$\DeclareMathOperator\Hom{Hom}$The question is in the title. If the isomorphism $\Hom(A, G) \cong \Hom(B, G)$ is natural in $G$ then this is just the Yoneda Lemma. If $A$ and $B$ are finitely generated this is also true by the structure theorem.

However this sounds like it should be false in general, else it would imply by Yoneda that if $\Hom(A, -)$ and $\Hom(B, -)$ are (a priori not naturally) isomorphic, then they are also isomorphic in a natural way (though possibly by a different set of isomorphisms).

The question of course immediately generalizes to $R$-modules.

Edit: Some context (that isn't really relevant for the question)
I'm interested in this question in light of the universal coefficient theorem for Cohomology. A positive answer would imply that knowing all Cohomology groups of a space, with arbitrary coefficients, would already determine its Homology (although I think it is conceivable that this topological statement can be proven in a different way).

  • $\begingroup$ I 'migrated' this question from math.stackexchange (I deleted the question there) since it had already been asked $\endgroup$ Jun 16, 2020 at 21:19
  • 3
    $\begingroup$ If $A=C_2^{(\aleph_0)}$ and $B=C_2^{(\aleph_1)}$, the these are isomorphic for every $G$ iff $\alpha^{\aleph_0}=\alpha^{\aleph_1}$ for all cardinal $\alpha$. This is false under CH (taking $\alpha=2$) but I don't know if it's consistent in general (=in ZFC). In case that yields a counterexample, one might restrict to $A,B$ countable, or assume GCH. $\endgroup$
    – YCor
    Jun 16, 2020 at 22:33
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    $\begingroup$ @YCor I think Eric Wofsey commented on mathSE (on a deleted answer to the linked question) that there always is a cardinal $\alpha$ such that $\alpha^\kappa \neq \alpha^\lambda$ for $\kappa \neq \lambda$ $\endgroup$ Jun 16, 2020 at 22:35
  • $\begingroup$ Ah, thanks indeed, I should have checked before. So the question has a positive answer when $A$ is an elementary abelian $p$-group for some prime $p$ (i.e., if for every $B$ the question has a positive answer for this given $A$); we can say that $A$ is "recognizable". $\endgroup$
    – YCor
    Jun 17, 2020 at 8:24
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    $\begingroup$ If $G=U(1)$ is a topological group, then this follows from Pontryagin duality (with $G=S^1=U(1)$, and the discrete topology on $A$ and $B$), where one remembers the topology on $Hom(A,G)$. en.wikipedia.org/wiki/… $\endgroup$
    – Ian Agol
    Jun 26, 2020 at 18:02

1 Answer 1


I just stumbled across the answer to this in Fuchs' 2015 book on Abelian Groups.

The papers

Hill, Paul, Two problems of Fuchs concerning tor and hom, J. Algebra 19, 379-383 (1971). ZBL0228.20027.

and a 1974 paper of Sebel'din that I've not been able to find, and seems to be in Russian, which means that I probably wouldn't be able to understand it even if I did:

Homomorphism groups of complete direct sums of torsion-free abelian groups of rank 1 [Russian]. Tomsk. Gos. Univ., Tomsk 1, 121–122 (1974)

give counterexamples where the groups $A$ and $B$ are $p$-groups and torsion-free groups respectively.

But from Fuchs' account, Sebel'din's example is fairly simple.

Let $S=\mathbb{Z}^{(\omega)}$ be the direct sum of countably many copies of $\mathbb{Z}$, and let $A=S\oplus\mathbb{Q}$ and $B=S\oplus\mathbb{Q}\oplus\mathbb{Q}$.

Then $A\not\cong B$, but $\operatorname{Hom}(A,G)\cong\operatorname{Hom}(B,G)$ for all abelian groups $G$.

It's a fun exercise to verify this, and for me it was an ISHTOT moment when I did, so I won't spoil it for you.


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