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 ?

• The hint suggests that to a torsion $A$, associate the group ${\rm Hom}(A,{\mathbf Q}/{\mathbf Z})$. To a profinite abelian group $G$, look at ${\rm Hom}(G,{\mathbf Q}/{\mathbf Z})$. (Maybe you should also review what Pontryagin duality is generally saying beyond the confines of the torsion and profinite cases.) – KCd Mar 31 '12 at 15:18
• Thanks, but I'm still stuck. I must admit, I'm not very good at this subject (I'm trying to solve as many exercises as I can about Galois Cohomology in this book, not always with success). Why is $\hom(G,\mathbb{Q}/\mathbb{Z})$ a torsion abelian group ? – KevinDL Apr 1 '12 at 8:48
• I am a little confused about your proof showing that a torsion abelian group is profinite. Don't you also need to show that they are equivalent as toplogical groups, since we are talking about category of profinite groups where morphsims are continuous homomorphisms. – user2902293 Jun 23 '15 at 12:54
• Possibly related: math.stackexchange.com/questions/1036377 – Watson Nov 26 '18 at 10:36

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.
• The third equality above is non-trivial, i.e., it is not immediately obvious that you can exchange the two limits. There is a natural injective (assuming $G\rightarrow G_i$ is surjective, which you can) map from each $\mathrm{Hom}(G_i,\mathbb{R}/\mathbb{Z})$ to $\mathrm{Hom}_{cts}(G,\mathbb{R}/\mathbb{Z})$, but to prove the induced map from the direct limit is surjective, you need to use that $G$ is profinite. Why must a continuous homomorphism from a profinite group $G$ to $\mathbb{R}/\mathbb{Z}$ factor through a finite quotient of $G$? This is because – Keenan Kidwell Apr 4 '12 at 3:30
• $\mathbb{R}/\mathbb{Z}$ has "no small subgroups," meaning that there is a neighborhood $U$ of the identity in $\mathbb{R}/\mathbb{Z}$ which contains no non-trivial subgroups. The inverse image in of $U$ in $G$ is an open neighborhood of the identity in $G$, so, because $G$ is profinite, there is an open normal subgroup $N$ of $G$ contained in the inverse image. It follows that $N$ is contained in the kernel of $G$ (its image is a subgroup of $U$, and so must be trivial), so your homomorphism factors through the finite quotient $G/N$. You might also want to check that the algebraic isomorphism – Keenan Kidwell Apr 4 '12 at 3:32
• In the second comment above this one, it should say "...kernel of the homomorphism" instead of "...kernel of $G$." – Keenan Kidwell Apr 4 '12 at 3:53