8.5.18 Internal Right Kan Lifts

    Let $A$, $B$, and $X$ be sets and let $R\colon A\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}B$ and $F\colon X\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}B$ be relations.

    We want to understand internal right Kan lifts in $\boldsymbol {\mathsf{Rel}}$, which look like this:

    Note in particular here that $F\colon B\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}X$ is a relation from $B$ to $X$. These will form a functor

    \[ \operatorname {\mathrm{Rift}}_{R}\colon \mathbf{Rel}(X,B)\to \mathbf{Rel}(X,A) \]

    that is right adjoint to the postcomposition by $R$ functor

    \[ R_{*}\colon \mathbf{Rel}(X,A)\to \mathbf{Rel}(X,B). \]

    The internal right Kan lift of $F$ along $R$ is the relation $\operatorname {\mathrm{Rift}}_{R}(F)$ described as follows:

    1. 1.

      Viewing relations from $X$ to $A$ as subsets of $X\times A$, we have

      \[ \operatorname {\mathrm{Rift}}_{R}(F)=\left\{ (x,a)\in X\times A\ \middle |\ \begin{aligned} & \text{for each $b\in B$, if $a\sim _{R}b$,}\\ & \text{then we have $x\sim _{F}b$}\end{aligned} \right\} . \]
    2. 2.

      Viewing relations as functions $X\times A\to \{ \mathsf{true},\mathsf{false}\} $, we have

      \begin{align*} (\operatorname {\mathrm{Rift}}_{R}(F))^{-_{1}}_{-_{2}} & = \int _{b\in B}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }(R^{b}_{-_{1}},F^{b}_{-_{2}})\\ & = \bigwedge _{b\in B}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }(R^{b}_{-_{1}},F^{b}_{-_{2}}),\end{align*}

      where the meet $\bigwedge $ is taken in the poset $(\{ \mathsf{true},\mathsf{false}\} ,\preceq )$ of Chapter 3: Sets, Definition 3.2.2.1.3.

    3. 3.

      Viewing relations as functions $X\to \mathcal{P}(A)$, we have

      \[ [\operatorname {\mathrm{Rift}}_{R}(F)](x)=\left\{ a\in A\ \middle |\ R(a)\subset F(x)\right\} \]

      for each $a\in A$.

    We have

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

    \[ \int _{b\in B}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }(R^{b}_{-_{1}},F^{b}_{-_{2}}) \]

    is right adjoint to the postcomposition functor $R\mathbin {\diamond }-$, being thus the right Kan lift 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 8.1.1.1.5.

    2. 2.

      Definition 8.1.3.1.1.

    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.

    Unresolved reference, Unresolved reference of Unresolved reference.

  • 7.

    Chapter 8: Relations, Item 1 of Proposition 8.1.1.1.5.

  • This finishes the proof.

    Here are some examples of internal right Kan lifts of relations.

    1. 1.

      Pullbacks. Let $p\colon A\to B$ and $f\colon X\to B$ be functions. We have

      \begin{align*} [\operatorname {\mathrm{Rift}}_{\operatorname {\mathrm{Gr}}(p)}(\operatorname {\mathrm{Gr}}(f))](x) & = \left\{ a\in A\ \middle |\ [\operatorname {\mathrm{Gr}}(p)](a)\subset [\operatorname {\mathrm{Gr}}(f)](x)\right\} \\ & = \left\{ a\in A\ \middle |\ p(a)=f(x)\right\} .\end{align*}

      Thus, as a subset of $X\times A$, the right Kan lift $\operatorname {\mathrm{Rift}}_{\operatorname {\mathrm{Gr}}(p)}(\operatorname {\mathrm{Gr}}(f))$ corresponds precisely to the pullback $X\times _{B}A$ of $X$ and $A$ along $p$ and $f$ of Chapter 4: Constructions With Sets, Section 4.1.4.

    Let $A$, $B$, $C$ and $X$ be sets and let $R\colon A\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}B$, $S\colon B\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}C$, and $F\colon X\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}B$ be relations.

    1. 1.

