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examples of equalizers
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
$C=\{x\in A\mid f(x)=g(x)\},$ 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 welldefined, 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 welldefined 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 $h_{1},h_{2}\in\{f,g\}$. 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 welldefined. 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 $h_{1},h_{2}\in\{f,g\}$. 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 $g$.
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