Show that the group structure on \(S^1\) coming from multiplication in \(\mathbb{C}\) induces a group structure on \(\langle X, S^1 \rangle\) such that the bijection \(\langle X, S^1 \rangle \to H^1(X; \mathbb{Z})\) of Theorem 4.57 is an isomorphism.
Solution
For maps \(f,g:X\to S^1\), define \[ (fg)(x)=f(x)g(x) \] using multiplication in \(S^1\subset\mathbb C\). Pointwise multiplication and inversion descend to homotopy classes, giving a group structure on \(\langle X,S^1\rangle\).
Let \[ u\in H^1(S^1;\mathbb Z) \] be the canonical generator. For the multiplication \[ \mu:S^1\times S^1\to S^1, \] restriction to each factor has degree \(1\), so by Künneth \[ \mu^*u=\operatorname{pr}_1^*u+\operatorname{pr}_2^*u. \] Therefore \[ (fg)^*u=(f,g)^*\mu^*u=f^*u+g^*u. \] Theorem 4.57 identifies \([f]\) with \(f^*u\), hence this bijection preserves addition. Thus \[ \boxed{\langle X,S^1\rangle\cong H^1(X;\mathbb Z)} \] as groups.