Let $f\colon X\to Y$ be a function and let $R$ be a relation on $X$.
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As a Coequaliser. We have an isomorphism of sets
\[ X/\mathord {\sim }^{\mathrm{eq}}_{R}\cong \text{CoEq}\webleft (R\hookrightarrow X\times X\stackrel{\stackrel{\text{pr}_{1}}{\rightarrow }}{\stackrel{\rightarrow }{\scriptsize \text{pr}_{2}}}X\webright ), \]
where $\mathord {\sim }^{\mathrm{eq}}_{R}$ is the equivalence relation generated by $\mathord {\sim }_{R}$.
- As a Pushout. We have an isomorphism of sets1 where $\mathord {\sim }^{\mathrm{eq}}_{R}$ is the equivalence relation generated by $\mathord {\sim }_{R}$.
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The First Isomorphism Theorem for Sets. We have an isomorphism of sets23
\[ X/\mathord {\sim }_{\mathrm{Ker}\webleft (f\webright )} \cong \mathrm{Im}\webleft (f\webright ). \]
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Descending Functions to Quotient Sets, I. Let $R$ be an equivalence relation on $X$. The following conditions are equivalent:
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There exists a map
\[ \overline{f}\colon X/\mathord {\sim }_{R}\to Y \]
making the diagram
commute.
- We have $R\subset \mathrm{Ker}\webleft (f\webright )$.
- For each $x,y\in X$, if $x\sim _{R}y$, then $f\webleft (x\webright )=f\webleft (y\webright )$.
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There exists a map
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Descending Functions to Quotient Sets, II. Let $R$ be an equivalence relation on $X$. If the conditions of Item 4 hold, then $\overline{f}$ is the unique map making the diagram
commute.
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Descending Functions to Quotient Sets, III. Let $R$ be an equivalence relation on $X$. We have a bijection
\[ \textup{Hom}_{\mathsf{Sets}}\webleft (X/\mathord {\sim }_{R},Y\webright )\cong \textup{Hom}^{R}_{\mathsf{Sets}}\webleft (X,Y\webright ), \]
natural in $X,Y\in \text{Obj}\webleft (\mathsf{Sets}\webright )$, given by the assignment $f\mapsto \overline{f}$ of Item 4 and Item 5, where $\textup{Hom}^{R}_{\mathsf{Sets}}\webleft (X,Y\webright )$ is the set defined by
\[ \textup{Hom}^{R}_{\mathsf{Sets}}\webleft (X,Y\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft\{ f\in \textup{Hom}_{\mathsf{Sets}}\webleft (X,Y\webright )\ \middle |\ \begin{aligned} & \text{for each $x,y\in X$,}\\ & \text{if $x\sim _{R}y$, then}\\ & \text{$f\webleft (x\webright )=f\webleft (y\webright )$}\end{aligned} \webright\} . \] -
Descending Functions to Quotient Sets, IV. Let $R$ be an equivalence relation on $X$. If the conditions of Item 4 hold, then the following conditions are equivalent:
- The map $\overline{f}$ is an injection.
- We have $R=\mathrm{Ker}\webleft (f\webright )$.
- For each $x,y\in X$, we have $x\sim _{R}y$ iff $f\webleft (x\webright )=f\webleft (y\webright )$.
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Descending Functions to Quotient Sets, V. Let $R$ be an equivalence relation on $X$. If the conditions of Item 4 hold, then the following conditions are equivalent:
- The map $f\colon X\to Y$ is surjective.
- The map $\overline{f}\colon X/\mathord {\sim }_{R}\to Y$ is surjective.
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Descending Functions to Quotient Sets, VI. Let $R$ be a relation on $X$ and let $\mathord {\sim }^{\mathrm{eq}}_{R}$ be the equivalence relation associated to $R$. The following conditions are equivalent:
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The map $f$ satisfies the equivalent conditions of Item 4:
- There exists a map
\[ \overline{f}\colon X/\mathord {\sim }^{\mathrm{eq}}_{R}\to Y \]
making the diagram
commute.
- For each $x,y\in X$, if $x\sim ^{\mathrm{eq}}_{R}y$, then $f\webleft (x\webright )=f\webleft (y\webright )$.
- There exists a map
- For each $x,y\in X$, if $x\sim _{R}y$, then $f\webleft (x\webright )=f\webleft (y\webright )$.
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The map $f$ satisfies the equivalent conditions of Item 4:
of $f$ are the underlying functors of (respectively) the induced monad and comonad of the adjunction
of Chapter 7: Constructions With Relations, Item 2 of Proposition 7.3.1.1.2.