Show that if \(h: S^1 \to S^1\) is nulhomotopic, then \(h\) has a fixed point and \(h\) maps some point \(x\) to its antipode \(-x\).
Solution
1. Degree of nullhomotopic maps:Since \(h: S^1 \to S^1\) is nullhomotopic (homotopic to a constant map), the induced map on fundamental groups \(h_*: \pi_1(S^1) \to \pi_1(S^1)\) is the zero homomorphism, so the winding number (topological degree) satisfies \(\deg(h) = 0\).
2. Existence of a fixed point:There exists \(x_0 \in S^1\) such that \(h(x_0) = x_0\).Proof:
2.1. Suppose for contradiction that \(h(x) \neq x\) for all \(x \in S^1\).
2.2. Then for all \(x \in S^1\) and \(t \in [0, 1]\), the convex combination \((1-t)h(x) + t(-x) \neq 0\).(If \((1-t)h(x) = tx\), taking norms gives \(1-t = t \implies t = 1/2 \implies h(x) = x\), a contradiction).
2.3. Define the straight-line homotopy \(H: S^1 \times [0, 1] \to S^1\) by:\[H(x, t) = \frac{(1-t)h(x) - tx}{\|(1-t)h(x) - tx\|}.\]
2.4. \(H\) is a continuous homotopy between \(h(x)\) and the antipodal map \(a(x) = -x\).
2.5. The antipodal map \(a(x) = -x\) on \(S^1\) is homotopic to the identity via the rotation homotopy \(R_t(x) = e^{i\pi t} x\), so \(\deg(a) = \deg(\operatorname{id}_{S^1}) = 1 \neq 0\).
2.6. Since homotopy preserves degree, \(\deg(h) = \deg(a) = 1\), strictly contradicting \(\deg(h) = 0\) from
1.
1.7. Thus there must exist \(x_0 \in S^1\) such that \(h(x_0) = x_0\).
3. Existence of an antipodal point:There exists \(x_1 \in S^1\) such that \(h(x_1) = -x_1\).Proof:
3.1. Suppose for contradiction that \(h(x) \neq -x\) for all \(x \in S^1\).
3.2. Then for all \(x \in S^1\) and \(t \in [0, 1]\), \((1-t)h(x) + tx \neq 0\).(If \((1-t)h(x) = -tx\), taking norms gives \(1-t = t \implies t = 1/2 \implies h(x) = -x\), a contradiction).
3.3. Define the straight-line homotopy \(K: S^1 \times [0, 1] \to S^1\) by:\[K(x, t) = \frac{(1-t)h(x) + tx}{\|(1-t)h(x) + tx\|}.\]
3.4. \(K\) is a continuous homotopy between \(h\) and the identity map \(\operatorname{id}_{S^1}\).
3.5. Thus \(\deg(h) = \deg(\operatorname{id}_{S^1}) = 1\), again strictly contradicting \(\deg(h) = 0\).
3.6. Thus there must exist \(x_1 \in S^1\) such that \(h(x_1) = -x_1\).
4. Q.E.D.Steps 2 and 3 give the fixed point and the point sent to its antipode.