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.

  • $\begingroup$ It highly depends on $B$. If $B = \mathbb{Q}/\mathbb{Z}$, for instance, then yes. Actually in this case $\nu$ is a natural isomorphism. $\endgroup$ – user 1987 Jul 8 '17 at 23:56
  • $\begingroup$ The title is about rings, but the question is about abelian groups. We get a ring (the endomorphism ring) ony for $A=B$. $\endgroup$ – Berci Jul 8 '17 at 23:56
  • $\begingroup$ @Berci Yes, thank you I was not thinking clearly. $\endgroup$ – K.Power Jul 8 '17 at 23:59
  • $\begingroup$ @user1987 what do you mean by natural isomorphism? $\endgroup$ – K.Power Jul 9 '17 at 0:00
  • $\begingroup$ @K.Power If you haven't learned that, then nevermind. Just keep in mind that $\nu$ is not only surjective, but also isomorphism. $\endgroup$ – user 1987 Jul 9 '17 at 0:01

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.

  • $\begingroup$ I can't claim that I fully understand your counter examples just, but I thank you for them. At least now I know I have to seek a different approach to the problem that inspired this question. $\endgroup$ – K.Power Jul 10 '17 at 18:39

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