Equivalence relations in arbitrary categories

Question

Let $C$ be a category and $A\in\mathrm{ob}(C)$. A relation is a subobject $q:Q\hookrightarrow A^{\times 2}$ and the quotient is defined as the coequalizer $$A/Q:=\mathrm{coeq}\left(Q\stackrel{q}{\hookrightarrow} A^{\times 2}\rightrightarrows A\right)$$ where the two maps are the two projections. Moreover, we can define what it means to be an equivalence relation: The inclusion $q$ should satisfy the following:

• Reflexivity: Consider the diagonal $\Delta:A\to A^{\times 2}$. Then there should be a map $i:A\to Q$ such that $q\circ i=\Delta$.
• Symmetry: Consider the flip $t:A^{\times 2}\to A^{\times 2}$. Then there should be an automorphism $s:Q\to Q$ such that $t\circ q=q\circ s$.
• Transitivity: Consider the map $q\times_A q: Q\times_AQ\to A^{\times 2}\times_AA^{\times 2}$ and the outer projections $p:A^{\times 2}\times_AA^{\times 2}\to A^{\times 2}$. Then there should be a map $j:Q\times_A Q\to Q$ such that $p\circ (q\times_A q)=q\circ j$.

Given an arbitrary relation $q:Q\hookrightarrow A^{\times 2}$, we can define $\overline{Q}\stackrel{\overline{q}}{\hookrightarrow} A^{\times 2}$ as the inital equivalence relation having a morphism $i:Q\to \overline{Q}$ such that $\overline{q}\circ i=q$. I claim that the canonical map $$A/Q\to A/\overline{Q}$$ is an isomorphism. In $\mathbf{Set}$, this is clear as building the coequalizer is the same as quotienting out the spanned equivalence relation. Is this also true in other categories?

Answer 1

Short answer :Yes, assuming $\overline{Q}$ exists and $C$ has kernel pairs (for example if it has finite limits).

For more details: The relation $\overline{Q}$ do not always exists, you need some assumption on the underlying categories, and there are various type of assumption that can work.

For example if $\mathcal{C}$ is a pretopos with a parametrized natural number object you can mimick the usual construction of the transitive closure of a relation internall to give a construction of $\overline{Q}$.

In a completely different style, if $Sub(A \times A)$ is small and order complete then a variant of the small object argument also allows to construct $\overline{Q}$.

But, assuming that $\overline{Q}$ exists, it is actually easy to see that assuming $C$ has Kernel pair, if $\overline{Q}$ exists, then the canonical map $A/Q \rightarrow A/\overline{Q}$ is always an isomorphisms.

Indeed, let $R$ the kernel pair of $A \rightarrow A/Q$, $R$ is an equivalence relation on $A$ containing $Q$, so $\overline{Q} \subset R$. It follows from the universal property of the quotient that $A \rightarrow A/Q $ factors as $A \rightarrow A/\overline{Q} \rightarrow A/Q$

the uniqueness part of both universal property shows that the map constructed $A/\overline{Q} \leftrightarrows A/Q$ are invers of each other.

The same argument also shows that $A/R$ is also isomorphic to these (but $R$ can actually be strictly bigger than $Q$).