Brouwer Fixed-Point Theorem (Dimension 2)

The Brouwer fixed-point theorem states that every continuous map from the closed disk to itself has a fixed point. The two-dimensional case can be proved using the fundamental group, making it one of the first applications of algebraic topology to a concrete analytical question.

Statement

Theorem9.11Brouwer Fixed-Point Theorem (2D)

Every continuous map $f: D^2 \to D^2$ has a fixed point: there exists $x \in D^2$ with $f(x) = x$ .

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

Proof via the Fundamental Group

Proof

The proof proceeds by contradiction, using two key lemmas.

Lemma 1 (No Retraction): There is no retraction $r: D^2 \to S^1$ , i.e., no continuous map $r$ with $r(x) = x$ for all $x \in S^1$ .

Proof of Lemma 1: Suppose $r: D^2 \to S^1$ is a retraction. Let $\iota: S^1 \hookrightarrow D^2$ be the inclusion. Then $r \circ \iota = \operatorname{id}_{S^1}$ , so $(r \circ \iota)_* = r_* \circ \iota_* = \operatorname{id}_{\pi_1(S^1)}$ .

Now $\iota_*: \pi_1(S^1) \to \pi_1(D^2)$ must be injective (if $r_* \circ \iota_*$ is the identity, $\iota_*$ is injective). But $\pi_1(S^1) \cong \mathbb{Z}$ and $\pi_1(D^2) = 0$ (since $D^2$ is contractible), so no injective homomorphism $\mathbb{Z} \to 0$ exists. Contradiction.

Lemma 2 (Fixed-point-free maps yield retractions): If $f: D^2 \to D^2$ has no fixed point, then there exists a retraction $r: D^2 \to S^1$ .

Proof of Lemma 2: For each $x \in D^2$ , since $f(x) \neq x$ , there is a unique ray starting at $f(x)$ and passing through $x$ . This ray hits $S^1$ at a unique point; call it $r(x)$ .

Explicitly: $r(x)$ is the point on $S^1$ obtained by starting at $f(x)$ , moving in the direction $x - f(x)$ , and hitting the boundary. Analytically: $r(x) = x + t(x)(x - f(x))$ where $t(x) \geq 0$ is chosen so that $\|r(x)\| = 1$ . This is the larger root of a quadratic in $t$ , which depends continuously on $x$ (since $f(x) \neq x$ ensures the ray is well-defined).

For $x \in S^1$ : since $\|x\| = 1$ and $f(x) \in D^2$ , the ray from $f(x)$ through $x$ exits $D^2$ at $x$ itself (with $t = 0$ giving $r(x) = x$ ). So $r|_{S^1} = \operatorname{id}_{S^1}$ .

Combining: If $f: D^2 \to D^2$ has no fixed point, Lemma 2 gives a retraction $r: D^2 \to S^1$ , contradicting Lemma 1. Therefore $f$ must have a fixed point.

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Applications

ExampleApplications of Brouwer's Theorem

Existence of equilibria: In game theory, Brouwer's theorem (and its generalization, Kakutani's theorem) is used to prove the existence of Nash equilibria.
Nonlinear equations: If $f: D^2 \to D^2$ is continuous, the equation $f(x) = x$ always has a solution. This is used in PDE theory (via Schauder's fixed-point theorem, which generalizes Brouwer to infinite dimensions).
The Ham Sandwich Theorem and other results in combinatorial geometry ultimately rely on fixed-point or degree-theoretic arguments.

ExampleNon-Constructive Nature

The Brouwer theorem guarantees existence of a fixed point but provides no algorithm for finding it. Consider $f: D^2 \to D^2$ defined by a complicated continuous formula. We know $f$ has a fixed point, but finding it may require iterative methods (not guaranteed to converge in general).

This non-constructive character led Brouwer himself to become an advocate of intuitionism, rejecting non-constructive existence proofs!

Generalizations

Theorem9.12Brouwer Fixed-Point Theorem (General)

Every continuous map $f: D^n \to D^n$ (the closed $n$ -ball in $\mathbb{R}^n$ ) has a fixed point, for every $n \geq 1$ .

RemarkProofs in Higher Dimensions

The proof for $n = 2$ uses $\pi_1$ . For general $n$ , one needs stronger tools:

Homology theory: The proof generalizes by replacing $\pi_1$ with $H_{n-1}$ (the $(n-1)$ -th homology group). The no-retraction lemma becomes: $H_{n-1}(S^{n-1}) \cong \mathbb{Z}$ does not inject into $H_{n-1}(D^n) = 0$ .
Degree theory: The degree of $\operatorname{id}_{S^{n-1}}$ is $1 \neq 0$ , but if a retraction existed, the degree would factor through $H_{n-1}(D^n) = 0$ .
Smooth methods: In the smooth category, Sard's theorem and differential forms provide alternative proofs.

Theorem9.13Borsuk--Ulam Theorem (2D)

For every continuous map $f: S^2 \to \mathbb{R}^2$ , there exists a point $x \in S^2$ with $f(x) = f(-x)$ . That is, some pair of antipodal points maps to the same value.

RemarkConnection to Brouwer

The Borsuk--Ulam theorem implies the Brouwer fixed-point theorem (in dimension 2) and is itself stronger. It has a memorable consequence: at any moment, there are two antipodal points on the Earth's surface with the same temperature and the same atmospheric pressure.