- Main Idea: Linear homotopy fails continuity without the condition from (2), so use complex embedding to avoid the origin at \(t=1/2\).
Suppose \(n\) is odd and define \(f:S^n \to S^n\) to be the antipodal map. Since \(n+1\) is even, we have \(n+1 =2m\) for some \(m\in \NN\), so identify \(S^n = S^{2m-1} \subset \RR^{2m} \cong\CC^m\)
Then \(z\in S^n\) can be written as a vector \(z \in \CC^m\) such that \(\norm{z} = 1\).
Then define \(P: \CC^m \to \CC^m\) by \(P(z) = z/\abs{z}\), the projection onto the complex unit sphere, and define \(H: \CC^m \cross I \to \CC^m\) by \(H(z, t) = P(e^{i\pi t}z)\).
This is a homotopy, since \(H(z, 0) = P(z) = z\) (since \(\norm{z} = 1\)), so this is the identity map. We also have \(H(z, 1) = P(-z) = -z\), the antipodal map.
This is well-defined, since \(e^{i\pi t} > 0\) and \(z \neq 0\), so the linear homotopy in ambient \(\CC^m\) avoids the origin and thus the denominator when taking the projection is never zero.
Solution
1. Write \(n=2m-1\) and identify \(S^n\) with the unit sphere in \(\mathbb C^m\).
\(n\) odd means \(n+1=2m\).
2. Define\[H(z,t)=e^{\pi i t}z.\]
Multiplication by a complex number of modulus \(1\) preserves the unit sphere, so \(H:S^n\times I\to S^n\) is continuous.
3. Then \(H(z,0)=z\) and \(H(z,1)=-z\).
\(e^0=1\) and \(e^{\pi i}=-1\).
4. Hence the antipodal map on \(S^n\) is homotopic to the identity for odd \(n\).
The homotopy is 2.