Proof. First, note that $\mathbb{Q}/\mathbb{Z}$ is an injective Abelian group, since it is divisible. For any Abelian group $A$ , let $A^* = \mathrm{Hom}(A, \mathbb{Q}/\mathbb{Z})$ .

We define $$ f: A \to \mathrm{Hom}(A^*, \mathbb{Q}/\mathbb{Z}), a \mapsto f_a, $$ where $f_a$ ist defined as $$ f_a : A^* \to \mathbb{Q}/\mathbb{Z}, \varphi \mapsto \varphi(a). $$

$f$ is one-to-one, for if $f_a = 0$ , i.e. $\varphi(a) = 0$ for all $\varphi \in A^*$ , it follows $a = 0$ . Indeed, if $a \neq 0$ , let the order of $a$ be denoted by $n$ , and for any $q \in \mathbb{Q}/\mathbb{Z}$ with order $n$ , the homomorphism defined by $a \mapsto q$ is well-defined on the subgroup generated by $a$ , and since $\mathbb{Q}/\mathbb{Z}$ is injective, it induces a homomorphism $A \to \mathbb{Q}/\mathbb{Z}$ which is different from zero.

Now, if we chose a presentation $\bigoplus_{i \in I} \mathbb{Z} \twoheadrightarrow A^*$ , we get an embedding $\mathrm{Hom}(A^*, \mathbb{Q}/\mathbb{Z}) \hookrightarrow \mathrm{Hom}(\bigoplus_{i \in I} \mathbb{Z}, \mathbb{Q}/\mathbb{Z})$ , where the latter is clearly isomorphic to the direct product $\prod_{i \in I} \mathbb{Q}/\mathbb{Z}$ . This last group is injective as a direct product of injectives.