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Prove or disprove:

If $M, N$ are R-module and for all $P$ R-module $Hom(M,P) \simeq Hom(N,P)$ then $M\simeq N$

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 This follows from Yoneda lemma if you assume this isomorphisms are natural. – Piotr Pstrągowski Jul 5 at 20:04
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 And if you do not assume the isomorphisms between Hom-sets to be natural, then for example over a field the question boils down to whether it is possible for two non-isomorphic vector spaces to have isomorphic duals. Over the field with two elements this is simply a question about the cardinality of power sets, which might very well be independant of ZFC. – Piotr Pstrągowski Jul 5 at 20:14
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 I suggest "Are modules isomorphic if their Hom-sets are all isomorphic?" (or something like that). – Mark Grant Jul 6 at 11:36
@Piotr: could you please explain the "boils down" a bit further? The implication "isomorphic duals" $\implies$ "all hom spaces isomorphic" seems to require some implication of the sort $2^\kappa=2^\lambda\implies \alpha^\kappa=\alpha^\lambda$ for all cardinals $\alpha$. Is this true? – a-fortiori Jul 6 at 11:52
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 I've cleaned up this comment thread; the removed comments are copied at meta.mathoverflow.net/discussion/1403/…. – Anton Geraschenko♦ Jul 6 at 17:11

closed as too localized by Fernando Muro, Vladimir Dotsenko, Martin Brandenburg, Dan Petersen, Mark Sapir Jul 6 at 20:50

1 Answer

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K. Bongartz, "A generalization of a theorem of Auslander":

Let R be a commutative ring and A an abelian R-linear category such that each morphism set in A has finite length as an R-module. Let C be a full subcategory of A closed under direct sums and kernels. Then two objects M and N of C are isomorphic if and only if the lengths of Hom(M, X) and Hom(N, X) coincide for all X in C.

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