8.7.4 Codirect Images

    Let $X$ and $Y$ be sets and let $R\colon X\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}Y$ be a relation.

    The codirect image function associated to $R$ is the function

    \[ R_{*}\colon \mathcal{P}(X)\to \mathcal{P}(Y) \]

    defined by1,2

    \begin{align*} R_{*}(U) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ b\in Y\ \middle |\ \begin{aligned} & \text{for each $a\in X$, if we have}\\ & \text{$b\in R(a)$, then $a\in U$}\end{aligned} \right\} \\ & = \left\{ b\in Y\ \middle |\ R^{-1}(b)\subset U\right\} \end{align*}

    for each $U\in \mathcal{P}(X)$.


    1. 1Further Terminology: The set $R_{*}(U)$ is called the codirect image of $U$ by $R$.
    2. 2We also have
      \[ R_{*}(U)=Y\setminus R_{!}(X\setminus U); \]
      see Item 17 of Proposition 8.7.4.1.4.

    Identifying $\mathcal{P}(X)$ with $\mathrm{Rel}(X,\mathrm{pt})$ via Chapter 4: Constructions With Sets, Item 3 of Proposition 4.4.1.1.4, we see that the codirect image function associated to $R$ is equivalently the function

    \[ R_{*}\colon \underbrace{\mathcal{P}(X)}_{\cong \mathrm{Rel}(X,\mathrm{pt})}\to \underbrace{\mathcal{P}(Y)}_{\cong \mathrm{Rel}(Y,\mathrm{pt})} \]

    defined by

    being explicitly computed by

    \begin{align*} R_{*}(U) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\operatorname {\mathrm{Ran}}_{R}(U)\\ & \cong \int _{a\in X}\operatorname {\mathrm{Hom}}_{\{ \mathsf{t},\mathsf{f}\} }(R^{-_{2}}_{a},U^{-_{1}}_{a}), \end{align*}

    where we have used Unresolved reference.

    We have

    \begin{align*} \operatorname {\mathrm{Ran}}_{R}(V)& \cong \int _{a\in X}\operatorname {\mathrm{Hom}}_{\{ \mathsf{t},\mathsf{f}\} }(R^{-_{2}}_{a},U^{-_{1}}_{a})\\ & =\left\{ b\in Y\ \middle |\ \int _{a\in X}\operatorname {\mathrm{Hom}}_{\{ \mathsf{t},\mathsf{f}\} }(R^{b}_{a},U^{\star }_{a})=\mathsf{true}\right\} \\ & = \left\{ b\in Y\ \middle |\ \begin{aligned} & \text{For each $a\in X$, at least one of the}\\ & \text{following conditions hold:}\\[2.5pt]& \mspace {25mu}\rlap {\text{1.}}\mspace {22.5mu}\text{We have $R^{b}_{a}=\mathsf{false}$.}\\ & \mspace {25mu}\rlap {\text{2.}}\mspace {22.5mu}\text{The following conditions hold:}\\ & \mspace {50mu}\rlap {\text{(a)}}\mspace {30mu}\text{We have $R^{b}_{a}=\mathsf{true}$.}\\ & \mspace {50mu}\rlap {\text{(b)}}\mspace {30mu}\text{We have $U^{\star }_{a}=\mathsf{true}$.}\\ \end{aligned} \right\} \\ & = \left\{ b\in Y\ \middle |\ \begin{aligned} & \text{For each $a\in X$, at least one of the}\\ & \text{following conditions hold:}\\[2.5pt]& \mspace {25mu}\rlap {\text{1.}}\mspace {22.5mu}\text{We have $b\not\in R(X)$.}\\ & \mspace {25mu}\rlap {\text{2.}}\mspace {22.5mu}\text{The following conditions hold:}\\ & \mspace {50mu}\rlap {\text{(a)}}\mspace {30mu}\text{We have $b\in R(a)$.}\\ & \mspace {50mu}\rlap {\text{(b)}}\mspace {30mu}\text{We have $a\in U$.}\\ \end{aligned} \right\} \\ & = \left\{ b\in Y\ \middle |\ \begin{aligned} & \text{for each $a\in X$, if we have}\\ & \text{$b\in R(a)$, then $a\in U$}\end{aligned} \right\} \\ & = \left\{ b\in Y\ \middle |\ R^{-1}(b)\subset U\right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}R^{-1}(U).\end{align*}
    This finishes the proof.

