8.1.5 The Converse of a Relation

    Let $A$, $B$, and $C$ be sets and let $R\subset A\times B$ be a relation.

    The converse of $R$1 is the relation $\smash {R^{\dagger }}$ defined as follows:

    • Viewing relations as subsets, we define

      \[ R^{\dagger } \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ \webleft (b,a\webright )\in B\times A\ \middle |\ \text{we have $b\sim _{R}a$}\right\} . \]
    • Viewing relations as functions $A\times B\to \{ \mathsf{true},\mathsf{false}\} $, we define

      \[ {\webleft [R^{\dagger }\webright ]}{}^{a}_{b} \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}R^{b}_{a} \]

      for each $\webleft (b,a\webright )\in B\times A$.

    • Viewing relations as functions $A\to \mathcal{P}\webleft (B\webright )$, we define

      \begin{align*} \webleft [R^{\dagger }\webright ]\webleft (b\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}R^{\dagger }\webleft (\left\{ b\right\} \webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ a\in A\ \middle |\ b\in R\webleft (a\webright )\right\} \end{align*}

      for each $b\in B$, where $R^{\dagger }\webleft (\left\{ b\right\} \webright )$ is the fibre of $R$ over $\left\{ b\right\} $.


    1. 1Further Terminology: Also called the opposite of $R$ or the transpose of $R$.

    Here are some examples of converses of relations.

    1. 1.

      Less Than Equal Signs. We have $\webleft (\mathord {\leq }\webright )^{\dagger }=\mathord {\geq }$.

  • 2.

    Greater Than Equal Signs. Dually to Item 1, we have $\webleft (\mathord {\geq }\webright )^{\dagger }=\mathord {\leq }$.

  • 3.

    Functions. Let $f\colon A\to B$ be a function. We have

    \begin{align*} \operatorname {\mathrm{Gr}}\webleft (f\webright )^{\dagger } & = f^{-1},\\ \webleft (f^{-1}\webright )^{\dagger } & = \operatorname {\mathrm{Gr}}\webleft (f\webright ). \end{align*}
  • Let $R\colon A\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}B$ and $S\colon B\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}C$ be relations.

    1. 1.

      Functoriality. The assignment $R\mapsto R^{\dagger }$ defines a functor (i.e. morphism of posets)

      \[ \webleft (-\webright )^{\dagger }\colon \mathbf{Rel}\webleft (A,B\webright )\to \mathbf{Rel}\webleft (B,A\webright ). \]

      In particular, given relations $R,S\colon A\mathrel {\rightrightarrows \kern -9.5pt\mathrlap {|}\kern 6pt}B$, we have:

      • (★)
      • If $R\subset S$, then $R^{\dagger }\subset S^{\dagger }$.
    2. 2.

      Interaction With Ranges and Domains. We have

      \begin{align*} \mathrm{dom}\webleft (R^{\dagger }\webright ) & = \mathrm{range}\webleft (R\webright ),\\ \mathrm{range}\webleft (R^{\dagger }\webright ) & = \mathrm{dom}\webleft (R\webright ). \end{align*}
    3. 3.

      Interaction With Composition I. We have

      \[ \webleft (S\mathbin {\diamond }R\webright )^{\dagger } = R^{\dagger }\mathbin {\diamond }S^{\dagger }. \]
    4. 4.

      Interaction With Composition II. We have

      \begin{align*} \chi _{B} & \subset R\mathbin {\diamond }R^{\dagger },\\ \chi _{A} & \subset R^{\dagger }\mathbin {\diamond }R. \end{align*}
    5. 5.

      Invertibility. We have

      \[ \webleft (R^{\dagger }\webright )^{\dagger } = R. \]
    6. 6.

      Identity. We have

      \[ \chi ^{\dagger }_{A} = \chi _{A}. \]

    Item 1: Functoriality
    Clear.

    Item 2: Interaction With Ranges and Domains
    Clear.

    Item 3: Interaction With Composition I
    Clear.

    Item 4: Interaction With Composition II
    Clear.

    Item 5: Invertibility
    Clear.

    Item 6: Identity
    Clear.


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