Here we go:
Let \(\alpha(t) = e^{2\pi it}\) where \(t \in [0, 1]\), be a loop in \(S^1\) parameterized by \(t\), which goes around \(S^1\) exactly once. Then under the map \(f: z \mapsto z^n\), we obtain \(f(\alpha(t)) = e^{2\pi n i t}\) where \(t \in [0,1]\). This resulting loop then goes around \(S^1\) \(n\) times, so the induced homomorphism on \(\pi_1(S^1) = \ZZ\) is the map \(f^*: \ZZ \into \ZZ\) given by \(f^*(a) = na\).
Define \(\alpha\) as above, and define \(f: S^1 \into S^1\) to be the antipodal map, so \(f(z) = -z\) for \(z \in S^1 \subset \CC\). We then left \(\alpha\) to the fundamental group, and define \(f_*([\alpha]) = [f \circ \alpha]\). Computing, we have \((f\circ \alpha)(t) = f(\alpha(t)) = -e^{2\pi i t}\). Where \(\alpha(0) = \alpha(1) = 1 + 0i\), we have \((f\circ \alpha)(0) = (f\circ \alpha)(1) = -1 + 0i\). But note that \(\alpha\) was a counter-clockwise loop in \(S^1\), and the image of \(\alpha\) is also a counter-clockwise loop. So this maps the generator \([\alpha] \in \pi_1(S^1, 1)\) to the generator \([\alpha'] \in \pi_1(S^1, -1)\). But since \(S^1\) is path-connected, the fundamental groups at these two base points are isomorphic. Alternatively: the antipodal map on \(S^1\) is homotopic to the identity map (since \(n=1\) is odd), so \([f\circ \alpha] = [f][\alpha] = [\id][\alpha] = [\alpha]\), so the induced homomorphism on \(\pi_1(S^1)\) is the identity map.
Let \(\alpha(t) = e^{it}\) where \(t\in [0, 2\pi]\) be a counter-clockwise loop in \(S^1\); then \([\alpha]\) generates the fundamental group. Then \(f^*([\alpha]) = [(f\circ \alpha) (t)] = [e^{it} \mapsto e^{2\pi i \sin t}]\). Then just consider how \(\sin\) behaves in each quadrant. In quadrant 1, as \(t\) ranges from \(0, \pi/2\) then \(\sin t\) ranges from 0 to 1, so \(\alpha\) is exactly traced out. In quadrant two, \(\bar\alpha\) is traced out, since \(\sin t\) decreases from 1 to 0. This happens again in the bottom quadrants, so we have \(f^*([\alpha]) = [\alpha\bar\alpha\alpha\bar\alpha] = [\alpha][\alpha]^{-1}[\alpha][\alpha]^{-1} = [\id]\). But the identity element in \(\ZZ\) is 0, so the induced homomorphism on \(\ZZ\) is \(f^*(a) = 0\), the homomorphism sending everything to 0.
Solution
Goal. Compute the induced maps on \(\pi_1(S^1) \cong \ZZ\) for the three maps \(z \mapsto z^n\), the antipodal map, and \(z \mapsto e^{2\pi i \sin(\arg z)}\).
1. \(f(z) = z^n\) induces \(f_*: \ZZ \to \ZZ\), \(a \mapsto na\).
1.1. The generator \([\alpha]\) of \(\pi_1(S^1)\) is the loop \(\alpha(t) = e^{2\pi i t}\).
this loop winds once around \(S^1\).
1.2. \(f \circ \alpha\) is the loop \(t \mapsto e^{2\pi i n t}\), which winds \(n\) times.
\(f(\alpha(t)) = (e^{2\pi i t})^n = e^{2\pi i n t}\).
1.3. Hence \(f_*([\alpha]) = n[\alpha]\), so \(f_*(a) = na\).
the winding number of \(f \circ \alpha\) is \(n\).
2. The antipodal map \(f(z) = -z\) induces the identity on \(\pi_1(S^1)\).
2.1. \(f(z) = -z = e^{i\pi} z\) is rotation by \(\pi\).
multiplication by \(-1\) on the unit circle is rotation through angle \(\pi\).
2.2. Rotation by \(\pi\) is homotopic to the identity.
2.3. Hence \(f_* = \id\), i.e. \(f_*(a) = a\).
homotopic maps induce the same homomorphism on \(\pi_1\).
3. \(f(e^{it}) = e^{2\pi i \sin t}\) induces the zero map on \(\pi_1(S^1)\).
3.1. As \(t\) runs \(0 \to \pi/2\), \(\sin t\) runs \(0 \to 1\), so \(f \circ \alpha\) traces the generator \(\alpha\) once.
\(e^{2\pi i \sin t}\) winds once counterclockwise.
3.2. As \(t\) runs \(\pi/2 \to \pi\), \(\sin t\) runs \(1 \to 0\), so \(f \circ \alpha\) traces \(\bar\alpha\) (the reverse loop).
the argument decreases from \(2\pi\) to \(0\).
3.3. On \([\pi,3\pi/2]\), \(\sin t\) decreases from \(0\) to \(-1\), so the image again winds once clockwise; on \([3\pi/2,2\pi]\), \(\sin t\) increases from \(-1\) to \(0\), so the image winds once counterclockwise. Thus the four quadrants contribute winding numbers \(+1,-1,-1,+1\).
The argument \(2\pi\sin t\) changes respectively by \(+2\pi,-2\pi,-2\pi,+2\pi\) on the four quarter-intervals.
3.4. Hence the total winding number is\[ 1-1-1+1=0, \]so \(f\circ\alpha\) is null-homotopic.
Homotopy classes of loops in \(S^1\) are classified by winding number.
3.5. Hence \(f_*([\alpha]) = 0\), so \(f_*(a) = 0\) for all \(a\).
the generator maps to the identity element \(0 \in \ZZ\).
4. Q.E.D.
1, 2, and 3 give the three induced maps.