Topology in the context of Pontryagin dual

Question

Let $A$ be an abelian group.

The definition of the Pontryagin dual of $A$ is $\text{Hom}_{\text{conti}}(A, \mathbb{Q}/\mathbb{Z})$.

In this context, what are the topologies on $A$ and $\mathbb{Q}/\mathbb{Z}$?

Is it the cofinite topology on $A$ and the discrete topology on $\mathbb{Q}/\mathbb{Z}$, or does that depend on the situation?

in the case where $A=\widehat{E(\mathbb{Q})}$($\widehat{E(\Bbb{Q})}$ is profinite completion of $E(\Bbb{Q})$ cf. Profinite completion of local Mordell-Weil group) and $E$ is an elliptic curve defined over $\mathbb{Q}$, $A$ is an infinite group. However, we usually consider its Pontryagin dual. I'm especially interested in this example.

Supplementary information in response to comments：

In the case where $A=\widehat{E(\mathbb{Q})}$ and $E$ is an elliptic curve defined over $\mathbb{Q}$, $A$ is an infinite group. However, we usually consider its Pontryagin dual. I'm especially interested in this example.

Answer 1

First, a clarification following the comments about Pontryagin duality. It indeed concerns a duality theory on locally compact abelian groups, and the dual of such a group $A$ is $\mathrm{Hom}_{C^0}(A,\mathbb{R}/\mathbb{Z})$ endowed with the compact open topology.

In particular, this duality preserves finite groups and exchanges compact groups and discrete groups.

Note that any continuous homomorphism from a pro-finite group $A$ into $\mathbb{R}/\mathbb{Z}$ must have open kernel (that is because $A$ has arbitrarily small open subgroups, while there is a neighborhood of $0$ in $\mathbb{R}/\mathbb{Z}$ containing no non-trivial subgroup). In particular, its image is finite, and the homomorphism factors through $\mathbb{Q}/\mathbb{Z}$ endowed with its discrete topology.

This is why we can say that the Pontryagin dual of a profinite group $A$ is exactly $\mathrm{Hom}_{C^0}(A,\mathbb{Q}/\mathbb{Z})$, where $\mathbb{Q}/\mathbb{Z}$ is discrete.

And indeed, it seems that the topology on $\widehat{E(\mathbb{Q})}$ should be the pro-finite topology, since there’s no other reasonable candidate.

Notes:

1. in particular, no weaker topology on $\widehat{E(\mathbb{Q})}$ is Hausdorff (let alone locally compact),

2. in the exact sequence, the only reasonable topology to put on the direct sum is discrete (see Arithmetic Duality Theorems, Corollary I.3.4), so the topology on $\widehat{E(\mathbb{Q})}^{\ast}$ is expected to be discrete, hence $\widehat{E(\mathbb{Q})}$ has to be compact.