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LSpice
LSpice
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If $A$A$, B$$B$ are abelian groups such that $\mathrm{Hom}(A, G) \cong \mathrm{Hom}(B, G)$ for all abelian groups $G$, must $A$ and $B$ be isomorphic?

The$\DeclareMathOperator\Hom{Hom}$The question is in the title. If the isomorphism $\mathrm{Hom}(A, G) \cong \mathrm{Hom}(B, G)$$\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 $\mathrm{Hom}(A, -)$$\Hom(A, -)$ and $\mathrm{Hom}(B, -)$$\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 immeadiatelyimmediately 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).

If $A, B$ are abelian groups such that $\mathrm{Hom}(A, G) \cong \mathrm{Hom}(B, G)$ for all abelian groups $G$, must $A$ and $B$ be isomorphic?

The question is in the title. If the isomorphism $\mathrm{Hom}(A, G) \cong \mathrm{Hom}(B, G)$ 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 $\mathrm{Hom}(A, -)$ and $\mathrm{Hom}(B, -)$ 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 immeadiately 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).

If $A$, $B$ are abelian groups such that $\mathrm{Hom}(A, G) \cong \mathrm{Hom}(B, G)$ for all abelian groups $G$, must $A$ and $B$ be isomorphic?

$\DeclareMathOperator\Hom{Hom}$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).

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Carlos Esparza
Carlos Esparza
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If $A, B$ are abelian groups such that $\mathrm{Hom}(A, G) \cong \mathrm{Hom}(B, G)$ for all abelian groups $G$, must $A$ and $B$ be isomorphic?

The question is in the title. If the isomorphism $\mathrm{Hom}(A, G) \cong \mathrm{Hom}(B, G)$ 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 $\mathrm{Hom}(A, -)$ and $\mathrm{Hom}(B, -)$ 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 immeadiately 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).