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Let $C$ be a category with finite hom-sets. Suppose that $X$ and $Y$ are objects in $C$ such that $C(Z,X)\cong C(Z,Y)$ for any Z (with no naturality condition). For which categories $C$ does it follow that $X \cong Y$? (Of course, it is true for posets).

A somewhat related question is the following.

Let $C$ be a symmetric monoidal closed category. Suppose that $X$ and $Y$ are objects in $C$ such that $[X,Z]\cong [Y,Z]$ for any Z (with no naturality condition). For which categories $C$ does it follow that $X \cong Y$?

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    $\begingroup$ mathoverflow.net/a/269745 seems to gesture towards an (unpublished) answer $\endgroup$
    – Martti Karvonen
    Commented Sep 29, 2020 at 20:21
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    $\begingroup$ after a bit of digging, I discovered that a sufficient condition based on similar factorization properties is proven in A. Pultr: Isomorphism types of objects in categories determined by numbers of morphisms. Acta Sci. Math. Szeged35(1973), 155–160 $\endgroup$
    – Martti Karvonen
    Commented Sep 29, 2020 at 20:32
  • $\begingroup$ @martti : thanks for the very useful references. In fact I had a similar idea of proof based on factorization. So it seems that it is true for instance for $C = Set_f^G$ for any finite category $G$ (as well as for their dual). $\endgroup$
    – Claudio Pisani
    Commented Sep 30, 2020 at 18:13

1 Answer 1

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The first question is the main topic of two recent papers:

As far as I understand, the theorems give orthogonal sufficient conditions: it's unclear whether there is a general theorem subsuming both.

I don't believe there are answers for enriched categories yet.

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  • $\begingroup$ One should also mention Lovàsz's hom-counting theorem by inclusion-exclusion principle by Shoma Fujino and Makoto Matsumoto (arxiv.org/abs/2206.01994). $\endgroup$
    – Ivan Di Liberti
    Commented Jul 13, 2022 at 20:08
  • $\begingroup$ And Locally-finite extensive categories, their semi-rings, and decomposition to connected objects by the same authors (arxiv.org/abs/2207.05702). $\endgroup$
    – Ivan Di Liberti
    Commented Jul 13, 2022 at 20:09
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    $\begingroup$ @IvanDiLiberti: thanks, I forgot about that paper – I've updated my answer to include it. $\endgroup$
    – varkor
    Commented Jul 13, 2022 at 21:07

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