    Let $U$ be a subset of $X$.1

    1. 1.

      The image part of the codirect image $R_{*}(U)$ of $U$ is the set $R^{\mathrm{im}}_{*}(U)$ defined by

      \begin{align*} R^{\mathrm{im}}_{*}(U) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}R_{*}(U)\cap \mathrm{Im}(R)\\ & = \left\{ y\in Y\ \middle |\ \begin{aligned} & \text{we have $R^{-1}(y)\subset U$}\\ & \text{and $R^{-1}(y)\neq \text{Ø}$.} \end{aligned}\right\} .\end{align*}
  • 2.

    The complement part of the codirect image $R_{*}(U)$ of $U$ is the set $R^{\mathrm{cp}}_{*}(U)$ defined by

    \begin{align*} R^{\mathrm{cp}}_{*}(U) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}R_{*}(U)\cap (Y\setminus \mathrm{Im}(R))\\ & = Y\setminus \mathrm{Im}(R)\\ & = \left\{ y\in Y\ \middle |\ \begin{aligned} & \text{we have $R^{-1}(y)\subset U$}\\ & \text{and $R^{-1}(y)=\text{Ø}$.} \end{aligned}\right\} \\ & = \left\{ y\in Y\ \middle |\ R^{-1}(y)=\text{Ø}\right\} .\end{align*}

    1. 1Note that we have
      \[ R_{*}(U)=R^{\mathrm{im}}_{*}(U)\cup R^{\mathrm{cp}}_{*}(U), \]
      as
      \begin{align*} R_{*}(U) & = R_{*}(U)\cap Y\\ & = R_{*}(U)\cap (\mathrm{Im}(R)\cup (Y\setminus \mathrm{Im}(R)))\\ & = (R_{*}(U)\cap \mathrm{Im}(R))\cup (R_{*}(U)\cap (Y\setminus \mathrm{Im}(R)))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}R^{\mathrm{im}}_{*}(U)\cup R^{\mathrm{cp}}_{*}(U). \end{align*}
      for each $U\in \mathcal{P}(X)$.

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

    1. 1.

      Functoriality. The assignment $U\mapsto R_{*}(U)$ defines a functor

      \[ R_{*}\colon (\mathcal{P}(X),\subset )\to (\mathcal{P}(Y),\subset ). \]

      In particular, for each $U,V\in \mathcal{P}(X)$, the following condition is satisfied:

      • If $U\subset V$, then $R_{*}(U)\subset R_{*}(V)$.

    2. 2.

      Adjointness. We have an adjunction

      witnessed by:

      1. (a)

        Units and counits of the form

        \begin{align*} \operatorname {\mathrm{id}}_{\mathcal{P}(Y)} \hookrightarrow R_{*}\circ R^{-1},\\ R^{-1}\circ R_{*} \hookrightarrow \operatorname {\mathrm{id}}_{\mathcal{P}(X)}. \end{align*}

        In particular:

        • For each $V\in \mathcal{P}(Y)$, we have $V\subset R_{*}(R^{-1}(V))$.

        • For each $U\in \mathcal{P}(X)$, we have $R^{-1}(R_{*}(U))\subset U$.