      Functoriality. The assignments $R,F,(R,F)\mapsto \operatorname {\mathrm{Rift}}_{R}(F)$ define functors

      \[ \begin{array}{ccc} \operatorname {\mathrm{Rift}}_{(-)}(F)\colon \mkern -15mu & \mathbf{Rel}(A,B)^{\mathsf{op}} \mkern -17.5mu& {}\mathbin {\to }\mathbf{Rel}(B,X),\\ \operatorname {\mathrm{Rift}}_{R}\colon \mkern -15mu & \mathbf{Rel}(A,X) \mkern -17.5mu& {}\mathbin {\to }\mathbf{Rel}(B,X),\\ \operatorname {\mathrm{Rift}}_{(-_{1})}(-_{2})\colon \mkern -15mu & \mathbf{Rel}(A,X)\times \mathbf{Rel}(A,B)^{\mathsf{op}} \mkern -17.5mu& {}\mathbin {\to }\mathbf{Rel}(B,X). \end{array} \]

      In other words, given relations

      if $R_{1}\subset R_{2}$ and $F_{1}\subset F_{2}$, then $\operatorname {\mathrm{Rift}}_{R_{2}}(F_{1})\subset \operatorname {\mathrm{Rift}}_{R_{1}}(F_{2})$.

    2. 2.

      Interaction With Composition. We have

      \[ \operatorname {\mathrm{Rift}}_{S\mathbin {\diamond }R}(F)= \operatorname {\mathrm{Rift}}_{R}(\operatorname {\mathrm{Ran}}_{S}(F)) \]

      and an equality

      of pasting diagrams in $\boldsymbol {\mathsf{Rel}}$.

    3. 3.

      Interaction With Converses. We have

      \[ \operatorname {\mathrm{Rift}}_{R}(F)^{\dagger } = \operatorname {\mathrm{Ran}}_{R^{\dagger }}(F^{\dagger }). \]
    4. 4.

      Interaction With Inverse Images. We have

      where $\operatorname {\mathrm{Ran}}_{\chi _{A}}\big (F^{\dagger }\big )$ is computed by the formula

      \begin{align*} [\operatorname {\mathrm{Ran}}_{\chi _{A}}\big (F^{\dagger }\big )](U) & \cong \int _{a\in A}\chi _{\mathcal{P}(B)^{\mathsf{op}}}(U,\chi _{a})\pitchfork F^{\dagger }(a)\\ & \cong \int _{a\in A}\chi _{\mathcal{P}(B)}(\chi _{a},U)\pitchfork F^{-1}(a)\\ & \cong \int _{a\in A}\chi _{U}(a)\pitchfork F(a)\\ & \cong \bigcap _{a\in A}\chi _{U}(a)\pitchfork F(a)\\ & \cong \bigcap _{a\in U}F(a) \end{align*}

      for each $U\in \mathcal{P}(A)$, so we have

      \[ [\operatorname {\mathrm{Rift}}_{R}(F)]^{-1}(a)=\bigcap _{b\in R(a)}F^{-1}(b) \]

      for each $a\in A$.

    Item 1: Functoriality
    We have

    \begin{align*} [\operatorname {\mathrm{Rift}}_{R_{2}}(F_{1})](x) & = \left\{ a\in A\ \middle |\ R_{2}(a)\subset F_{1}(x)\right\} \\ & \subset \left\{ a\in A\ \middle |\ R_{1}(a)\subset F_{1}(x)\right\} \\ & \subset \left\{ a\in A\ \middle |\ R_{1}(a)\subset F_{2}(x)\right\} \\ & = \operatorname {\mathrm{Rift}}_{R_{1}}(F_{2}) \end{align*}

    for each $x\in X$, so we therefore have $\operatorname {\mathrm{Rift}}_{R_{2}}(F_{1})\subset \operatorname {\mathrm{Rift}}_{R_{1}}(F_{2})$.

    Item 2: Interaction With Composition
    This holds in a general bicategory with the necessary right Kan lifts, being therefore a special case of Unresolved reference.

    Item 3: Interaction With Converses
    This follows from Item 3 of Proposition 8.5.17.1.4 by duality.

    Item 4: Interaction With Inverse Images
    We proceed in a few steps.

    • We have $x\in \operatorname {\mathrm{Rift}}_{R}(F)^{\dagger }(a)$ iff $a\in \operatorname {\mathrm{Rift}}_{R}(F)(x)$.

    • This holds iff $R(a)\subset F(x)$.

    • This holds iff, for each $b\in R(a)$, we have $b\in F(x)$.

    • This holds iff, for each $b\in R(a)$, we have $x\in F^{-1}(b)$.

    • This holds iff $x\in \bigcap _{b\in R(a)}F^{-1}(b)$.

    This finishes the proof.


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