Calculating $\text{Hom}(\mathbb{Z}^n, \mathbb{Z})$

Question

I'm aware that $\text{Hom}(\mathbb{Z}, \mathbb{Z}) \cong \mathbb{Z}$ and that calculating $\text{Hom}$ is dependent on where the generators of the domain get mapped to. I'm struggling to determine Hom in the particular case of $$\text{Hom}(\mathbb{Z}^n, \mathbb{Z}),$$ where $\mathbb{Z}^n = \bigoplus_{i=1}^n \mathbb{Z}_i$.

Answer 1

$\mathbb{Z}^n$ is the free Abelian group on $n$ generators. Thus for any Abelian group $A$, $$\text{Hom}(\mathbb{Z}^n,A)=A^n$$ Since the $n$ generators can be mapped to any elements.

Answer 2

Alternatively to the other answer, if you want to apply your knowledge that $\operatorname{Hom} (\mathbb{Z},\mathbb{Z}) \cong \mathbb{Z}$, note that $\operatorname{Hom} (-,-)$ is additive in both arguments (= preserves binary "$\oplus$"):

\begin{align*} \operatorname{Hom} (A\oplus B, C) & \cong \operatorname{Hom} (A, C) \oplus \operatorname{Hom} (B, C),\\ \operatorname{Hom} (A, B\oplus C) & \cong \operatorname{Hom} (A, B) \oplus \operatorname{Hom} (A, C). \end{align*}

In your case, $$\operatorname{Hom} (\mathbb{Z}^{\oplus n}, A) \cong \operatorname{Hom} (\mathbb{Z}, A)^{\oplus n} \cong A^{\oplus n}.$$

(This argument is not really necessary here, but often helps to calculate other $\operatorname{Hom}$'s.)