Mathematics

Brouwer's Fixed-Point Theorem: A Proof via the Fundamental Group

We prove Brouwer's Fixed-Point Theorem for the closed disk using the fundamental group and the non-existence of a retraction, deriving an algebraic contradiction.

Topology

The Theorem

Theorem (Brouwer, 1911)

Every continuous function $f: D^2 \to D^2$ has at least one fixed point $x_0$ such that: $\\f(x_0) = x_0$

Here $D^2 = \{(x,y) \in \mathbb{R}^2 : x^2 + y^2 \leq 1\}$ is the closed unit disk and $S^1 = \partial D^2$ is its boundary circle.

Intuition

Before the formal proof, here are two everyday examples that capture the idea.

The coffee cup. Stir a cup of coffee, then let it settle. No matter how you stirred, at least one point of the liquid ends up exactly where it started.

The map. Place a map of a city on a table inside that same city. There is always one point on the map that lies directly above the real location it represents.

Both examples reflect the same fact: you cannot continuously move every point of a compact convex region — something always stays fixed.

Background

The Fundamental Group

The fundamental group $\pi_1(X)$ classifies loops in a space $X$ up to continuous deformation. A loop based at $x_0$ is a continuous map

$\gamma: [0,1] \to X \quad \text{with} \quad \gamma(0) = \gamma(1) = x_0$

Two loops $\gamma_0$ and $\gamma_1$ are homotopic (written $\gamma_0 \simeq \gamma_1$ ) if one can be continuously deformed into the other while keeping the basepoint fixed.

The Two Groups We Need

Disk $D^2$

$\pi_1(D^2) = 0$

Every loop shrinks to a point via the straight-line homotopy

$H(t,s) = (1-s)\,\gamma(t) + s \cdot x_0$

Circle $S^1$

$\pi_1(S^1) = \mathbb{Z}$

Loops are classified by their winding number: how many times they wrap around the circle.

Key Lemma

Lemma. Let $f_1, f_2 : X \to Y$ and $g_1, g_2 : Y \to Z$ be continuous maps. If $f_1 \simeq f_2$ and $g_1 \simeq g_2$ , then $\\g_1 \circ f_1 \simeq g_2 \circ f_2$

Homotopy is preserved under composition.

Proof. Let $H_f$ be the homotopy from $f_1$ to $f_2$ and $H_g$ from $g_1$ to $g_2$ . Define

$H(t,s) = H_g(H_f(t,s),\, s)$

This is a valid homotopy from $g_1 \circ f_1$ to $g_2 \circ f_2$ . $\square$

Proof

Proof (by contradiction)

Step 1 — Assume no fixed point.

Suppose $f: D^2 \to D^2$ is continuous and satisfies $f(x) \neq x$ for all $x \in D^2$ .

Step 2 — Construct a retraction $r: D^2 \to S^1$ .

Since $f(x) \neq x$ everywhere, we can draw a ray from $f(x)$ through $x$ and extend it until it hits the boundary $S^1$ . Define $r(x)$ to be that intersection point.

The map $r$ satisfies:

$r$ is continuous.
$r(x) = x$ for all $x \in S^1$ .

This makes $r$ a retraction of $D^2$ onto $S^1$ .

Step 3 — Apply the Key Lemma.

Let $h_0, h_1 : [0,1] \to D^2$ be any two paths inside $D^2$ . Since $\pi_1(D^2) = 0$ , any two paths with the same endpoints are homotopic:

$h_0 \simeq h_1 \quad \text{in } D^2$

Composing with $r$ and applying the Key Lemma:

$r \circ h_0 \simeq r \circ h_1 \quad \text{in } S^1$

Step 4 — Reach a contradiction.

Choose paths carefully:

$h_0(t) = x_0$ constant $\Rightarrow$ $r \circ h_0$ is a single point, winding number $= 0$
$h_1(t) = (\cos 2\pi t,\, \sin 2\pi t)$ loops around $S^1$ once, winding number $= 1$

But $r \circ h_0 \simeq r \circ h_1$ forces equal winding numbers:

$0 = 1 \quad \text{in } \pi_1(S^1) = \mathbb{Z}$

This is impossible. $\square$

The Algebraic Picture

The contradiction has a clean algebraic form. The retraction $r$ gives the chain:

$S^1 \xrightarrow{\;i\;} D^2 \xrightarrow{\;r\;} S^1$

where $i$ is the inclusion and $r \circ i = \mathrm{id}_{S^1}$ . Applying $\pi_1$ :

$\mathbb{Z} \xrightarrow{\;i_*\;} 0 \xrightarrow{\;r_*\;} \mathbb{Z}$

The composition $r_* \circ i_*$ must equal $\mathrm{id}_\mathbb{Z}$ , but any map factoring through $0$ is the zero map:

$1 \;\longmapsto\; 0 \;\longmapsto\; 0 \;\neq\; 1$

Why Every Condition Is Necessary

Without continuity. The map $f: [0,1] \to [0,1]$ with $f(x) = 1$ for $x < \tfrac{1}{2}$ and $f(x) = 0$ for $x \geq \tfrac{1}{2}$ has no fixed point.

Without compactness. The map $f: (0,1) \to (0,1)$ with $f(x) = x/2$ has no fixed point since $x = 0 \notin (0,1)$ .

Without convexity. The map $f: S^1 \to S^1$ with $f(z) = -z$ has no fixed point since $-z \neq z$ for all $z \in S^1$ .

The General Theorem

General Brouwer Fixed-Point Theorem.

Let $K \subset \mathbb{R}^n$ be a nonempty compact convex set. Every continuous function $f: K \to K$ has at least one fixed point.

The proof is identical in every dimension: build a retraction $r: B^n \to S^{n-1}$ and derive

$\pi_{n-1}(S^{n-1}) = \mathbb{Z} \xrightarrow{} \pi_{n-1}(B^n) = 0 \xrightarrow{} \mathbb{Z}$

which forces $1 = 0$ in $\mathbb{Z}$ .

Applications

Economics. Arrow and Debreu used Brouwer's theorem to prove that competitive markets always have a price equilibrium — work that earned them the Nobel Prize in Economics.

Game Theory. Nash used a fixed-point theorem to prove every finite game has at least one Nash equilibrium, the foundation of modern game theory.

Differential Equations. The theorem guarantees existence of solutions to certain ODEs via the Peano existence theorem.

Proof at a Glance

\begin{aligned} &\textbf{Assume } f: D^2 \to D^2 \text{ has no fixed point} \\[6pt] &\Downarrow \\[6pt] &\textbf{Build retraction } r: D^2 \to S^1,\quad r|_{S^1} = \mathrm{id} \\[6pt] &\Downarrow \\[6pt] &\pi_1(D^2) = 0 \implies \text{all paths in } D^2 \text{ are homotopic} \\[6pt] &\Downarrow \text{ Key Lemma} \\[6pt] &r \circ h_0 \simeq r \circ h_1 \text{ in } S^1 \\[6pt] &\Downarrow \\[6pt] &0 = 1 \text{ in } \mathbb{Z} \quad \textbf{Contradiction!} \\[6pt] &\Downarrow \\[6pt] &\therefore\ f \text{ must have a fixed point.} \quad \square \end{aligned}

References

Hatcher, A., Algebraic Topology, Cambridge University Press, 2002. Free PDF
Munkres, J., Topology, 2nd edition, Prentice Hall, 2000.
Wikipedia — Brouwer Fixed-Point Theorem
3Blue1Brown — YouTube
MIT OpenCourseWare — Algebraic Topology
MIT OpenCourseWare — Algebraic Topology

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