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LemmaStatement: AI-adaptedProof: AI-generatedprecheck 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 D2=B‾2(0,1) and S1=S2(0,1). Every continuous fixed-point-free map f:D2→D2 determines a continuous retraction D2→S1.

Facts & Assumptions

Given: A continuous map f:D2→D2 such that f(x)≠x for every x∈D2.

[F1]

The closed unit disk and unit circle are D2={x:∥x∥2≤1} and S1={x:∥x∥2=1} (Euclidean spheres and closed balls as subspaces of Rn).

[L1]

The Euclidean inner product is bilinear and positive definite, with ∥u∥22=⟨u,u⟩ (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn).

[L2]

A map into R2 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).

[F2]

A continuous map r:D2→S1 is a retraction when r(x)=x for every x∈S1 (Retractions and deformation retracts, with a deformation retraction required to fix the retract pointwise).

Proof

technique · constructive
1.1givenF1L1L4construct

For x∈D2, put a=f(x), v=x−a, A=⟨v,v⟩, B=⟨a,v⟩, and C=⟨a,a⟩. Fixed-point-freeness gives v≠0 and A>0. Define λ(x)=−B+B2+A(1−C)A,r(x)=a+λ(x)v.

2.1step 1.1F1L1algebra

The equation ∥a+tv∥22=1 is At2+2Bt+C−1=0, whose discriminant is 4(B2+A(1−C))≥0 because C=∥f(x)∥22≤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)≥1 and is the unique intersection parameter of the ray a+tv with S1 for t≥1.

3.1step 1.1step 2.1L2L3L4

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.

4.1step 2.1step 3.1F1F2discharge-construct∎

Step 2.1 gives ∥r(x)∥2=1, so r maps D2 into S1. If x∈S1, then t=1 is a root and is the larger root because 1≥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 D2 onto S1.

Depends on

Used by

Dependency tree · two levels

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Sources