      2. (b)

        A bijections of sets

        \[ \operatorname {\mathrm{Hom}}_{\mathcal{P}(X)}(R^{-1}(U),V)\cong \operatorname {\mathrm{Hom}}_{\mathcal{P}(X)}(U,R_{*}(V)), \]

        natural in $U\in \mathcal{P}(X)$ and $V\in \mathcal{P}(Y)$. In particular:

        • (★)
        • The following conditions are equivalent:
          • We have $R^{-1}(U)\subset V$.

          • We have $U\subset R_{*}(V)$.

    3. 3.

      Interaction With Unions of Families of Subsets. The diagram

      commutes, i.e. we have

      \[ \bigcup _{V\in \mathcal{V}}R_{*}(V)=\bigcup _{U\in R_{*}(\mathcal{U})}U \]

      for each $\mathcal{V}\in \mathcal{P}(Y)$, where $R_{*}(\mathcal{V})\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}(R_{*})_{!}(\mathcal{V})$.

    4. 4.

      Interaction With Intersections of Families of Subsets. The diagram

      commutes, i.e. we have

      \[ \bigcap _{V\in \mathcal{V}}R_{*}(V)=\bigcap _{U\in R_{*}(\mathcal{U})}U \]

      for each $\mathcal{V}\in \mathcal{P}(Y)$, where $R_{*}(\mathcal{V})\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}(R_{*})_{!}(\mathcal{V})$.

    5. 5.

      Interaction With Binary Unions. We have a natural transformation

      with components

      \[ R_{*}(U)\cup R_{*}(V)\subset R_{*}(U\cup V) \]

      indexed by $U,V\in \mathcal{P}(Y)$.

    6. 6.

      Interaction With Binary Intersections. The diagram

      commutes, i.e. we have

      \[ R_{*}(U\cap V)=R_{*}(U)\cap R_{*}(V) \]

      for each $U,V\in \mathcal{P}(Y)$.

    7. 7.

      Interaction With Differences. We have a natural transformation

      with components

      \[ R_{*}(U\setminus V)\subset R_{*}(U)\setminus R_{*}(V) \]

      indexed by $U,V\in \mathcal{P}(Y)$.

    8. 8.

      Interaction With Complements. The diagram

      commutes, i.e. we have

      \[ R_{*}(U^{\textsf{c}})=R^{-1}(U)^{\textsf{c}} \]

      for each $U\in \mathcal{P}(X)$.

    9. 9.

      Interaction With Symmetric Differences. The diagram

      does not commute in general, i.e. we may have

      \[ R_{*}(U)\mathbin {\triangle }R_{*}(V)\neq R_{*}(U\mathbin {\triangle }V) \]

      in general, where $U,V\in \mathcal{P}(Y)$.

    10. 10.

      Interaction With Internal Homs of Powersets. We have a natural transformation

      with components

      \[ [R_{!}(U),R_{*}(V)]_{X}\subset R_{*}([U,V]_{Y}), \]

      indexed by $U,V\in \mathcal{P}(Y)$.

    11. 11.

      Lax Preservation of Colimits. We have an inclusion of sets

      \[ \bigcup _{i\in I}R_{*}(U_{i})\subset R_{*}\left(\bigcup _{i\in I}U_{i}\right), \]

      natural in $\left\{ U_{i}\right\} _{i\in I}\in \mathcal{P}(X)^{\times I}$. In particular, we have inclusions

      \[ \begin{gathered} R_{*}(U)\cup R_{*}(V) \subset R_{*}(U\cup V),\\ \text{Ø}\subset R_{*}(\text{Ø}), \end{gathered} \]

      natural in $U,V\in \mathcal{P}(X)$.

    12. 12.

      Preservation of Limits. We have an equality of sets

      \[ R_{*}\left(\bigcap _{i\in I}U_{i}\right)=\bigcap _{i\in I}R_{*}(U_{i}), \]

      natural in $\left\{ U_{i}\right\} _{i\in I}\in \mathcal{P}(X)^{\times I}$. In particular, we have equalities

      \[ \begin{gathered} R_{*}(U\cap V) = R_{*}(U)\cap R_{*}(V),\\ R_{*}(X) = Y, \end{gathered} \]

      natural in $U,V\in \mathcal{P}(X)$.

