots of $f$. Freezing the Jacobian at $a$ gives the map used in the proof; updating it each step is what upgrades linear convergence to quadratic.
Practice 5 · Determinant zero

Does $f(x)=x^3$ on $\mathbb{R}$ contradict the theorem at $0$?

Solution
No. $f'(0)=0$, so the hypothesis fails; the theorem simply does not apply. That $f$ happens to be a bijection anyway shows the condition is sufficient, not necessary.
Practice 6 · Smoothness transfers

If $f$ is $C^\infty$ with invertible derivative, is $f^{-1}$ smooth?

Solution
Yes. From $D(f^{-1})(y)=Jf(f^{-1}(y))^{-1}$, the right side is a composition of smooth maps whenever $f^{-1}$ is $C^k$, so $f^{-1}$ is $C^{k+1}$. Induct from $k=1$.
Practice 7 · Spherical coordinates

Compute $\det Jg$ for $g(\rho,\varphi,\theta)=(\rho\sin\varphi\cos\theta,\rho\sin\varphi\sin\theta,\rho\cos\varphi)$ and say where $g$ is a local diffeomorphism.

Solution
$\det Jg=\rho^2\sin\varphi$. It is nonzero when $\rho>0$ and $\varphi\notin\{0,\pi\}$, so $g$ is a local diffeomorphism away from the polar axis. That determinant is the $\rho^2\sin\varphi$ factor in every spherical integral.
Practice 8 · Injective on a compact set

If $f$ is $C^1$ and injective on a compact $K$ with invertible derivative throughout, must $f^{-1}$ be continuous on $f(K)$?

Solution
Yes, and compactness alone gives it: a continuous bijection from a compact space to a Hausdorff space is a homeomorphism, because closed subsets of $K$ are compact and their images are compact hence closed.

Expanded topic checklist

Course coverage. This unit is part of the complete MAT257Y1 topic map. It should cover the following items before you move on:

Textbook/source alignment

Spivak proves the theorem with a compactness argument rather than a contraction; Munkres uses the fixed point route given here. Both are worth reading, and the syllabus decides which one is examinable.

Core theorem map

Completeness gives the contraction principle, which gives the inverse function theorem, which gives the implicit function theorem in Unit 7, which gives the manifold charts in Unit 13. This is the longest single chain of dependence in the course.

Study routine. Be able to reproduce the reduction to $Df(0)=I$ and the definition of $T_y$ from memory. Everything else in the proof is estimate-chasing that follows once those two lines are on the page.

Deep practice extensions

Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.

Extension 1 · One-sentence statement

State the theorem in one sentence.

Guide
A $C^1$ map with invertible derivative at a point is a $C^1$ bijection between some neighbourhood of that point and some neighbourhood of its image.
Extension 2 · The contraction in disguise

Which map is contracted in the proof, and why does its fixed point solve the equation?

Guide
$T_y(x)=x+Df(a)^{-1}(y-f(x))$. Its fixed points are exactly the $x$ with $f(x)=y$, since $Df(a)^{-1}$ is injective.
Extension 3 · Failure of globality

Give the standard map with invertible derivative everywhere that is not injective.

Guide
$(x,y)\mapsto(e^x\cos y,e^x\sin y)$, periodic in $y$.
Extension 4 · Dimension mismatch

What replaces the theorem when the derivative has constant rank but is not square?

Guide
The rank theorem: near such a point $f$ looks, in suitable local coordinates on both sides, like a linear projection followed by an inclusion.
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