For each of the following \(f: S^1 \rightarrow S^1\), identify the corresponding \(f_*: \mathbb{Z} \to \mathbb{Z}\):
\(z\mapsto z^n\)
\(\bar{x} \mapsto -\bar{x}\)
\(e^{i\theta} \mapsto e^{2\pi i\sin\theta}\)
Solution
1. For \(f(z)=z^n\), the induced map is\[ \boxed{f_*:\mathbb Z\to\mathbb Z,\qquad k\mapsto nk.} \]
The map \(z\mapsto z^n\) has degree \(n\). Under the standard identification \(\pi_1(S^1)\cong\mathbb Z\), a degree-\(n\) map sends the generator to \(n\) times the generator.
2. For the antipodal map \(f(z)=-z\), the induced map is the identity:\[ \boxed{f_*(k)=k.} \]
The map \(z\mapsto-z=e^{i\pi}z\) is a rotation of the circle, hence is homotopic to the identity through rotations \(H_t(z)=e^{i\pi t}z\). Therefore it has degree \(1\) and induces the identity on \(\pi_1(S^1)\).
3. For\[ f(e^{i\theta})=e^{2\pi i\sin\theta}, \]the induced map is zero:\[ \boxed{f_*(k)=0.} \]
The map has a global real-valued lift through the universal covering \(\mathbb R\to S^1\), namely \[ \widetilde f(e^{i\theta})=2\pi\sin\theta. \] Equivalently, the homotopy \[ H_t(e^{i\theta})=e^{2\pi i t\sin\theta} \] contracts \(f\) to the constant map \(1\). Hence \(f\) has degree \(0\) and induces the zero homomorphism on \(\pi_1(S^1)\).