10.6.2 Quotients of Sets by Equivalence Relations

Let $A$ be a set and let $R$ be a relation on $A$.

The quotient of $X$ by $R$ is the set $X/\mathord {\sim }_{R}$ defined by

\[ X/\mathord {\sim }_{R}\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ \webleft [a\webright ]\in \mathcal{P}\webleft (X\webright )\ \middle |\ a\in X\right\} . \]

The reason we define quotient sets for equivalence relations only is that each of the properties of being an equivalence relation—reflexivity, symmetry, and transitivity—ensures that the equivalences classes $\webleft [a\webright ]$ of $X$ under $R$ are well-behaved:

  • Reflexivity. If $R$ is reflexive, then, for each $a\in X$, we have $a\in \webleft [a\webright ]$.

  • Symmetry. The equivalence class $\webleft [a\webright ]$ of an element $a$ of $X$ is defined by

    \[ \webleft [a\webright ]\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ x\in X\ \middle |\ x\sim _{R}a\right\} , \]

    but we could equally well define

    \[ \webleft [a\webright ]'\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ x\in X\ \middle |\ a\sim _{R}x\right\} \]

    instead. This is not a problem when $R$ is symmetric, as we then have $\webleft [a\webright ]=\webleft [a\webright ]'$.1

  • Transitivity. If $R$ is transitive, then $\webleft [a\webright ]$ and $\webleft [b\webright ]$ are disjoint iff $a\nsim _{R}b$, and equal otherwise.


  1. 1When categorifying equivalence relations, one finds that $\webleft [a\webright ]$ and $\webleft [a\webright ]'$ correspond to presheaves and copresheaves; see Unresolved reference, Unresolved reference.

Let $f\colon X\to Y$ be a function and let $R$ be a relation on $X$.

  1. 1.

    As a Coequaliser. We have an isomorphism of sets

    \[ X/\mathord {\sim }^{\mathrm{eq}}_{R}\cong \operatorname {\mathrm{CoEq}}\webleft (R\hookrightarrow X\times X\stackrel{\stackrel{\operatorname {\mathrm{\mathrm{pr}}}_{1}}{\rightarrow }}{\stackrel{\rightarrow }{\scriptsize \operatorname {\mathrm{\mathrm{pr}}}_{2}}}X\webright ), \]

    where $\mathord {\sim }^{\mathrm{eq}}_{R}$ is the equivalence relation generated by $\mathord {\sim }_{R}$.

  2. 2.

    As a Pushout. We have an isomorphism of sets1

    where $\mathord {\sim }^{\mathrm{eq}}_{R}$ is the equivalence relation generated by $\mathord {\sim }_{R}$.

  3. 3.

    The First Isomorphism Theorem for Sets. We have an isomorphism of sets2,3

    \[ X/\mathord {\sim }_{\mathrm{Ker}\webleft (f\webright )} \cong \mathrm{Im}\webleft (f\webright ). \]
  4. 4.

    Descending Functions to Quotient Sets, I. Let $R$ be an equivalence relation on $X$. The following conditions are equivalent:

    1. (a)

      There exists a map

      \[ \overline{f}\colon X/\mathord {\sim }_{R}\to Y \]

      making the diagram

      commute.

    2. (b)

      We have $R\subset \mathrm{Ker}\webleft (f\webright )$.

    3. (c)

      For each $x,y\in X$, if $x\sim _{R}y$, then $f\webleft (x\webright )=f\webleft (y\webright )$.

  5. 5.

    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.

  6. 6.

    Descending Functions to Quotient Sets, III. Let $R$ be an equivalence relation on $X$. We have a bijection

    \[ \operatorname {\mathrm{Hom}}_{\mathsf{Sets}}\webleft (X/\mathord {\sim }_{R},Y\webright )\cong \operatorname {\mathrm{Hom}}^{R}_{\mathsf{Sets}}\webleft (X,Y\webright ), \]

    natural in $X,Y\in \operatorname {\mathrm{Obj}}\webleft (\mathsf{Sets}\webright )$, given by the assignment $f\mapsto \overline{f}$ of Item 4 and Item 5, where $\operatorname {\mathrm{Hom}}^{R}_{\mathsf{Sets}}\webleft (X,Y\webright )$ is the set defined by

    \[ \operatorname {\mathrm{Hom}}^{R}_{\mathsf{Sets}}\webleft (X,Y\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ f\in \operatorname {\mathrm{Hom}}_{\mathsf{Sets}}\webleft (X,Y\webright )\ \middle |\ \begin{aligned} & \text{for each $x,y\in X$,}\\ & \text{if $x\sim _{R}y$, then}\\ & \text{$f\webleft (x\webright )=f\webleft (y\webright )$}\end{aligned} \right\} . \]
  7. 7.

