The horizontal composition functor of $\mathsf{Rel}^{\mathsf{dbl}}$ is the functor

\[ \mathbin {\odot }^{\mathsf{Rel}^{\mathsf{dbl}}} \colon \mathsf{Rel}^{\mathsf{dbl}}_{1}\operatorname*{\mathbin {\times }}_{\mathsf{Rel}^{\mathsf{dbl}}_{0}}\mathsf{Rel}^{\mathsf{dbl}}_{1} \to \mathsf{Rel}^{\mathsf{dbl}}_{1} \]

of $\mathsf{Rel}^{\mathsf{dbl}}$ is the functor where

  • Action on Objects. For each composable pair $\smash {A\mathbin {\overset {R}{\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}}}B\mathbin {\overset {S}{\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}}}C}$ of horizontal morphisms of $\mathsf{Rel}^{\mathsf{dbl}}$, we have

    \[ S\mathbin {\odot }R\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}S\mathbin {\diamond }R, \]

    where $S\mathbin {\diamond }R$ is the composition of $R$ and $S$ of Chapter 6: Constructions With Relations, Definition 6.3.12.1.1.

  • Action on Morphisms. For each horizontally composable pair
    of $2$-morphisms of $\mathsf{Rel}^{\mathsf{dbl}}$, i.e. for each pair
    of inclusions of relations, the horizontal composition

    of $\alpha $ and $\beta $ is the inclusion of relations[1]


Footnotes

[1] This is justified by noting that, given $\webleft (a,c\webright )\in A\times C$, the statement
  • We have $a\sim _{\webleft (U\mathbin {\diamond }T\webright )\circ \webleft (f\times h\webright )}c$, i.e. $f\webleft (a\webright )\sim _{U\mathbin {\diamond }T}h\webleft (c\webright )$, i.e. there exists some $y\in Y$ such that:
    1. We have $f\webleft (a\webright )\sim _{T}y$;
    2. We have $y\sim _{U}h\webleft (c\webright )$;
is implied by the statement
  • We have $a\sim _{S\mathbin {\diamond }R}c$, i.e. there exists some $b\in B$ such that:
    1. We have $a\sim _{R}b$;
    2. We have $b\sim _{S}c$;
since:
  • If $a\sim _{R}b$, then $f\webleft (a\webright )\sim _{T}g\webleft (b\webright )$, as $T\circ \webleft (f\times g\webright )\subset R$;
  • If $b\sim _{S}c$, then $g\webleft (b\webright )\sim _{U}h\webleft (c\webright )$, as $U\circ \webleft (g\times h\webright )\subset S$.

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