9.2.3 Right Kan Extensions in $\boldsymbol {\mathsf{Rel}}$

    Let $R\colon A\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}B$ be a relation.

    The right Kan extension

    \[ \operatorname {\mathrm{Ran}}_{R}\colon \mathrm{Rel}\webleft (A,X\webright )\to \mathrm{Rel}\webleft (B,X\webright ) \]

    along $R$ in $\boldsymbol {\mathsf{Rel}}$ exists and is given by

    \[ \operatorname {\mathrm{Ran}}_{R}\webleft (S\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\int _{a\in A}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (R^{-_{2}}_{a},S^{-_{1}}_{a}\webright ) \]

    for each $S\in \mathrm{Rel}\webleft (A,X\webright )$, so that the following conditions are equivalent:

    1. 1.

      We have $b\sim _{\operatorname {\mathrm{Ran}}_{R}\webleft (S\webright )}x$.

    2. 2.

      For each $a\in A$, if $a\sim _{R}b$, then $a\sim _{S}x$.

    We have

    \begin{align*} \operatorname {\mathrm{Hom}}_{\mathbf{Rel}\webleft (A,X\webright )}\webleft (S\mathbin {\diamond }R,T\webright ) & \cong \int _{a\in A}\int _{x\in X}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (\webleft (S\mathbin {\diamond }R\webright )^{x}_{a},T^{x}_{a}\webright )\\ & \cong \int _{a\in A}\int _{x\in X}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (\webleft (\int ^{b\in B}S^{x}_{b}\times R^{b}_{a}\webright ),T^{x}_{a}\webright )\\ & \cong \int _{a\in A}\int _{x\in X}\int _{b\in B}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (S^{x}_{b}\times R^{b}_{a},T^{x}_{a}\webright )\\ & \cong \int _{a\in A}\int _{x\in X}\int _{b\in B}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (S^{x}_{b},\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (R^{b}_{a},T^{x}_{a}\webright )\webright )\\ & \cong \int _{b\in B}\int _{x\in X}\int _{a\in A}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (S^{x}_{b},\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (R^{b}_{a},T^{x}_{a}\webright )\webright )\\ & \cong \int _{b\in B}\int _{x\in X}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (S^{x}_{b},\int _{a\in A}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (R^{b}_{a},T^{x}_{a}\webright )\webright )\\ & \cong \operatorname {\mathrm{Hom}}_{\mathbf{Rel}\webleft (B,X\webright )}\webleft (S,\int _{a\in A}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (R^{-_{2}}_{a},T^{-_{1}}_{a}\webright )\webright )\end{align*}
    naturally in each $S\in \mathbf{Rel}\webleft (B,X\webright )$ and each $T\in \mathbf{Rel}\webleft (A,X\webright )$, showing that

    \[ \int _{a\in A}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (R^{-_{2}}_{a},T^{-_{1}}_{a}\webright ) \]

    is right adjoint to the precomposition functor $-\mathbin {\diamond }R$, being thus the right Kan extension along $R$. Here we have used the following results, respectively (i.e. for each $\cong $ sign):

    1. 1.

      Chapter 8: Relations, Item 1 of Proposition 8.1.1.1.8.

    2. 2.

      Unresolved reference.

    3. 3.

      Unresolved reference, Unresolved reference of Unresolved reference.

    4. 4.

      Chapter 3: Sets, Proposition 3.2.2.1.5.

    5. 5.

      Unresolved reference, Unresolved reference of Unresolved reference.

    6. 6.

      Unresolved reference, Unresolved reference of Unresolved reference.

  • 7.

    Chapter 8: Relations, Item 1 of Proposition 8.1.1.1.8.

  • This finishes the proof.


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