Show that every homomorphism \(\pi_1(S^1) \to \pi_1(S^1)\) can be realized as the induced homomorphism \(\varphi_*\) of a map \(\varphi: S^1 \to S^1\).
Solution
1. \(\pi_1(S^1) = \ZZ\), so a homomorphism \(\pi_1(S^1) \to \pi_1(S^1)\) is multiplication by some integer \(n\).
\(\operatorname{Hom}(\ZZ, \ZZ) = \ZZ\).
2. Define \(\varphi_n: S^1 \to S^1\) by \(\varphi_n(z) = z^n\) (viewing \(S^1 \subset \CC\)).
definition.
3. \(\varphi_n\) has degree \(n\), so \((\varphi_n)_*\) is multiplication by \(n\) on \(\pi_1(S^1) = \ZZ\).
the map \(z \mapsto z^n\) has degree \(n\), and the induced map on \(\pi_1\) is multiplication by the degree.
4. Hence every homomorphism (multiplication by \(n\)) is realized by \(\varphi_n\).
1 and 3.
5. Q.E.D.
4.