    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:

    1. (a)

      The map $\overline{f}$ is an injection.

    2. (b)

      We have $R=\mathrm{Ker}\webleft (f\webright )$.

    3. (c)

      For each $x,y\in X$, we have $x\sim _{R}y$ iff $f\webleft (x\webright )=f\webleft (y\webright )$.

  8. 8.

    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:

    1. (a)

      The map $f\colon X\to Y$ is surjective.

    2. (b)

      The map $\overline{f}\colon X/\mathord {\sim }_{R}\to Y$ is surjective.

  9. 9.

    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:

    1. (a)

      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 $f\webleft (x\webright )=f\webleft (y\webright )$.

    2. (b)

      For each $x,y\in X$, if $x\sim _{R}y$, then $f\webleft (x\webright )=f\webleft (y\webright )$.


  1. 1Dually, we also have an isomorphism of sets
  2. 2Further Terminology: The set $X/\mathord {\sim }_{\mathrm{Ker}\webleft (f\webright )}$ is often called the coimage of $f$, and denoted by $\mathrm{CoIm}\webleft (f\webright )$.
  3. 3In a sense this is a result relating the monad in $\boldsymbol {\mathsf{Rel}}$ induced by $f$ with the comonad in $\boldsymbol {\mathsf{Rel}}$ induced by $f$, as the kernel and image
    \begin{gather*} \mathrm{Ker}\webleft (f\webright )\colon X\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}X,\\ \mathrm{Im}\webleft (f\webright )\subset Y \end{gather*}
    of $f$ are the underlying functors of (respectively) the induced monad and comonad of the adjunction
    of Chapter 9: Constructions With Relations, Unresolved reference of Unresolved reference.

Item 1: As a Coequaliser
Omitted.

Item 2: As a Pushout
Omitted.

Item 3: The First Isomorphism Theorem for Sets
Clear.

Item 4: Descending Functions to Quotient Sets, I
See [Proof Wiki Contributors, Condition For Mapping From Quotient Set To Be Well-Defined — Proof Wiki].

Item 5: Descending Functions to Quotient Sets, II
See [Proof Wiki Contributors, Mapping From Quotient Set When Defined Is Unique — Proof Wiki].

Item 6: Descending Functions to Quotient Sets, III
This follows from Item 5 and Item 6.

Item 7: Descending Functions to Quotient Sets, IV
See [Proof Wiki Contributors, Condition For Mapping From Quotient Set To Be An Injection— Proof Wiki].

Item 8: Descending Functions to Quotient Sets, V
See [Proof Wiki Contributors, Condition For Mapping from Quotient Set To Be A Surjection — Proof Wiki].

Item 9: Descending Functions to Quotient Sets, VI
The implication Item 8a$\implies $Item 8b is clear.

Conversely, suppose that, for each $x,y\in X$, if $x\sim _{R}y$, then $f\webleft (x\webright )=f\webleft (y\webright )$. Spelling out the definition of the equivalence closure of $R$, we see that the condition $x\sim ^{\mathrm{eq}}_{R}y$ unwinds to the following:

  • (★)
  • There exist $\webleft (x_{1},\ldots ,x_{n}\webright )\in R^{\times n}$ satisfying at least one of the following conditions:
    • The following conditions are satisfied:

      • We have $x\sim _{R}x_{1}$ or $x_{1}\sim _{R}x$;

      • We have $x_{i}\sim _{R}x_{i+1}$ or $x_{i+1}\sim _{R}x_{i}$ for each $1\leq i\leq n-1$;

      • We have $y\sim _{R}x_{n}$ or $x_{n}\sim _{R}y$;

    • We have $x=y$.

Now, if $x=y$, then $f\webleft (x\webright )=f\webleft (y\webright )$ trivially; otherwise, we have

\begin{align*} f\webleft (x\webright ) & = f\webleft (x_{1}\webright ),\\ f\webleft (x_{1}\webright ) & = f\webleft (x_{2}\webright ),\\ & \vdots \\ f\webleft (x_{n-1}\webright ) & = f\webleft (x_{n}\webright ),\\ f\webleft (x_{n}\webright ) & = f\webleft (y\webright ), \end{align*}

and $f\webleft (x\webright )=f\webleft (y\webright )$, as we wanted to show.


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