No retraction and the disk fixed-point theorem

Try

Can a disk be continuously pushed onto its boundary while leaving every boundary point fixed? Radial projection seems tempting, but it has no continuous value at the center. This is not merely a defect of that formula: the fundamental group proves that no such retraction exists. That obstruction also forces fixed points of disk self-maps.

Predict

Suppose a continuous map f from the closed unit disk to itself had no fixed point. For each x, draw the ray starting at f(x), passing through x and continuing to the boundary. Predict what happens when x is already on the boundary. The ray’s outward exit is x itself, so this construction would be a forbidden retraction if it is continuous.

Experiment

Inspect three fixed-point-free local pairs; these are sample inputs to the ray construction, not a globally defined counterexample map.

x	f(x)	Outward boundary point
(0,0)	(1/2,0)	(−1,0)
(1/2,0)	(0,0)	(1,0)
(0,1)	(0,0)	(0,1)

The first two rows show that the ray direction matters. Extending from x toward f(x) would choose the wrong endpoint. The third checks the retraction condition.

Understand

Let i:S¹→D² be boundary inclusion. If r:D²→S¹ were a retraction, r∘i would be the identity. With a boundary basepoint, functoriality would give r_*i_*=id on π₁(S¹)=Z. But D² is convex and has trivial fundamental group, so this composite factors through the trivial group and sends every class to zero. It cannot fix the integer one. This proves no retraction.

Now assume f:D²→D² is continuous and f(x)≠x everywhere. Put a=f(x) and v=x−a, so v≠0. Points on the forward ray have form a+tv. The boundary equation is ||v||²t²+2(a·v)t+||a||²−1=0. Its outward root is

t(x)=(−a·v+sqrt((a·v)²+||v||²(1−||a||²)))/||v||².

The discriminant is nonnegative because a lies in the disk. The point at t=1 is x, also in the disk, so the outward root is at least one. The nonzero denominator and the continuous square-root expression show t and r(x)=a+t(x)v are continuous. By construction r(x) lies on S¹.

If x is already on S¹, t=1 is the outward exit. Indeed the segment from a to x stays in the disk, and strict convexity of the Euclidean disk puts its interior in the interior whenever a and x are distinct boundary points; the case of interior a is even more direct. Continuing beyond x leaves the disk. Therefore r(x)=x on the boundary. The resulting retraction contradicts the preceding theorem, so every continuous disk self-map has a fixed point. This is Brouwer’s theorem in dimension two.

Apply

The proof does not identify a fixed point or provide an algorithm with a numerical error bound. It proves existence by ruling out a continuous fixed-point-free map. Also distinguish the closed disk from an open disk: a translation-like map on an open region can avoid fixed points while approaching a missing boundary. The compact closed domain is part of the statement.

Continue

The same ray construction works in higher-dimensional balls, but the fundamental-group obstruction does not: the two-sphere has trivial π₁. A higher-dimensional no-retraction proof needs a stronger invariant. This bridge proves the disk case and keeps that boundary explicit rather than presenting a two-dimensional argument as the full theorem.

Continue to Van Kampen and gluing loops.