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Let $G$, $H$ be abelian groups and $f: G\to H$ a homomorphism. Assume that $f^*: \mathrm{Hom}(H, R) \to \mathrm{Hom}(G, R)$ (as morphisms of abelian groups, taking $R$ with its additive group structure) is an isomorphism for all commutative rings (with 1) $R$. Is then $f$ an isomorphism as well?

The question is motivated by a question on homological algebra, see here.

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    $\begingroup$ What is $\mathrm{Hom}(G, R)$ when $G$ is a group and $R$ is a ring? Does one implicitly take the additive group pf $R$? $\endgroup$ May 12, 2016 at 12:51
  • $\begingroup$ Yes, that is what I wanted to express. I edited the question accordingly. $\endgroup$
    – blue
    May 12, 2016 at 12:56

1 Answer 1

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For any abelian group $A$, take $R=\mathbb{Z}\oplus A$ with multiplication given by $(m,a)\cdot(n,b)=(mn,mb+na)$. Then there are isomorphisms $\operatorname{Hom}(G,R)\cong\operatorname{Hom}(G,\mathbb{Z})\oplus\operatorname{Hom}(G,A)$ which are natural in $G$. Thus a map $f:G\to H$ induces an isomorphism $\operatorname{Hom}(H, R) \to \operatorname{Hom}(G, R)$ iff it induces isomorphisms $\operatorname{Hom}(H, \mathbb{Z}) \to \operatorname{Hom}(G, \mathbb{Z})$ and $\operatorname{Hom}(H, A) \to \operatorname{Hom}(G, A)$.

In particular, a map satisfying your hypotheses induces isomorphisms $\operatorname{Hom}(H, A) \to \operatorname{Hom}(G, A)$ for all abelian groups $A$. By Yoneda, it follows that such a map is an isomorphism.

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