Quotient Mapping is Coequalizer

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Theorem

Let $\mathbf{Set}$ be the category of sets.

Let $S$ be a Set, and let $\mathcal R \subseteq S \times S$ be an equivalence relation on $S$.

Let $r_1, r_2: \mathcal R \to S$ be the projections corresponding to the inclusion mapping $\mathcal R \hookrightarrow S \times S$.

Let $q: S \to S / \mathcal R$ be the quotient mapping induced by $\mathcal R$.


Then $q$ is a coequalizer of $r_1$ and $r_2$ in $\mathbf{Set}$.


Proof

Let $f: S \to T$ be a mapping as in the following commutative diagram:

$\begin{xy}\xymatrix{ \mathcal R \ar[r]<2pt>^*{r_1} \ar[r]<-2pt>_*{r_2} & S \ar[r]^*{q} \ar[rd]_*{f} & S / \mathcal R \[email protected]{.>}[d]^*{\bar f} \\ & & T }\end{xy}$

This translates to, for $s_1, s_2 \in S$ with $s_1 \mathcal R s_2$:

$f \circ r_1 \left({s_1, s_2}\right) = f \circ r_2 \left({s_1, s_2}\right)$

i.e. $f \left({s_1}\right) = f \left({s_2}\right)$.


The commutativity of the diagram implies that we must define $\bar f: S / \mathcal R \to T$ by:

$\bar f \left({\left[\!\left[{s_1}\right]\!\right]}\right) = f \left({s_1}\right)$

since $q \left({s_1}\right) = \left[\!\left[{s_1}\right]\!\right]$.


The above condition precisely states that $\bar f$ is well-defined.

In conclusion, for any $f$ with $f \circ r_1 = f \circ r_2$, there is a unique $\bar f$ making the diagram commute.


That is, $q$ is a coequalizer of $r_1$ and $r_2$.

$\blacksquare$


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