(This is a somewhat lazy question which came up as I'm reading about Pontryagin duality for the first time)

For a locally compact abelian topological group $G$, its Pontryagin dual is the group of continuous homomorphisms $$G^* := \text{Hom}(G,\mathbb{R}/\mathbb{Z})$$ with the compact-open topology ($\mathbb{R}/\mathbb{Z}$ is given the usual topology).

If $G$ is profinite abelian, then any such homomorphism cannot be surjective, and hence its image is a proper closed subgroup, hence finite, and hence there is a natural isomorphism $$\text{Hom}(G,\mathbb{R}/\mathbb{Z}) \cong \text{Hom}(G,\mathbb{Q}/\mathbb{Z})$$ where $\mathbb{Q}/\mathbb{Z}$ is given the topology induced from the inclusion $\mathbb{Q}/\mathbb{Z}\subset\mathbb{R}/\mathbb{Z}$. The same isomorphism holds if $G$ is discrete torsion abelian.

This topology on $\mathbb{Q}/\mathbb{Z}$ is clearly totally disconnected, but neither discrete nor compact, hence not profinite.

On the other hand, there is another appearance of $\mathbb{Q}/\mathbb{Z}$, now viewed as an ind-finite abelian topological group, hence a discrete abelian group, namely under the identification $\widehat{\mathbb{Z}}^*\cong \mathbb{Q}/\mathbb{Z}$.

Can some wise people say a word about these two topologies on $\mathbb{Q}/\mathbb{Z}$ and how they're related?

Certainly if $G$ is discrete abelian, then $\text{Hom}(G,\mathbb{Q}/\mathbb{Z})$ does not depend on the topology on $\mathbb{Q}/\mathbb{Z}$, and if $G = \varprojlim G_i$ is profinite, then we have $$\text{Hom}(\varprojlim G_i,\mathbb{Q}/\mathbb{Z}) = \varinjlim\text{Hom}(G_i,\mathbb{Q}/\mathbb{Z}) = \varinjlim\text{Hom}(G_i,\mathbb{Z}/|G_i|\mathbb{Z})$$ and hence the result seems to also be independent of the choice of topology on $\mathbb{Q}/\mathbb{Z}$.

Is there any situation where one must be careful which topology to consider on $\mathbb{Q}/\mathbb{Z}$?

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    $\begingroup$ If $G$ is a locally compact group, every closed subgroup of countable index is open. It follows that every continuous homomorphism into $(Q/Z)_i$ (endowed with the topology of inclusion in $R/Z$) is continuous into $(Q/Z)_d$ (discrete topology). However, the topology (uniform convergence on compact subsets) of $Hom(G,(Q/Z)_i)$ and $Hom(G,(Q/Z)_d)$ can differ, e.g., when $G$ is discrete and infinite cyclic. When $G$ is profinite, or more generally locally elliptic, however, all homomorphisms $Hom(G,(Q/Z)_d)\to Hom(G,(Q/Z)_i)\to Hom(G,R/Z)$ are topological isomorphisms. $\endgroup$ – YCor Oct 3 '18 at 22:16
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    $\begingroup$ Thanks for this, @YCor. I was disquieted by the idea that differing topologies on $\Bbb Q/\Bbb Z$ could matter, but l lacked the clarity to see why. $\endgroup$ – Lubin Oct 3 '18 at 22:33

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