Prove that for any \(f: S^2 \to \mathbb{R}^2\), there exists \(x\in S^2\) such that \(f(x) = f(-x)\).
Solution
Goal: Prove the Borsuk-Ulam theorem in dimension 2: for every continuous map \(f \colon S^2 \to \mathbb{R}^2\), there exists a point \(x \in S^2\) such that \(f(x) = f(-x)\).
1. Reduce to showing there is no odd continuous map \(g \colon S^2 \to S^1\).
1.1. Suppose for contradiction that there exists a continuous map \(f \colon S^2 \to \mathbb{R}^2\) such that \(f(x) \neq f(-x)\) for all \(x \in S^2\).
1.2. Define \(F \colon S^2 \to \mathbb{R}^2 \setminus \{0\}\) by \(F(x) = f(x) - f(-x)\).
1.3. \(F\) is continuous and antipodal-preserving (odd): \(F(-x) = f(-x) - f(x) = -F(x)\).
1.4. Define \(g \colon S^2 \to S^1\) by \(g(x) = \frac{F(x)}{\|F(x)\|}\).
1.5. The map \(g\) is continuous and odd: \(g(-x) = \frac{-F(x)}{\|-F(x)\|} = -g(x)\) for all \(x \in S^2\).
1.6. The map \(g\) is the composition of the continuous map \(F\) with the continuous normalization \(y \mapsto y/\|y\|\), so it is continuous; the oddness follows from \(F(-x) = -F(x)\).
2. Restrict \(g\) to the equatorial circle \(S^1 \subset S^2\).
2.1. Let \(h = g|_{S^1} \colon S^1 \to S^1\).
2.2. Since \(g\) is odd on \(S^2\), \(h\) is an odd map on \(S^1\): \(h(-x) = -h(x)\) for all \(x \in S^1\).
2.3. Every continuous odd map \(h \colon S^1 \to S^1\) has odd degree \(\deg(h) \equiv 1 \pmod 2\).
2.3.1. Parameterize \(S^1\) by \([0, 1] / (0 \sim 1)\) via \(t \mapsto e^{2\pi i t}\).Antipodal points correspond to \(t\) and \(t + 1/2\).
2.3.2. Lift \(h\) to a continuous map \(\widetilde{h} \colon \mathbb{R} \to \mathbb{R}\) via the covering map \(p(t) = e^{2\pi i t}\).
2.3.3. The condition \(h(t + 1/2) = -h(t) = e^{i\pi} h(t)\) means \(p(\widetilde{h}(t + 1/2)) = p(\widetilde{h}(t) + 1/2)\).
2.3.4. Thus \(\widetilde{h}(t + 1/2) - \widetilde{h}(t) = k + 1/2\) for some fixed integer \(k \in \mathbb{Z}\) (by connectedness of \(\mathbb{R}\)).
2.3.5. The degree of \(h\) is given by the total shift over the period 1: \[\deg(h) = \widetilde{h}(1) - \widetilde{h}(0) = (\widetilde{h}(1) - \widetilde{h}(1/2)) + (\widetilde{h}(1/2) - \widetilde{h}(0)) = 2(k + 1/2) = 2k + 1.\]
2.3.6. Thus \(\deg(h) = 2k + 1\) is an odd integer, and in particular \(\deg(h) \neq 0\).
2.3.7. The lift \(\widetilde{h}\) exists by the lifting criterion for the universal cover \(\mathbb{R} \to S^1\); the oddness condition forces the half-period shift to be a half-integer, and the degree is the total shift over one period, which is therefore odd.
3. Obtain a contradiction via the upper hemisphere disk.
3.1. Let \(D_+^2 = \{(x_1, x_2, x_3) \in S^2 \mid x_3 \ge 0\}\) be the closed upper hemisphere.
3.2. \(D_+^2\) is homeomorphic to the closed 2-disk \(D^2\), and its boundary is \(\partial D_+^2 = S^1\) (the equator).
3.3. The restriction \(g|_{D_+^2} \colon D_+^2 \to S^1\) is a continuous extension of \(h = g|_{S^1}\) to the entire disk \(D_+^2\).
3.4. If a continuous map \(h \colon S^1 \to S^1\) extends to a continuous map \(D^2 \to S^1\), then \(h\) is nullhomotopic, so \(\deg(h) = 0\).
3.5. The disk \(D^2\) is contractible, so the inclusion \(\iota \colon S^1 \hookrightarrow D^2\) induces the zero map on \(\pi_1(S^1)\); hence \(h_* = (g|_{D^2} \circ \iota)_* = 0\), so \(\deg(h) = 0\).
4. Derive the final contradiction.
4.1. By 2, \(\deg(h)\) is odd (so \(\deg(h) \neq 0\)).
4.2. By 3, \(\deg(h) = 0\).
4.3. This contradiction shows no such odd map \(g \colon S^2 \to S^1\) exists.
4.4. Hence for any continuous \(f \colon S^2 \to \mathbb{R}^2\), there exists \(x \in S^2\) such that \(f(x) = f(-x)\).
4.5. 2 shows \(\deg(h)\) is odd, while 3 shows \(\deg(h) = 0\); this contradiction rules out the existence of an odd map \(g \colon S^2 \to S^1\), and hence of a map \(f\) with \(f(x) \neq f(-x)\) everywhere.
5. Q.E.D.
5.1. 4 establishes the result.