Set of homomorphisms on a free abelian group is a free abelian group.

Question

If $G$ is a free abelian group with rank $n$, I need to show that ${\rm Hom}(G,\mathbb{Z})$, set of all homomorphisms is also free abelian group of rank $n$,

My work:

Since $G$ is free abelian group then $G$ can be generated by the set $\{g_1,g_2, \cdots,g_n\}$. Let $f \in{\rm Hom}(G,\mathbb{Z})$, then $f(x)=\sum_{i=1}^n{c_if(g_i)}$, where $x=\sum_{i=1}^n{c_ig_i}$, but this gives a generator set of ${\rm Im}(f)$ and I need to get generators for ${\rm Hom}(G,\mathbb{Z})$, so for arbitrary $f$ we have the same set of generators which is in $\mathbb{Z}$.

Answer 1

The idea is the same used to get a basis for the dual space of a vector space $V$ of finite dimension.

Let us define the homomorphisms $g_i^*\colon G\to \mathbb{Z}$ such that $g_i^*(g_j):=\delta_{ij}$. Then given $f\in Hom(G,\mathbb{Z})$ you have that

$ f=\sum_{g_i}f(g_i)\cdot g_i^*$

Now suppose that $\sum_{g_i}a_ig_i^*=0$. Then $\left(\sum_{g_i}a_ig_i^*\right)(g_j)=a_j=0$ for each $j$. Hence $Hom(G,\mathbb{Z})$ is a free abelian group of rank $n$.