Building a Boundary Retraction from a Fixed-Point-Free Map

From a Fixed-Point-Free Map to a Boundary Map

Let \(D^2=\{x\in\mathbb{R}^2:\|x\|\leq 1\}\), with the subspace topology inherited from the standard topology on \(\mathbb{R}^2\), and let \(S^1=\{x\in\mathbb{R}^2:\|x\|=1\}\), also with its subspace topology. Suppose \(f:D^2\to D^2\) is continuous and has no fixed point, meaning \(f(x)\ne x\) for every \(x\in D^2\). For each \(x\), start at \(f(x)\) and move along the ray through \(x\), in the direction from \(f(x)\) toward \(x\). The construction below finds where that ray reaches \(S^1\).

The no-fixed-point hypothesis is essential: it ensures the direction vector \(x-f(x)\) is never zero. The disk constraint is essential as well: both \(f(x)\) and \(x\) lie in the disk, so the ray begins inside it and passes through a point of the disk before reaching its exit. We will show that the exit point depends continuously on \(x\), and that when \(x\) is already on the boundary, the exit point is \(x\) itself.

The Unique Forward Exit Point

First fix two points \(a,x\in D^2\) with \(a\ne x\). The ray from \(a\) through \(x\) is parametrized by \(a+t(x-a)\) for \(t\geq 0\). The point \(x\) occurs at \(t=1\), so the boundary point in the forward direction must correspond to a parameter \(t\geq1\).

Proof. Put \(v=x-a\), \(A=v\cdot v\), \(b=a\cdot v\), and \(c=a\cdot a-1\), where \(\cdot\) is the Euclidean dot product. Since \(a\ne x\), \(A=\|v\|^2>0\); since \(a\in D^2\), \(c\leq0\). The equation for an intersection with \(S^1\) is \(\|a+tv\|^2-1=At^2+2bt+c=0\). Its discriminant after dividing the linear coefficient by two is \(\Delta=b^2-Ac\), which is nonnegative because \(b^2\geq0\), \(A>0\), and \(c\leq0\).

If \(c<0\), then \(\Delta>b^2\), so the two roots \((-b-\sqrt{\Delta})/A\) and \((-b+\sqrt{\Delta})/A\) have opposite signs: their product is \(c/A<0\). Thus exactly one is positive. If \(c=0\), then \(\|a\|=1\). In this case \(b=a\cdot x-1<0\). Indeed, \(a\cdot x\leq\|a\|\|x\|\leq1\) by the Cauchy–Schwarz inequality; equality \(a\cdot x=1\) would imply \(\|x-a\|^2=\|x\|^2+1-2a\cdot x=\|x\|^2-1\leq0\), and hence \(x=a\), contrary to the assumption. The roots when \(c=0\) are \(0\) and \(-2b/A>0\), so again there is exactly one positive root. In both cases the unique positive root is \(t_+=(-b+\sqrt{\Delta})/A\).

Finally, at \(t=1\) the point \(a+v=x\) lies in \(D^2\), so \(At^2+2bt+c\leq0\) at \(t=1\). The other root is nonpositive: it is negative when \(c<0\), and zero when \(c=0\). Factoring the quadratic as \(A(t-t_-)(t-t_+)\), with \(t_-\leq0\), shows that its value at \(1\) can be nonpositive only if \(1\leq t_+\). Thus the unique positive root is at least \(1\), as required. \(\square\)

This proposition includes the case where \(a\) is already on \(S^1\). Then \(t=0\) is one intersection parameter, but it is the starting point, not the forward exit; the proposition identifies the unique positive intersection.

The Formula Defines a Retraction

Apply the proposition separately at each input \(x\), taking \(a=f(x)\). To make the dependence on \(x\) explicit, define \(A(x)=\|x-f(x)\|^2\), \(b(x)=f(x)\cdot(x-f(x))\), and \(c(x)=\|f(x)\|^2-1\). Then \(A(x)>0\) and \(c(x)\leq0\) for every \(x\in D^2\). The ray parameter and its boundary point are given by the following formulas.

Proof. We check the claims in turn. First, the denominator \(A(x)\) is positive because \(x-f(x)\ne0\). The expression under the square root is nonnegative by the proposition, applied with \(a=f(x)\). Thus \(t(x)\) is defined for every \(x\), and the proposition shows it is the unique positive parameter for which \(\|f(x)+t(x)(x-f(x))\|=1\). Consequently \(r(x)\in S^1\), so \(r\) has the stated codomain.

For continuity, the maps \(x\mapsto f(x)\) and \(x\mapsto x\) are continuous. Dot products, subtraction, and squared norms are continuous operations on \(\mathbb{R}^2\); therefore \(A(x)\), \(b(x)\), and \(c(x)\) are continuous. The square-root function is continuous on \([0,\infty)\), and its argument is nonnegative at every input. Division by \(A(x)\) preserves continuity because \(A(x)>0\) everywhere. Hence \(t\) is continuous, and the formula for \(r\), using continuous sums and scalar multiplication, shows that \(r\) is continuous.

It remains to check the boundary condition. If \(x\in S^1\), then \(f(x)+1\cdot(x-f(x))=x\), whose norm is \(1\). Thus \(t=1\) is a positive root of the ray-intersection equation. The proposition says there is only one positive root, so \(t(x)=1\). Substitution gives \(r(x)=f(x)+(x-f(x))=x\). Therefore \(r\) restricts to the identity on \(S^1\), and is a retraction. \(\square\)

Worked Ray Calculations

The following examples compute the construction from a single pair \(a=f(x)\), \(x\). They illustrate the pointwise formula; they do not assert that the specified pair extends to a global fixed-point-free map on the disk.

Why the Positive Root Matters

A common error is to solve the quadratic correctly but select the wrong root. For the first example, the intersection equation is \((1/2-t/2)^2=1\). Expanding gives \(t^2-2t-3=0\), which factors as \((t-3)(t+1)=0\). The roots are \(3\) and \(-1\). Choosing the smaller root \(-1\) gives \(a+(-1)v=(1/2,0)-(-1/2,0)=(1,0)\). This point is on \(S^1\), but it lies in the direction opposite to the ray from \(a\) through \(x\): its parameter is negative, not \(t\geq1\). The required exit point is the one at \(t=3\), namely \((-1,0)\).

The distinction matters for the retraction condition. The construction must travel from \(f(x)\) through \(x\) and onward, not backward from \(f(x)\). The positive-root formula enforces that direction, while the proof that \(t(x)=1\) on \(S^1\) ensures the map fixes boundary inputs. These two facts—choosing the forward exit and proving the boundary identity—are both necessary to obtain a retraction. The no-retraction theorem for the disk then rules out the assumed fixed-point-free map: if such a map existed, this construction would produce a retraction.