    13. 13.

      Symmetric Lax Monoidality With Respect to Unions. The codirect image function of Item 1 has a symmetric lax monoidal structure

      \[ (R_{*},R^{\otimes }_{*},R^{\otimes }_{!|\mathbb {1}}) \colon (\mathcal{P}(X),\cup ,\text{Ø}) \to (\mathcal{P}(Y),\cup ,\text{Ø}), \]

      being equipped with inclusions

      \[ \begin{gathered} R^{\otimes }_{!|U,V} \colon R_{*}(U)\cup R_{*}(V) \subset R_{*}(U\cup V),\\ R^{\otimes }_{!|\mathbb {1}} \colon \text{Ø}\subset R_{*}(\text{Ø}), \end{gathered} \]

      natural in $U,V\in \mathcal{P}(X)$.

    14. 14.

      Symmetric Strict Monoidality With Respect to Intersections. The direct image function of Item 1 has a symmetric strict monoidal structure

      \[ (R_{*},R^{\otimes }_{*},R^{\otimes }_{!|\mathbb {1}}) \colon (\mathcal{P}(X),\cap ,X) \to (\mathcal{P}(Y),\cap ,Y), \]

      being equipped with equalities

      \[ \begin{gathered} R^{\otimes }_{!|U,V} \colon R_{*}(U\cap V) \mathbin {\overset {=}{\rightarrow }}R_{*}(U)\cap R_{*}(V),\\ R^{\otimes }_{!|\mathbb {1}} \colon R_{*}(X) \mathbin {\overset {=}{\rightarrow }}Y, \end{gathered} \]

      natural in $U,V\in \mathcal{P}(X)$.

    15. 15.

      Interaction With Coproducts. Let $R\colon X\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}X'$ and $S\colon Y\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}Y'$ be relations. The diagram

      commutes, i.e. we have

      \[ (R\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}S)_{*}(U'\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}V')=R_{*}(U')\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}S_{*}(V') \]

      for each $U'\in \mathcal{P}(X')$ and each $V'\in \mathcal{P}(Y')$.

    16. 16.

      Interaction With Products. Let $f\colon X\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}X'$ and $g\colon Y\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}Y'$ be relations. The diagram

      commutes, i.e. we have

      \[ (R\boxtimes _{X'\times Y'}S)_{*}(U'\boxtimes _{X'\times Y'}V')=R_{*}(U')\boxtimes _{X\times Y}S_{*}(V') \]

      for each $U'\in \mathcal{P}(X')$ and each $V'\in \mathcal{P}(Y')$.

    17. 17.

      Relation to Direct Images. We have

      \[ R_{*}(U)=Y\setminus R_{!}(X\setminus U) \]

      for each $U\in \mathcal{P}(X)$.

    18. 18.

      Interaction With Injections. If $R$ is injective, then we have

      \begin{align*} R^{\mathrm{im}}_{*}(U) & = R_{!}(U),\\ R^{\mathrm{cp}}_{*}(U) & = Y\setminus \mathrm{Im}(R), \end{align*}

      so

      \begin{align*} R_{*}(U) & = R^{\mathrm{im}}_{*}(U)\cup R^{\mathrm{cp}}_{*}(U)\\ & = R_{!}(U)\cup (Y\setminus \mathrm{Im}(R)) \end{align*}

      for each $U\in \mathcal{P}(X)$.

    19. 19.

      Interaction With Surjections. If $R$ is surjective, then we have

      \begin{align*} R^{\mathrm{im}}_{*}(U) & \subset R_{!}(U),\\ R^{\mathrm{cp}}_{*}(U) & = \text{Ø},\\ \end{align*}

      so

      \[ R_{*}(U)\subset R_{!}(U) \]

      for each $U\in \mathcal{P}(X)$.

