Suppose you are given the fact that for each \(n\), no continuous antipode-preserving map \(h: S^n \to S^n\) is nulhomotopic. (This result can be proved using more advanced techniques of algebraic topology.) Prove the following:
There is no retraction \(r: B^{n+1} \to S^n\).
There is no continuous antipode-preserving map \(g: S^{n+1} \to S^n\).
(Borsuk-Ulam theorem) Given a continuous map \(f: S^{n+1} \to \mathbb{R}^{n+1}\), there is a point \(x\) of \(S^{n+1}\) such that \(f(x) = f(-x)\).
If \(A_1, \ldots, A_{n+1}\) are bounded measurable sets in \(\mathbb{R}^{n+1}\), there exists an \(n\)-plane in \(\mathbb{R}^{n+1}\) that bisects each of them.
Solution
Goal: Prove the non-retraction theorem, the non-existence of odd maps \(S^{n+1} \to S^n\), the Borsuk-Ulam theorem, and the Ham Sandwich theorem using the fact that odd self-maps of \(S^n\) are never nullhomotopic.
1. Part (a): Non-existence of a retraction \(r: B^{n+1} \to S^n\).Proof:
1.1. Suppose \(r: B^{n+1} \to S^n\) is a continuous retraction, so \(r(x) = x\) for all \(x \in \partial B^{n+1} = S^n\).
1.2. Since \(B^{n+1}\) is contractible, the inclusion map \(i: S^n \hookrightarrow B^{n+1}\) is nullhomotopic.
1.3. Therefore the composition \(r \circ i = \operatorname{id}_{S^n}: S^n \to S^n\) is nullhomotopic.
1.4. The identity map \(\operatorname{id}_{S^n}\) satisfies \(\operatorname{id}_{S^n}(-x) = -x = -\operatorname{id}_{S^n}(x)\), so it is an antipode-preserving map.
1.5. This contradicts the given fact that no continuous antipode-preserving map \(S^n \to S^n\) is nullhomotopic.
1.6. Thus no retraction \(r: B^{n+1} \to S^n\) exists.
2. Part (b): Non-existence of continuous antipode-preserving maps \(g: S^{n+1} \to S^n\).Proof:
2.1. Suppose \(g: S^{n+1} \to S^n\) is continuous and satisfies \(g(-x) = -g(x)\) for all \(x \in S^{n+1}\).
2.2. Embed \(S^n\) as the equator of \(S^{n+1}\): \(S^n = \{(x, 0) \in S^{n+1} : x \in S^n\}\).
2.3. The restriction \(h = g|_{S^n}: S^n \to S^n\) is continuous and antipode-preserving on \(S^n\).
2.4. Let \(E_+^{n+1} = \{(x, z) \in S^{n+1} : z \ge 0\}\) be the upper hemisphere. \(E_+^{n+1}\) is homeomorphic to \(B^{n+1}\), with boundary \(\partial E_+^{n+1} = S^n\).
2.5. The map \(g|_{E_+^{n+1}}: E_+^{n+1} \to S^n\) is a continuous extension of \(h\) to the disk \(B^{n+1}\).
2.6. A continuous map from \(S^n\) to \(S^n\) extends to \(B^{n+1}\) if and only if it is nullhomotopic, so \(h\) is nullhomotopic.
2.7. This contradicts the given fact that \(h\) cannot be nullhomotopic. Hence no such \(g\) exists.
3. Part (c): Borsuk-Ulam theorem.Proof:
3.1. Suppose for contradiction that \(f(x) \neq f(-x)\) for all \(x \in S^{n+1}\).
3.2. Define \(v: S^{n+1} \to \mathbb{R}^{n+1} \setminus \{0\}\) by \(v(x) = f(x) - f(-x)\).
3.3. Define \(g: S^{n+1} \to S^n\) by \(g(x) = \frac{v(x)}{\|v(x)\|}\).
3.4. \(g\) is continuous, and \(v(-x) = f(-x) - f(x) = -v(x)\) implies:\[g(-x) = \frac{-v(x)}{\|-v(x)\|} = -\frac{v(x)}{\|v(x)\|} = -g(x).\]
3.5. Thus \(g: S^{n+1} \to S^n\) is a continuous antipode-preserving map, contradicting
2.
2.6. Therefore, there exists \(x \in S^{n+1}\) such that \(f(x) = f(-x)\).
4. Part (d): Ham Sandwich theorem.Proof:
4.1. Embed \(\mathbb{R}^{n+1}\) into \(\mathbb{R}^{n+2}\) via \(y \mapsto (y, 1)\). For each \(i \in \{1, \dots, n+1\}\), let \(A_i' = \{(y, 1) : y \in A_i\}\).
4.2. For each unit vector \(u = (u_1, \dots, u_{n+2}) \in S^{n+1} \subset \mathbb{R}^{n+2}\), define the open half-space \(H_u^+ = \{z \in \mathbb{R}^{n+2} : \langle z, u \rangle > 0\}\).
4.3. Define \(f: S^{n+1} \to \mathbb{R}^{n+1}\) by \(f(u) = (f_1(u), \dots, f_{n+1}(u))\) where \(f_i(u) = \mu(A_i' \cap H_u^+)\).
4.4. By the Dominated Convergence Theorem, each \(f_i\) is continuous, so \(f\) is continuous.
4.5. By 3 (Borsuk-Ulam), there exists \(u \in S^{n+1}\) such that \(f(u) = f(-u)\).
4.6. Since the boundary hyperplane \(P_u = \{z : \langle z, u \rangle = 0\}\) has Lebesgue measure zero, \(\mu(A_i) = f_i(u) + f_i(-u) = 2f_i(u)\), which gives \(f_i(u) = \frac{1}{2}\mu(A_i)\) for all \(i\).
4.7. If \(u' = (u_1, \dots, u_{n+1}) \neq 0\), the equation \(\langle y, u' \rangle + u_{n+2} = 0\) defines an affine hyperplane (\(n\)-plane) in \(\mathbb{R}^{n+1}\) bisecting every \(A_i\). If \(u' = 0\), then \(u_{n+2} = \pm 1\), giving \(f_i(u) \in \{0, \mu(A_i)\}\), so \(\mu(A_i) = 0\) for all \(i\); in this case every \(A_i\) has measure zero, so any \(n\)-plane bisects each \(A_i\) (both halves have measure \(0 = \frac{1}{2}\mu(A_i)\)). Q.E.D.