Suppose $E$ is an equivalence relation on a set $X$. Show that the usual set $X/E$ of equivalence classes can be described by a coequalizer in $\mathbf{\text{Set}}$.

Hints

Recall that an equivalence relation on $X$ is a subset $E\beta \x8a\x86X\Gamma \x97X$ satisfying various properties (e.g., transitive, etc.). Two elements ${x}_{1},{x}_{2}\beta \x88\x88X$ are said to be equivalent exactly when $({x}_{1},{x}_{2})\beta \x88\x88E$. As such, the set $E$ comes with two projection maps to $X$. These are your parallel arrows that the quotient set $X/E$ will "coequalize." Speaking of which, the set $X/E$ is defined to be the collection of equivalence classes in $X$. Each element in $X/E$ is a subset $S\beta \x8a\x86X$ consisting of all elements equivalent to each other. Each such subset $S$ is usually denoted $[x]$, where $x\beta \x88\x88X$ is any representative of that subset. In other words, we have $[{x}_{1}]=[{x}_{2}]$ exactly when $({x}_{1},{x}_{2})\beta \x88\x88E$.

The quotient set $X/E$ also comes with a projection map $\mathrm{{\rm O}\x80}:X\beta \x86\x92X/E$. This is what is claimed to be the coequalizer of your parallel arrows.