I encountered the following situation:

Let C be an abelian category and X, Y be objects of C verifying $ Hom_C(X, Z)\cong Hom_C(Y, Z) $ For any object Z of C.

Does it follow that $ X \cong Y$?

My belief is that in general this not true. The real question is what additional hypothesis have to be satisfied by C in order to obtain isomorphic objects?

For example, if C closed then probably these two objects are isomorphic.

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    $\begingroup$ If the isomorphisms are natural, then you can use Yoneda to get $X\simeq Y$. $\endgroup$
    – David Roberts
    May 9, 2013 at 6:38
  • $\begingroup$ Yes, they are naturals and nowI realize that you are right, $Nat(h^X, h^Y)$ is in bijection with $ Hom_C(X, Y)$. Since this bijection behaves well with composition then it follows that to my natural transformation should correspond an isomorphism. Therefore I should impose only the conditions necessary to be satisfied Yoneda' s lemma, i.e the hom spaces to be sets. Thank you very much! $\endgroup$
    – Stef
    May 9, 2013 at 6:53
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    $\begingroup$ If you're dealing with a non-locally small abelian category, then your problems are bigger than deciding whether two objects are isomorphic :-) $\endgroup$
    – David Roberts
    May 9, 2013 at 7:46
  • $\begingroup$ I would be interested whether there exists a counterexample if the isomorphism is not assumed natural. if C doesn't have to be abelian it is easy: Take C to be the category freely generated by two objects and two arrows between them (in different directions). Then for any $X$,$Y$ (possibly $X=Y$) in C we have $Hom(X,Y)\cong \aleph_0$, but the two objects in C are not isomorphic. $\endgroup$
    – Toink
    May 9, 2013 at 9:17
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    $\begingroup$ For the "non-natural" case: Every $A$-module, for $A=k[x]/(x^2)$ ($k$ a field), is a direct sum of copies of $k$ and $A$. Taking $X$ to be a countable direct sum of copies of $k$ and $Y$ a countable direct sum of copies of $A$ gives a counterexample in the category of $A$-modules. Though this feels a bit like cheating. If $C$ is the category of finitely-generated modules for a finite-dimensional algebra (or, more generally, an Artin algebra), then there is no counterexample. I think this is due to Auslander, and in any case is an easy consequence of the existence of almost split sequences. $\endgroup$ May 9, 2013 at 12:52

1 Answer 1


Let $C$ be a category, and $X$ an object in $C$. Then $h_{X} = \textrm{Hom}(X, \_)$ is a functor from $C$ to $\textrm{Set}$.

If for two objects $X,Y$ in $C$ the functors $h_{X}$ and $h_{Y}$ are naturally isomorphic, then so are $X$ and $Y$. This is called the Yoneda lemma.

In fancy categorical terms it says that the functor $h \colon C^{\textrm{opp}} \to \textrm{Func}(C, \textrm{Set})$ is fully faithful. The slogan is "tell me who your friends are, and I will tell you who you are". So if you know all the arrows from (or to) an object, then you can determine the object up to isomorphism.

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    $\begingroup$ Let's be honest here: category theorists are their own friends, which is why the slogan works. $\endgroup$ May 9, 2013 at 7:01
  • 1
    $\begingroup$ Probably the slogan should be "tell me how you relate to others, and I will tell you who you are", but I went for a pretty concordant translation of the Dutch saying. $\endgroup$
    – jmc
    May 9, 2013 at 9:34
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    $\begingroup$ The modern version is: tell me who your Facebook friends are and I will tell you not only who you are, but also how you use your credit cards. $\endgroup$ May 9, 2013 at 10:09
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    $\begingroup$ Another apposite phrase: If it walks like a duck and quacks like a duck, it probably is a duck. What the proof of the Yoneda lemma reveals is that the best judge of whether something's a duck is a duck. $\endgroup$ May 9, 2013 at 13:37
  • 1
    $\begingroup$ ...or a European swallow. $\endgroup$
    – David Roberts
    May 10, 2013 at 0:58

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