Show that there is no map \(f: S^2 \to S^1\) that commutes with the antipodal map.
Solution
Goal: Prove that there does not exist any continuous map \(f \colon S^2 \to S^1\) such that \(f(-x) = -f(x)\) for all \(x \in S^2\) (i.e. commuting with the antipodal map).
1. Restrict \(f\) to the equatorial circle \(S^1 \subset S^2\).
1.1. Let \(i \colon S^1 \hookrightarrow S^2\) be the equatorial inclusion \(x \mapsto (x_1, x_2, 0)\), and let \(g = f|_{S^1} = f \circ i \colon S^1 \to S^1\).
1.2. Since \(f(-x) = -f(x)\) for all \(x \in S^2\), the restriction \(g\) satisfies \(g(-x) = -g(x)\) for all \(x \in S^1\).
1.3. Every continuous odd map \(g \colon S^1 \to S^1\) has odd degree \(\deg(g) \equiv 1 \pmod 2\).
1.3.1. Using the standard covering map \(p \colon \mathbb{R} \to S^1\) given by \(t \mapsto e^{2\pi i t}\), lift \(g\) to a continuous map \(\widetilde{g} \colon \mathbb{R} \to \mathbb{R}\).
1.3.2. The condition \(g(t + 1/2) = -g(t) = e^{i\pi} g(t)\) implies \(\widetilde{g}(t + 1/2) - \widetilde{g}(t) = k + 1/2\) for some fixed integer \(k \in \mathbb{Z}\).
1.3.3. The degree of \(g\) is \(\deg(g) = \widetilde{g}(1) - \widetilde{g}(0) = (\widetilde{g}(1) - \widetilde{g}(1/2)) + (\widetilde{g}(1/2) - \widetilde{g}(0)) = 2(k + 1/2) = 2k + 1\).
1.3.4. Thus \(\deg(g) \neq 0\).
1.3.5. Proof: Standard degree computation of odd maps on the circle.Q.E.D.
2. Extend \(g\) to the upper hemisphere disk \(D_+^2 \subset S^2\).
2.1. The upper hemisphere \(D_+^2 = \{(x_1, x_2, x_3) \in S^2 \mid x_3 \ge 0\}\) is homeomorphic to the 2-disk \(D^2\), with boundary \(\partial D_+^2 = S^1\).
2.2. The map \(F = f|_{D_+^2} \colon D_+^2 \to S^1\) is a continuous map extending \(g\) to the entire disk \(D_+^2\).
2.3. If a continuous map \(g \colon S^1 \to S^1\) extends to a continuous map \(D^2 \to S^1\), then \(g\) is nullhomotopic, which implies \(\deg(g) = 0\).
2.4. Proof: The inclusion \(S^1 = \partial D^2 \hookrightarrow D^2\) is nullhomotopic since \(D^2\) is contractible, so the induced homomorphism \(g_* \colon \pi_1(S^1) \to \pi_1(S^1)\) is zero, meaning \(\deg(g) = 0\).Q.E.D.
3. Derive contradiction.
3.1. 1 shows \(\deg(g) = 2k + 1 \neq 0\).
3.2. 2 shows \(\deg(g) = 0\).
3.3. This contradiction proves no continuous odd map \(f \colon S^2 \to S^1\) can exist.
3.4. 1 gives \(\deg(g) = 2k + 1 \neq 0\), while 2 gives \(\deg(g) = 0\); these contradict each other, so the assumed odd map \(f\) cannot exist.
4. Q.E.D.
4.1. 3 establishes the result.