Questions: Brouwer Fixed Point Theorem (Homological Proof)

Question 1

The key step in the homological proof of Brouwer's theorem is showing that no retraction r: D^n → S^{n-1} exists. What algebraic contradiction does a retraction produce?

• A. It would make H_n(D^n) nonzero
• B. The composition S^{n-1} →^i D^n →^r S^{n-1} would be the identity, forcing id_*: H_{n-1}(S^{n-1}) → H_{n-1}(S^{n-1}) to factor through H_{n-1}(D^n) = 0, which means id = 0 on Z — a contradiction
• C. It would make π_1(D^n) non-trivial
• D. It would violate the excision theorem

Answer: (B)

If r: D^n → S^{n-1} is a retraction (r ∘ i = id where i: S^{n-1} ↪ D^n), then on homology: r_* ∘ i_* = id_* on H_{n-1}. But i_*: H_{n-1}(S^{n-1}) → H_{n-1}(D^n) has target H_{n-1}(D^n) = 0 (since D^n is contractible), so i_* = 0, and r_* ∘ i_* = r_* ∘ 0 = 0. But id_* on H_{n-1}(S^{n-1}) ≅ Z is the identity, not zero. Contradiction. Therefore no retraction exists.

Question 2

The Brouwer fixed point theorem fails for open disks — there exist continuous maps from the open disk to itself with no fixed points.

• T. True
• F. False

Answer: True

The open disk is homeomorphic to all of R^n, and the translation map x ↦ x + (1,0,...,0) on R^n has no fixed points. The Brouwer theorem requires the domain to be the CLOSED disk D^n, which is compact. Compactness is essential: the retraction argument uses the fact that S^{n-1} is the boundary of D^n and that D^n is contractible while S^{n-1} has nontrivial homology. The open disk is also contractible but has no boundary in the relevant sense.

Question 3

How does one construct the hypothetical retraction r: D^n → S^{n-1} from a fixed-point-free map f: D^n → D^n?

Model answer: For each x ∈ D^n, since f(x) ≠ x, consider the ray starting at f(x) and passing through x. This ray hits the boundary sphere S^{n-1} at a unique point r(x). Formally, r(x) = x + t(x - f(x)) where t ≥ 0 is chosen so |r(x)| = 1. This map r is continuous (because f is continuous and the ray direction varies continuously) and satisfies r(x) = x for x ∈ S^{n-1} (since the ray from f(x) through x, with x already on the boundary, hits the boundary at x). So r is a retraction of D^n onto S^{n-1}.

The geometric picture is clear: for each interior point x, draw a ray from f(x) through x and extend it to the boundary. If x is on the boundary, the ray starts at f(x) (which is in the interior since the map goes D^n → D^n) and passes through x, which is already on the boundary — so r(x) = x. The no-fixed-point assumption ensures f(x) ≠ x, so the ray direction is well-defined.

Question 4

The Brouwer fixed point theorem can be proved using the fundamental group for n = 2 but requires homology for n ≥ 3.

• T. True
• F. False

Answer: False

The fundamental group proof works for n = 2 (the retraction r: D^2 → S^1 would give a surjection r_*: π_1(D^2) → π_1(S^1), but π_1(D^2) = 0 and π_1(S^1) = Z, contradiction). However, this argument does NOT work for n = 3: π_1(D^3) = π_1(S^2) = 0, so the fundamental group sees no contradiction. The homological proof works uniformly in all dimensions by using H_{n-1}(S^{n-1}) = Z and H_{n-1}(D^n) = 0. The statement in the question is wrong because the fundamental group proof works for n = 2 (via π_1) but fails to even address n ≥ 3 — it's not that it 'requires' homology for large n, but that only homology provides a uniform proof.