Recall that given an equivalence relation $R$ on a set $A$ , we can form the quotient $A/R$ of $A$ by $R$ . Elements of $A/R$ are the equivalence classes under $R$ . There are two functional properties of $A/R$ :

• If $p_1,p_2$ are projections of $R$ onto $A$ , given by $p_1(a,b)=a$ and $p_2(a,b)=b$ , then the canonical surjection $q:A\to A/R$ is the coequalizer of $p_1$ and $p_2$ .
  Proof. First, $q\circ p_1(a,b) = q(a)=[a]=[b]= q(b)=q\circ p_2$ . Suppose that $r:A\to B$ is another function with $r\circ p_1=r\circ p_2$ . Define $f:A/R\to B$ by $f([a])=r(a)$ . This is a well-defined function because $[a]=[b]$ implies that $r(a)=r\circ p_1(a,b)=r\circ p_2(a,b)=r(b)$ . This shows that $f\circ q=r$ , which also implies that $f$ is uniquely determined.
• $p_1$ and $p_2$ form a kernel pair of $q$ .
  Proof. Again, $q\circ p_1 = q\circ p_2$ , as was just shown previously. Now suppose $g,h:C\to A$ are functions with $q\circ g = q\circ h$ . For any $c\in C$ , we see that $[g(c)]=q(g(c))=q(h(c))=[h(c)]$ , so that $(g(c),h(c))\in R$ . Define $s:C\to R$ by $s(c)=(g(c),h(c))$ . Then $p_1\circ s=g$ and $p_2\circ s=h$ . It is again easy to see that $s$ is uniquely determined by $g$ and $h$ . Hence, $p_1,p_2$ are a kernel pair of $g$ .

Definition. An equivalence relation object $(R,p_1,p_2)$ on an object $A$ in a category $\mathcal{C}$ is said to be an effective equivalence relation object if

• the projections $p_1,p_2$ has a coequalizer $q:A\to A/R$ , and
• $p_1,p_2$ form the kernel pair of $q$ .

In Set, the category of sets, every equivalence relation object (which is just an equivalence relation on a set) is effective, as we have just shown above. However, this is not true in general. For example, not every equivalence relation object is effective in Top, the category of topological spaces (and continuous functions).

More to come...