Let $S^1 = \{z \in C | |z| = 1\}$ be the unit circle in the complex field with well-known group structure. Let G be an abelian group and $\hat{G} = \{\chi \in Hom(G, S^1)\}$ be the character group. I know there is a homomorphism of groups
$$ \tau : G \longrightarrow \hat{\hat{G}} $$ $$ g \mapsto \tau(g) : \chi \mapsto \chi(g) $$ I know that $\tau$ is an isomorphism if G is finite. But what if G is an arbitrary abelian group?
Pontryagin duality is pretty restrictive ($G$ has a topology making it locally compact and $\widehat{G}$ is the group of continuous characters, with its own locally compact topology).
In general, with the axiom of choice and without restricting to continuous characters, it fails.
Try with $G=\bigoplus_{n=1}^\infty\Bbb{F}_p$ an infinite but countable $\Bbb{F}_p$ vector space.
$\widehat{G} \cong Hom(G,\Bbb{F}_p)\cong \prod_{n=1}^\infty\Bbb{F}_p$ is uncountable.
And $\widehat{\widehat{G}}$ is uncountable as well.
Without the axiom of choice we can't construct any character of $\Bbb{R/Q}$ nor prove that there are none and so we are stuck.