In which categories does $|\text{Hom}(A,B)| = |\text{Hom}(B,A)| \neq 0$ and $|\text{Hom}(A,A)| = |\text{Hom}(B,B)|$ imply that $A$ and $B$ are isomorphic?
When does $|\mathrm{Hom}(A,B)| = |\mathrm{Hom}(B,A)|$ imply $A \simeq B$?
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category-theory
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5Is there some intuition which leads you to ask this question? It's not true in some of the easy categories one normally thinks of: in the category of sets, take $A = 4$ and $B = 2$, and in the category of groups, take $A = \mathbb{Z}/2\mathbb{Z}$ and $B = \mathbb{Z}/3\mathbb{Z}$. It is true in preorder categories though. – 2011-10-30
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0@Zhen Lin: Indeed, I had in mind as an intuition the category of sets, but your example shows that the preconditions are too weak. For my intuitions's sake it's OK to strengthen it by $|\text{Hom}(A,A)| = |\text{Hom}(B,B)|$ (as I did above). – 2011-10-31
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1Category theory is not enriched set theory. Consider the category which has two objects $A$ and $B$, each with a single absorbing non-identity morphism $0$ and morphisms $A \to B$ and $B \to A$ whose composition either way is $0$. By construction there are no non-trivial isomorphisms, but all your conditions are satisfied. – 2011-10-31
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0OK, this is a counter-example, but I asked for categories where the implication *does* hold. – 2011-10-31
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1I'm not even convinced it holds for infinite sets. Asaf probably knows. If you have natural isomorphisms $\textrm{Hom}(A, -) \cong \textrm{Hom}(B, -)$ then $A \cong B$ by Yoneda's lemma. I think that's the best one could hope for. – 2011-10-31