    20. 20.

      Interaction With Bijections I. If $R$ is bijective, then we have

      \begin{align*} R^{\mathrm{im}}_{*}(U) & = R_{!}(U),\\ R^{\mathrm{cp}}_{*}(U) & = \text{Ø}, \end{align*}

      so $R_{!}=R_{*}$.

    21. 21.

      Interaction With Bijections II. If $R_{!}=R_{*}$, then $R$ is bijective.

    Item 1: Functoriality
    Omitted.

    Item 2: Adjointness
    This follows from Unresolved reference, Unresolved reference of Unresolved reference.

    Item 3: Interaction With Unions of Families of Subsets
    We have

    \begin{align*} \bigcup _{U\in R_{*}(\mathcal{V})}U & = \bigcup _{U\in \left\{ R_{*}(V)\in \mathcal{P}(X)\ \middle |\ V\in \mathcal{V}\right\} }U\\ & = \bigcup _{V\in \mathcal{V}}R_{*}(V).\end{align*}

    This finishes the proof.

    Item 4: Interaction With Intersections of Families of Subsets
    We have

    \begin{align*} \bigcap _{U\in R_{*}(\mathcal{V})}U & = \bigcap _{U\in \left\{ R_{*}(V)\in \mathcal{P}(X)\ \middle |\ V\in \mathcal{V}\right\} }U\\ & = \bigcap _{V\in \mathcal{V}}R_{*}(V).\end{align*}

    This finishes the proof.

    Item 5: Interaction With Binary Unions
    We have

    \begin{align*} R_{*}(U)\cup R_{*}(V) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ b\in Y\ \middle |\ R^{-1}(b)\subset U\right\} \cup \left\{ b\in Y\ \middle |\ R^{-1}(b)\subset V\right\} \\ & = \left\{ b\in Y\ \middle |\ \text{$R^{-1}(b)\subset U$ or $R^{-1}(b)\subset V$}\right\} \\ & \subset \left\{ b\in Y\ \middle |\ R^{-1}(b)\subset U\cup V\right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}R_{-1}(U\cup V). \end{align*}

    Since $\mathcal{P}(X)$ is posetal, naturality is automatic (Chapter 11: Categories, Item 4 of Proposition 11.2.7.1.2). This finishes the proof.

    Item 6: Interaction With Binary Intersections
    We have

    \begin{align*} R_{*}(U)\cap R_{*}(V) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ b\in X\ \middle |\ R^{-1}(b)\subset U\right\} \cap \left\{ b\in Y\ \middle |\ R^{-1}(b)\subset V\right\} \\ & = \left\{ b\in Y\ \middle |\ \text{$R^{-1}(b)\subset U$ and $R^{-1}(b)\subset V$}\right\} \\ & = \left\{ b\in Y\ \middle |\ R^{-1}(b)\subset U\cap V\right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}R_{*}(U\cap V). \end{align*}

    This finishes the proof.

    Item 7: Interaction With Differences
    We have

    \begin{align*} R_{*}(U\setminus V) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ b\in Y\ \middle |\ R^{-1}(b)\subset U\setminus V\right\} \\ & \subset \left\{ b\in Y\ \middle |\ \text{$R^{-1}(b)\subset U$ and $R^{-1}(b)\centernot {\subset }V$}\right\} \\ & = \left\{ b\in Y\ \middle |\ R^{-1}(b)\subset U\right\} \setminus \left\{ b\in Y\ \middle |\ R^{-1}(b)\subset V\right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}R_{*}(U)\setminus R_{*}(V)\\ \end{align*}

    Since $\mathcal{P}(X)$ is posetal, naturality is automatic (Chapter 11: Categories, Item 4 of Proposition 11.2.7.1.2). This finishes the proof.

