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Let $\mathcal{D}$ be a small category and let $A=A\left(\mathcal{D}\right)$ be its set of arrows. Define $P$ on $A$ by: $fPg\Leftrightarrow\left[f\text{ and }g\text{ are parallel}\right]$ and let $R\subseteq P$.

Now have a look at equivalence relations $C$ on $A$. Let's say that:

  • $C\in\mathcal{C}_{s}$ iff $R\subseteq C\subseteq P$ and $fCg\Rightarrow\left(h\circ f\circ k\right)C\left(h\circ g\circ k\right)$ whenever these compositions are defined;
  • $C\in\mathcal{C}_{w}$ iff $R\subseteq C\subseteq P$ but now combined with $fCg\wedge f'Cg'\Rightarrow\left(f\circ f'\right)C\left(g\circ g'\right)$ whenever these compositions are defined.

Then $P\in\mathcal{C}_{s}$ and $P\in\mathcal{C}_{w}$ so both are not empty. For $C_{s}:=\cap\mathcal{C}_{s}$ and $C_{w}:=\cap\mathcal{C}_{w}$ it is easy to verify that $C_{s}\in\mathcal{C}_{s}$ and $C_{w}\in\mathcal{C}_{w}$.

My question is:

Do we have $C_{w}=C_{s}$ here?

It is in fact the question whether two different definitions of 'congruences' both result in the same smallest 'congruence' that contains relation $R\subseteq P$. I ask it here for small categories so that I can conveniently speak of 'relations' (small sets), but for large categories I have the same question.

Mac Lane works in CWM with $C_{s}$, but is $C_{w}$ also an option?

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I didn't realise until after editing how much the introduction of some spacing and other formatting can improve clarity and readability. I hope you agree and will take this into consideration for future questions. Cheers and welcome to Math.SE! – Lord_Farin May 16 '13 at 11:00
up vote 2 down vote accepted

They are identical. I will suppress the composition symbol for brevity and convenience.

Suppose first that $C \in \mathcal C_w$, and that $f C g$. Since $h C h$ and $k C k$, we have $f C g$ implies $hf C hg$, which in turn implies $hfk C hgk$. Thus $C \in \mathcal C_s$.

Suppose now that $C \in \mathcal C_s$, and that $f C g, f' C g'$. Then we have $ff' C gf'$ (take $h = \operatorname{id}, k = f'$) and $gf'Cgg'$ (take $h = g, k = \operatorname{id}$). By transitivity, $ff'Cgg'$. Thus $C \in \mathcal C_w$.

Therefore, $\mathcal C_s = \mathcal C_w$, and we conclude $C_s = C_w$.

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Thank you very much for helping me out! It is much simpler than I thought. – drhab May 16 '13 at 17:42

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