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We know that the fundamental theorem of equivalence relations can be stated without reference to sets by using congruence as a conjunction of two propositions as following:

  1. If $R$ together with projections $p_0,p_1:R\rightarrow A$ is a congruence on an object $A$, then it is the kernel pair of the coequalizer of $p_0$ with $p_1$.

  2. If $p:A\rightarrow X$ is any partition (any epimorphism) of object $A$, then $p$ is the coequalizer of its cokernel pair.

It is straightforward to prove 2. by assuming just the axiom of choice (that every epimorphism has a section), while for the 1. I could only prove it using specific properties of abstract sets as following: If $R'$ together with projections $p_0',p_1':R'\rightarrow A$ is the kernel of $p_R$ and $R\not\equiv_{X\times X} R'$, then we can construct an equivalence class whose characteristic map cannot be factored through the coequalizer $p_R$ of $R$.

I'm wondering if there is a more general (than the one sketched above) proof of the 1st statement because the author (F.W. Lawvere and R. Rosebrugh) of the book did not mention if one would need to use any specific properties of $Set$ preceding the question as follows:

Now if we start with an equivalence relation $R$ on $X$, we may form the coequalizer of $p_0,p_1:R\rightarrow X$, which we denote by $p_R:X\rightarrow P_R$. This is a partition of $X$. (We obtain a partition of X by form the coequalizer of any two mappings with codomain $X$ - it is the special properties of equivalence relations that allow the next result.) Taking the equivalence relation of a partition of $X$ and taking the partition from an equivalence relation on $X$ are inverse processes:

Proposition: If $p$ is a partition of $X$, then $p=p_{R_p}$. If $R$ is an equivalence relation on $X$ then $R=R_{p_R}$.

I was wondering if I don't need to use all of the constructions I've used. Perhaps, there is a cleaner proof.

Thanks!

Jaspreet
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    For example, these two axioms hold in any abelian category, such as the category of abelian groups. – Qiaochu Yuan Jan 03 '21 at 00:58
  • @QiaochuYuan Thank you, but I am such an amateur that I haven't come across the definition, let alone understand Abelian categories. I suppose I will look into that. Thanks so much again. – Jaspreet Jan 03 '21 at 04:48
  • @QiaochuYuan Is the category of abstract sets is not abelian (It does not have a zero object for example)? I should have said I was wondering more along the lines if I can weaken hypothesis in way that it still holds for abstract sets. – Jaspreet Jan 03 '21 at 04:55
  • I think you're looking for the setting of an exact category (in the sense of Barr). You will probably want to restrict attention to regular epimorphisms rather than all epimorphisms, though. – tcamps Feb 15 '21 at 19:16

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