Pontrjagin duality for profinite and torsion abelian groups

I'm having trouble proving exercise 6.11.3 of "Introduction to homological algebra" by Weibel. I need to show that the category of torsion abelian groups is dual to the category of profinite abelian groups. It also gives a hint to show that $A$ is a torsion abelian group iff $\hom(A,\mathbb{Q}/\mathbb{Z})$ is a profinite group.

I'm stuck with the hint. I've proved that the torsion abelian group part implies that $$\hom(A,\mathbb{Q}/\mathbb{Z}) = \lim_{\leftarrow} \hom(H,\mathbb{Q}/\mathbb{Z}),$$ with $H$ going through all finite subgroups of $A$ with restriction maps as homomorphisms in the obvious way. I have absolutely no idea how to proof the other implication. I also don't see how this is going to help to associate a torsion abelian group to a profinite abelian group to make the duality.

Any thoughts ?

I've done some reading about Pontryagin duality and this is what I've come up with: Pontryagin duality says that $\hom(\hom(G,\mathbb{R}/\mathbb{Z}),\mathbb{R}/\mathbb{Z}) = G$ in case of $G$ being a locally compact abelian group and $\hom$ standing for all continuous group homomorphisms. Now, in case of $G$ being a torsion abelian group or a profinite abelian group, we can change $\mathbb{R}$ by $\mathbb{Q}$ (If $G$ is torsion, this is trivial, since every image of $g \in G$ has to have finite order. If $G$ is profinite, this follows from $$ \hom(G,\mathbb{R}/\mathbb{Z}) = \hom(\varprojlim G_i,\mathbb{R}/\mathbb{Z}) = \varinjlim\hom(G_i,\mathbb{R}/\mathbb{Z}) = \varinjlim\hom(G_i,\mathbb{Q}/\mathbb{Z})\\ = \hom(G,\mathbb{Q}/\mathbb{Z})$$, since all $G_i$ are finite). Now, since $\varinjlim\hom(G_i,\mathbb{Q}/\mathbb{Z})$ is a quotient $\oplus_i\hom(G_i,\mathbb{Q}/\mathbb{Z})$, which is clearly a torsion abelian group, we are done.

Does this seem correct ?