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The following takes place in the category of sets.

Suppose $(p_0,p_1):R \rightarrow X \times X$ in an equivalence relation on $X$. Let $p_R: X \rightarrow P_R$ be the coequalizer of $p_0,p_1:R \rightarrow X$. Let $x,x': T \rightarrow X$.

I want to prove the following are equivalent:

$1)$ $p_R (x) = p_R (x')$ in $P_R$.

$2)$ There exists an $r: T \rightarrow R$ such that $p_0 r = x$ and $p_1 r = x'$

$3)$ $(x,x') \in R$

My definition of an equivalence relation is as follows:

1) $(p_0,p_1): R \rightarrow X \times X$ is reflexive iff there is an $r: X \rightarrow R$ such that $p_0 r = 1_X = p_1 r$.

2) $(p_0,p_1):R \rightarrow X \times X$ is symmetric iff $R^{\text{op}} \subseteq_{X \times X} R$.

3) $(p_0,p_1):R \rightarrow X \times X$ is transitive iff for all $(x,y),(y,z): T \rightarrow X \times X$, if $(x,y) \in R$ and $(y,z) \in R$ then $(x,z) \in R$.

$R^{\text{op}}=(p_1,p_0): R \rightarrow X \times X$ is the opposite relation on $X \times X$ with the same domain $R$ as $(p_0,p_1)$.

So far my work on $1) \Rightarrow 2)$ goes as follows:

Suppose $p_R x = p_R x'$ for some $x,x'$. Since $R$ is reflexive, there is an $s:X \rightarrow R$ such that $p_0 s = 1_X = p_1 s$. Define $r=s x'$..

Now my problem is that, I find it hard to see how to define $r$ with out running into trouble, since it is not obvious to me that $s x' = s x$ and I suspect it is not true, but I don't know how else to define a working $r$. Proving 2) implies 3) and 3) implies 1) are easy enough.

Any help is much appreciated!

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    What sort of categories are you working with? If your category is regular, then 1) and 2) are only equivalent in an exact category (by definition). If I remember correctly, the standard example of regular but not exact category is the category of torsion-free abelian groups (but it doesn't have coequalizers). On the other hand, 2) and 3) are equivalent by definition of $(x,x')\in R$. – 2017-02-28
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    My category is the category of sets. I have edited the question to make it explicit. – 2017-02-28
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    Notice that $sx'=sx$ implies that $x'=x$ since $s$ is a split monomorphism, so it need not be true. Here you have to use an explicit description of the coequalizer as the map taking an element to its equivalence class (and then 1)$\Leftrightarrow $ 3) is trivial). – 2017-02-28
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    Hmm.. what do you mean by explicit? All I know about $p_R$ and $P_R$ is that $p_R: X \rightarrow P_R$ is the coequalizer of $p_0$ and $p_1$ from $R \rightarrow X$. – 2017-02-28
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    What I mean is that you cannot prove 1) $\Rightarrow $ 2) using only universal properties, because in general it is not true. You need to know something more about how the coequalizer is formed in your category. – 2017-02-28
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    Okay, I know that $p_R:X \rightarrow P_R$ is a surjection in SETS and that the $p_0$ and $p_1$ above comes from the equivalence relation $R$ by taking the projections to $X$. The textbook I use doesn't say much more? – 2017-02-28
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    Is $T$ here a terminal object (one element set), or any object? – 2017-02-28
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    $T$ is any object, though I guess it would not do much if $T$ was the terminal object since the terminal object in Sets separates. – 2017-03-01
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    @ArnaudD. The axiom of choice gives that there is an $\bar{x}:T \rightarrow X$ such that $p_R \bar{x} = p_R x = p_R x'$, and since $R$ is reflexive, there is an $s:X \rightarrow R$ such that $p_0 s = p_1 s = 1_X$, would $r=s \bar{x}$ do the job? – 2017-03-01
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    @Bartuc You don't need the axiom of choice to prove that such an $\bar{x}$ exists, since you can simply take $\bar{x}=x$ (or $x'$); and then your new suggestion is the same as the old one. – 2017-03-01
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    Right, but then $T$ is the terminal object right, that is what you are using as domain for $x$ and $x'$? – 2017-03-01

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