The right Kan extension

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

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

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

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

  1. We have $b\sim _{\text{Ran}_{R}\webleft (S\webright )}x$.
  2. For each $a\in A$, if $a\sim _{R}b$, then $a\sim _{S}x$.

We have

\begin{align*} \textup{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 )^{a}_{x},T^{a}_{x}\webright )\\ & \cong \int _{a\in A}\int _{x\in X}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (\webleft (\int ^{b\in B}S^{b}_{x}\times R^{a}_{b}\webright ),T^{a}_{x}\webright )\\ & \cong \int _{a\in A}\int _{x\in X}\int _{b\in B}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (S^{b}_{x}\times R^{a}_{b},T^{a}_{x}\webright )\\ & \cong \int _{a\in A}\int _{x\in X}\int _{b\in B}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (S^{b}_{x},\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (R^{a}_{b},T^{a}_{x}\webright )\webright )\\ & \cong \int _{b\in B}\int _{x\in X}\int _{a\in A}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (S^{b}_{x},\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (R^{a}_{b},T^{a}_{x}\webright )\webright )\\ & \cong \int _{b\in B}\int _{x\in X}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (S^{b}_{x},\int _{a\in A}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (R^{a}_{b},T^{a}_{x}\webright )\webright )\\ & \cong \textup{Hom}_{\mathbf{Rel}\webleft (B,X\webright )}\webleft (S,\int _{a\in A}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (R^{a}_{-_{1}},T^{a}_{-_{2}}\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

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

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. Chapter 6: Relations, Item 1 of Proposition 6.1.1.1.5;
  2. Definition 7.3.12.1.1;
  3. of ;
  4. Chapter 1: Sets, Proposition 1.2.2.1.5;
  5. of ;
  6. of ;
  7. Chapter 6: Relations, Item 1 of Proposition 6.1.1.1.5.
This finishes the proof.


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