I am trying to figure out the following easy example of Pontryagin duality (see Pontrjagin duality for profinite and torsion abelian groups).
For an abelian group $A$ (torsion or profinite), let its dual $A^D$ be the group of homomorphisms $\operatorname{Hom}_\mathbb{Z} (A, \mathbb{Q}/\mathbb{Z})$.
For example, the dual of $\mathbb{Q}/\mathbb{Z}$ are the profinite integers $\hat{\mathbb{Z}}$: $$\operatorname{Hom}_\mathbb{Z} (\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \cong \operatorname{Hom}_\mathbb{Z} (\varinjlim_m \mathbb{Z}/m\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \cong \varprojlim_m \operatorname{Hom}_\mathbb{Z} (\mathbb{Z}/m\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \cong \varprojlim_m \mathbb{Z}/m\mathbb{Z} \cong \hat{\mathbb{Z}},$$ as the $m$-torsion part of $\mathbb{Q}/\mathbb{Z}$ is given by $\left[\frac{i}{m}\right]$ for $i = 0, 1, \ldots, m-1$. So we have $$\mathbb{Q}/\mathbb{Z}^D \cong \hat{\mathbb{Z}}.$$
Then we should have $\operatorname{Hom}_\mathbb{Z} (\hat{\mathbb{Z}}, \mathbb{Q}/\mathbb{Z}) \cong \mathbb{Q}/\mathbb{Z}$, but what is the right and easy proof? We just saw that $\hat{\mathbb{Z}}^D \cong (\mathbb{Q}/\mathbb{Z}^D)^D$, but how do I see directly that $(\mathbb{Q}/\mathbb{Z}^D)^D \cong \mathbb{Q}/\mathbb{Z}$ without invoking the Pontryagin duality?