This entry illustrates some common examples of equalizers and coequalizers.

Examples of Equalizers

• In Set, the category of sets, the equalizer of a pair of functions $f,g:A\to B$ is given by the following: a set $$C=\lbrace x\in A\mid f(x)=g(x)\rbrace,$$ and function $i:C\to A$ the canonical injection. Clearly $i$ equalizes $f$ and $g$ by construction: $f\circ i = g\circ i$ . Now, if $j:D\to A$ also equalizes $f$ and $g$ , then define $k:D\to C$ by $k(d)=j(d)$ . To see that this is well-defined, we need to show that $j(d)\in C$ . Since $j$ equalizes $f$ and $g$ , we have $f(j(d))=g(j(d))$ , so that $j(d)\in C$ . Therefore $k$ is a well-defined function from $D$ into $C$ . In addition, $i\circ k(d)=i(j(d))=j(d)$ . Finally, it is easy to see that if $i\circ t=j$ , then $t=k$ . Therefore, $(C,i)$ is the equalizer of $f$ and $g$ .
• In fact, most concrete categories (concrete over Sets), the equalizer of a pair of morphisms is given by the object $C$ above with $i$ the corresponding injective mapping.

Examples of Coequalizers

• In Set, the coequalizer of a pair of functions $f,g:A\to B$ can be found as follows: define a binary relation $\sim$ on $B$ such that for any $x,y\in B$ , $x\sim y$ iff either $x=y$ , or there is an $a\in A$ such that $h_1(a)=x$ and $h_2(a)=y$ , where $h_1,h_2\in \lbrace f,g\rbrace$ . Then $\sim$ is easily seen to be a reflexive symmetric relation. Now, take the transitive closure $\sim^*$ of $\sim$ . So $\sim^*$ is an equivalence relation on $B$ . Let $B/\sim^*$ be the set of all the equivalence classes, and $p:B\to B/\sim^*$ the canonical projection. Then $(B/\sim^*, p)$ is the coequalizer of $f$ and $g$ . First, $p\circ f(a)=[f(a)]=[g(a)]=p\circ g(a)$ , since $f(a)\sim g(a)$ . Suppose now that $q:B\to D$ is another function that coequalizes $f$ and $g$ . Define $r:B/\sim^* \to D$ by $r([b])=q(b)$ . We want to show that $r$ is well-defined. In other words, if $b\sim^* c$ , then $q(b)=q(c)$ . First, assume $b\sim c$ . Then either $b=c$ (in which case, $q(b)=q(c)$ is immediate), or there is $a\in A$ such that $h_1(a)=b$ and $h_2(a)=c$ , with $h_1,h_2\in \lbrace f,g\rbrace$ . In this case, $q(b)=q(h_1(a))=q(h_2(a))=q(c)$ since $q\circ f=q\circ g$ . Now, if $b\sim^* c$ , then there are $d_1,\ldots, d_n \in B$ such that $b=d_1\sim d_2 \sim \cdots \sim d_n = c$ . As a result, $q(b)=q(d_1)=q(d_2)=\cdots = q(d_n)=q(c)$ . In addition, $r\circ p(b)=r([b])=q(b)$ , and that $r$ is uniquely determined this way. Therefore, $(B/\sim^*,p)$ is the coequalizer of $f$ and $g$ .