LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-24

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A fixed-point-free self-map of the disk produces a continuous retraction onto the unit circle

Statement

Let $D^{2} = {\overline{B}}_{2} (0, 1)$ and $S^{1} = S_{2} (0, 1)$ . Every continuous fixed-point-free map $f : D^{2} \to D^{2}$ determines a continuous retraction $D^{2} \to S^{1}$ .

Facts & Assumptions

Given: A continuous map $f : D^{2} \to D^{2}$ such that $f (x) \neq x$ for every $x \in D^{2}$ .

[F1]

The closed unit disk and unit circle are $D^{2} = {x : ∥ x ∥_{2} \leq 1}$ and $S^{1} = {x : ∥ x ∥_{2} = 1}$ (Euclidean spheres and closed balls as subspaces of $R^{n}$ ).

[L1]

The Euclidean inner product is bilinear and positive definite, with $∥ u ∥_{2}^{2} = ⟨ u, u ⟩$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[L2]

A map into $R^{2}$ is continuous exactly when its coordinate functions are continuous; finite sums, scalar multiples, inner products, and norms of continuous vector-valued functions are continuous (A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions).

[L3]

Finite sums and products of continuous real-valued maps are continuous, and a quotient is continuous wherever its denominator is nonzero (Sums, products, absolute values, finite maxima and minima, and quotients of continuous real-valued maps on a topological space are continuous where defined).

[L4]

Every nonnegative real has a unique nonnegative square root, and the square-root function is continuous on $[0, \infty)$ as the inverse of the continuous strictly increasing square map there (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Continuous inverse theorem: a continuous injective $f$ on an interval $I$ is a bijection onto the order-convex set $f [I]$ , and the inverse $g : f [I] \to I$ is continuous and strictly monotone in the same sense as $f$ ).

[F2]

A continuous map $r : D^{2} \to S^{1}$ is a retraction when $r (x) = x$ for every $x \in S^{1}$ (Retractions and deformation retracts, with a deformation retraction required to fix the retract pointwise).

Proof

technique · constructive

1.1

For $x \in D^{2}$ , put $a = f (x)$ , $v = x - a$ , $A = ⟨ v, v ⟩$ , $B = ⟨ a, v ⟩$ , and $C = ⟨ a, a ⟩$ . Fixed-point-freeness gives $v \neq 0$ and $A > 0$ . Define $λ (x) = \frac{- B + \sqrt{B^{2} + A (1 - C)}}{A}, r (x) = a + λ (x) v .$

givenF1L1L4construct

2.1

The equation $∥ a + t v ∥_{2}^{2} = 1$ is $A t^{2} + 2 B t + C - 1 = 0$ , whose discriminant is $4 (B^{2} + A (1 - C)) \geq 0$ because $C = ∥ f (x) ∥_{2}^{2} \leq 1$ ; thus $λ (x)$ is its larger root. Since the upward-opening quadratic is nonpositive at both $t = 0$ and $t = 1$ , its larger root satisfies $λ (x) \geq 1$ and is the unique intersection parameter of the ray $a + t v$ with $S^{1}$ for $t \geq 1$ .

step 1.1F1L1algebra

3.1

The maps $a, v, A, B, C$ are continuous; the radicand is nonnegative, its square root is continuous, and the denominator $A$ never vanishes. Hence $λ$ is continuous, and componentwise continuity makes $r (x) = a + λ (x) v$ continuous.

step 1.1step 2.1L2L3L4

4.1

Step 2.1 gives $∥ r (x) ∥_{2} = 1$ , so $r$ maps $D^{2}$ into $S^{1}$ . If $x \in S^{1}$ , then $t = 1$ is a root and is the larger root because $1 \geq 0$ lies in the interval on which the quadratic is nonpositive; hence $λ (x) = 1$ and $r (x) = a + v = x$ . Therefore $r$ is a continuous retraction of $D^{2}$ onto $S^{1}$ .

step 2.1step 3.1F1F2discharge-construct∎

Depends on

Used by

Brouwer fixed-point theorem for the closed disk Theorem

Dependency tree · two levels

59 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Allen Hatcher, Algebraic Topology, proof of Theorem 1.9 (standard reference, not scraped)
J. Peter May, A Concise Course in Algebraic Topology, Chapter 1, §6 (standard reference, not scraped)