    Item 8: Interaction With Complements
    Applying Item 17 to $X\setminus U$, we have

    \begin{align*} R_{*}(U^{\textsf{c}}) & = R_{*}(X\setminus U)\\ & = Y\setminus R_{!}(X\setminus (X\setminus U))\\ & = Y\setminus R_{!}(U)\\ & = R_{!}(U)^{\textsf{c}}. \end{align*}

    This finishes the proof.

    Item 9: Interaction With Symmetric Differences
    Omitted.

    Item 10: Interaction With Internal Homs of Powersets
    We have

    \begin{align*} [R_{!}(U),R_{*}(V)]_{Y} & = R_{!}(U)^{\textsf{c}}\cup R_{*}(V)\\ & = R_{*}(U^{\textsf{c}})\cup R_{*}(V)\\ & \subset R_{*}(U^{\textsf{c}}\cup V)\\ & = R_{*}([U,V]_{X}).\end{align*}

    Since $\mathcal{P}(X)$ is posetal, naturality is automatic (Chapter 11: Categories, Item 4 of Proposition 11.2.7.1.2). This finishes the proof.

    Item 11: Lax Preservation of Colimits
    Omitted.

    Item 12: Preservation of Limits
    This follows from Item 2 and Unresolved reference, Unresolved reference of Unresolved reference.

    Item 13: Symmetric Lax Monoidality With Respect to Unions
    This follows from Item 11.

    Item 14: Symmetric Strict Monoidality With Respect to Intersections
    This follows from Item 12.

    Item 15: Interaction With Coproducts
    Omitted.

    Item 16: Interaction With Products
    Omitted.

    Item 17: Relation to Direct Images
    This follows from Item 17 of Proposition 8.7.1.1.4. Alternatively, we may prove it directly as follows, with the proof proceeding in the same way as in the case of functions (Chapter 4: Constructions With Sets, Item 16 of Proposition 4.6.3.1.7).

    We claim that $R_{*}(U)=Y\setminus R_{!}(X\setminus U)$:

    • Proof of $R_{*}(U)\subset Y\setminus R_{!}(X\setminus U)$: We proceed in a few steps:

      • Let $b\in R_{*}(U)$. We need to show that $b\not\in R_{!}(X\setminus U)$, i.e. that there is no $a\in X\setminus U$ such that $b\in R(a)$.

      • This is indeed the case, as otherwise we would have $a\in R^{-1}(b)$ and $a\not\in U$, contradicting $R^{-1}(b)\subset U$ (which holds since $b\in R_{*}(U)$).

      • Thus $b\in Y\setminus R_{!}(X\setminus U)$.

    • Proof of $Y\setminus R_{!}(X\setminus U)\subset R_{*}(U)$: We proceed in a few steps:

      • Let $b\in Y\setminus R_{!}(X\setminus U)$. We need to show that $b\in R_{*}(U)$, i.e. that $R^{-1}(b)\subset U$.

      • Since $b\not\in R_{!}(X\setminus U)$, there exists no $a\in X\setminus U$ such that $b\in R(a)$, and hence $R^{-1}(b)\subset U$.

      • Thus $b\in R_{*}(U)$.

    This finishes the proof.

    Item 18: Interaction With Injections
    If $R$ is injective, then the condition $R^{-1}(y)\subset U$ is equivalent to $R^{-1}(y)\cap U\neq \text{Ø}$ when $y\in \mathrm{Im}(R)$. Thus $R^{\mathrm{im}}_{*}(U)=R_{!}(U)$.

    Item 19: Interaction With Surjections
    Indeed, we have

    \begin{align*} R^{\mathrm{cp}}_{*}(U) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}Y\setminus \mathrm{Im}(R)\\ & = Y\setminus Y\\ & = \text{Ø}. \end{align*}

    Since $R^{-1}(y)\subset U$ is a strictly stronger condition than $R^{-1}(y)\cap U\neq \text{Ø}$ when $R^{-1}(y)$ is nonempty, we have $R^{\mathrm{im}}_{*}(U)\subset R_{!}(U)$. Thus $R_{*}(U)\subset R_{!}(U)$.

