if $G$ is a free abelian group with basis {${a_\alpha}$} then given the elements {${y_\alpha}$} of an abelian group $H$, there are homomorphisms $h_\alpha : G_\alpha \to H$ such that $h(a_\alpha)=y_\alpha$ (because $G_\alpha$ is infinite cyclic).
which $G_\alpha$ is the subgroup generated by $a_\alpha$
the above is part of a proof from Munkres Topology( page 411), can someone explain how the infinite cyclicness of these subgroups would imply the existence of such homomorphisms
Let $G = \langle a_\alpha \rangle_\alpha$ be free abelian and for fixed $\alpha$, let $G_\alpha = \langle a_\alpha \rangle \subseteq G$. Define $h_\alpha: G_\alpha \to H$ by $h_\alpha({a_\alpha}^n) = {y_\alpha}^n$ for $n \in \mathbb{Z}$. Then $h_\alpha$ is well-defined as a mapping between sets because $G_\alpha$ is infinite cyclic and it is a group homomorphism because for $m, n \in \mathbb{Z}$ we have $h_\alpha({a_\alpha}^n \cdot {a_\alpha}^m) = h_\alpha({a_\alpha}^{n+m}) \overset{\text{Def. of } h_\alpha}{=} {y_\alpha}^{n+m} = {y_\alpha}^n \cdot {y_\alpha}^m \overset{\text{Def. of } h_\alpha}{=} h_\alpha({a_\alpha}^n) \cdot h_\alpha({a_\alpha}^m)$.