Let $A$, $B$ be abelian groups. Define, for convenience, $A^*:=\operatorname{Hom}(A,B)$. We can define a map $\nu:A\to \operatorname{Hom}(A^*,B)$ by $a\mapsto \nu_a$, where $\nu_a(\phi)=\phi(a)$ for all $a\in A$ and $\phi\in A^*$. It is easy to see that this map is a group homomorphism, but is it surjective? I know it is not injective, picking $B$ to be the trivial group is a counterexample for that, but I suspect it may be surjective. I suspect, however, that defining $\eta:B\to \operatorname{Hom}(A^*,B)$ analogously to $\nu$ gives an injective map that need not be surjective. I would appreciate any counterexamples or proofs to my above two suspicions.
EDIT: As User1987 has correctly deduced, I am particularly interested in the map $\nu:A\to\operatorname{Hom}(A^*,B)$ where $A$ is arbitrary but $B=\mathbb Q/\mathbb Z$. In fact ideally I'd like to show that this map is a group isomorphism. User1987 has kindly given a link to a related result about characters of finite abelian groups, the proof of which I believe I could, with a nontrivial amount of effort, adapt to prove that $\nu$ is an isomorphism (perhaps using the answer to this question. I'd also need to figure out a nice relationship between $\mathbb C^\times$ and $\mathbb Q/\mathbb Z$). Unfortunately I am not at all well acquainted with characters of groups, so I would also appreciate a proof not adapted from the theory of characters of finite abelian groups.
No, the homomorphism is neither injective nor surjective, in general.
In what follows, group will mean abelian group.
In order for it to be injective, the condition is that, for $x\in A$, $x\ne0$, there exists a homomorphism $f\colon A\to B$ with $f(x)\ne0$. For instance, if $B$ is torsionfree and $A$ is torsion, $\nu_A$ cannot be injective.
A simple example when $\nu_A$ is not surjective is $A=\mathbb{Z}$ and $B=\mathbb{Q}/\mathbb{Z}$, because in this case $A^*=\mathbb{Q}/\mathbb{Z}$ and $A^{**}=\hat{\mathbb{Z}}$, the completion of $\mathbb{Z}$ in its natural topology or, equivalently, the product of the groups of $p$-adic integers $\mathbb{Z}_p$ over the set of primes. Such a group is not countable.
Actually, the structure theorem of finitely generated abelian groups tells you that, for $A$ finitely generated and $B=\mathbb{Q}/\mathbb{Z}$, $\nu_A$ is surjective if and only if $A$ is torsion.
Note that if you want that $\nu_A$ is injective for every $A$, you need that $B$ has a direct summand isomorphic to $\mathbb{Q}/\mathbb{Z}$, which is the minimal cogenerator.