    Item 20: Interaction With Bijections I
    This follows from Item 18 and Item 19.

    Item 21: Interaction With Bijections II
    First, recall from Unresolved reference that we have

    \[ R_{*}(U)=Y\setminus R_{!}(X\setminus U) \]

    for all $U\in \mathcal{P}(A)$. We now claim that $R$ is bijective:

    • Injectivity: We proceed in a few steps:

      • Choosing $U=\left\{ a\right\} $, we obtain $R(a)=Y\setminus R(X\setminus \left\{ a\right\} )$.

      • This implies $R(a)\cap R(X\setminus \left\{ a\right\} )=\text{Ø}$.

      • Thus there can be no $a'\in X$ other than $a$ with $R(a')\cap R(a')\neq \text{Ø}$.

    • Surjectivity: Choosing $U=X$ gives $R_{*}(X)=Y$. But since $R_{!}=R_{*}$, we have $R_{!}(X)=Y$, so $R$ must be surjective.

    This finishes the proof.

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

    1. 1.

      Functionality I. The assignment $R\mapsto R_{*}$ defines a function

      \[ (-)_{*}\colon \mathsf{Sets}(X,Y) \to \mathsf{Sets}(\mathcal{P}(X),\mathcal{P}(Y)). \]
    2. 2.

      Functionality II. The assignment $R\mapsto R_{*}$ defines a function

      \[ (-)_{*}\colon \mathsf{Sets}(X,Y) \to \operatorname {\mathrm{Hom}}_{\mathsf{Pos}}((\mathcal{P}(X),\subset ),(\mathcal{P}(Y),\subset )). \]
    3. 3.

      Interaction With Identities. For each $X\in \operatorname {\mathrm{Obj}}(\mathsf{Sets})$, we have

      \[ (\operatorname {\mathrm{id}}_{X})_{*}=\operatorname {\mathrm{id}}_{\mathcal{P}(X)}. \]
    4. 4.

      Interaction With Composition. For each pair of composable relations $R\colon X\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}Y$ and $S\colon Y\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}C$, we have

    5. 5.

      Interaction With Converses. We have

      \[ R^{\dagger }_{*}=R_{-1}. \]

    Item 1: Functionality I
    There is nothing to prove.

    Item 2: Functionality II
    This follows from Item 1 of Proposition 8.7.4.1.4.

    Item 3: Interaction With Identities
    Indeed, we have

    \begin{align*} (\chi _{X})_{*}(U) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ a\in X\ \middle |\ \chi ^{-1}_{X}(a)\subset U\right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ a\in X\ \middle |\ \left\{ a\right\} \subset U\right\} \\ & = U \end{align*}

    for each $U\in \mathcal{P}(X)$. Thus $(\chi _{X})_{*}=\operatorname {\mathrm{id}}_{\mathcal{P}(X)}$.

    Item 4: Interaction With Composition
    Indeed, we have

    \begin{align*} (S\mathbin {\diamond }R)_{*}(U) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ c\in C\ \middle |\ [S\mathbin {\diamond }R]^{-1}(c)\subset U\right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ c\in C\ \middle |\ S^{-1}(R^{-1}(c))\subset U\right\} \\ & = \left\{ c\in C\ \middle |\ R^{-1}(c)\subset S_{*}(U)\right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}R_{*}(S_{*}(U))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[R_{*}\circ S_{*}](U) \end{align*}

    for each $U\in \mathcal{P}(C)$, where we used Item 2 of Proposition 8.7.4.1.4, which implies that the conditions

    • We have $S^{-1}(R^{-1}(c))\subset U$.

    • We have $R^{-1}(c)\subset S_{*}(U)$.

    are equivalent. Thus $(S\mathbin {\diamond }R)_{*}=S_{*}\circ R_{*}$.

    Item 5: Interaction With Converses
    Omitted.


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