The Clowder Project

is faithful, i.e. iff the morphism

is injective for each $S,T\in \operatorname {\mathrm{Obj}}\webleft (\mathbf{Rel}\webleft (X,A\webright )\webright )$. However, $\operatorname {\mathrm{Hom}}_{\mathbf{Rel}\webleft (X,A\webright )}\webleft (S,T\webright )$ is either empty or a singleton, in either case of which the map $R_{*|S,T}$ is necessarily injective.

• •

  Item 2a$\iff $Item 2b: This is simply a matter of unwinding definitions: The relation $R$ is a representably full morphism in $\boldsymbol {\mathsf{Rel}}$ iff , for each $X\in \operatorname {\mathrm{Obj}}\webleft (\boldsymbol {\mathsf{Rel}}\webright )$, the functor

  \[ R_{!}\colon \mathbf{Rel}\webleft (X,A\webright )\to \mathbf{Rel}\webleft (X,B\webright ) \]

  is full, i.e. iff the morphism

  \[ R_{*|S,T}\colon \operatorname {\mathrm{Hom}}_{\mathbf{Rel}\webleft (X,A\webright )}\webleft (S,T\webright )\to \operatorname {\mathrm{Hom}}_{\mathbf{Rel}\webleft (X,B\webright )}\webleft (R\mathbin {\diamond }S,R\mathbin {\diamond }T\webright ) \]

  is surjective for each $S,T\in \operatorname {\mathrm{Obj}}\webleft (\mathbf{Rel}\webleft (X,A\webright )\webright )$, i.e. iff , whenever $R\mathbin {\diamond }S\subset R\mathbin {\diamond }T$, we also have $S\subset T$.

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  Item 2c$\iff $Item 2d: This is also simply a matter of unwinding definitions: The functor

  \[ R_{!}\colon \webleft (\mathcal{P}\webleft (A\webright ),\subset \webright )\to \webleft (\mathcal{P}\webleft (B\webright ),\subset \webright ) \]

  is full iff , for each $U,V\in \mathcal{P}\webleft (A\webright )$, the morphism

  \[ R_{*|U,V}\colon \operatorname {\mathrm{Hom}}_{\mathcal{P}\webleft (A\webright )}\webleft (U,V\webright )\to \operatorname {\mathrm{Hom}}_{\mathcal{P}\webleft (B\webright )}\webleft (R_{!}\webleft (U\webright ),R_{!}\webleft (V\webright )\webright ) \]

  is surjective, i.e. iff whenever $R_{!}\webleft (U\webright )\subset R_{!}\webleft (V\webright )$, we also necessarily have $U\subset V$.

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  Item 2e$\iff $Item 2f: This is once again simply a matter of unwinding definitions, and proceeds exactly in the same way as in the proof of the equivalence between Item 2c and Item 2d given above.

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  Item 2d$\implies $Item 2f: Suppose that the following condition is true:

  • (★)
  For each $U,V\in \mathcal{P}\webleft (A\webright )$, if $R_{!}\webleft (U\webright )\subset R_{!}\webleft (V\webright )$, then $U\subset V$.

  We need to show that the condition

  • (★)
  For each $U,V\in \mathcal{P}\webleft (A\webright )$, if $R_{*}\webleft (U\webright )\subset R_{*}\webleft (V\webright )$, then $U\subset V$.

  is also true. We proceed step by step:

  • •

    Suppose we have $U,V\in \mathcal{P}\webleft (A\webright )$ with $R_{*}\webleft (U\webright )\subset R_{*}\webleft (V\webright )$.

  • •

    By Chapter 9: Constructions With Relations, of , we have

    \begin{align*} R_{*}\webleft (U\webright ) & = B\setminus R_{!}\webleft (A\setminus U\webright ),\\ R_{*}\webleft (V\webright ) & = B\setminus R_{!}\webleft (A\setminus V\webright ). \end{align*}
  • •

    By Chapter 4: Constructions With Sets, Item 1 of Proposition 4.3.10.1.2 we have $R_{!}\webleft (A\setminus V\webright )\subset R_{!}\webleft (A\setminus U\webright )$.

  • •

    By assumption, we then have $A\setminus V\subset A\setminus U$.

  • •

    By Chapter 4: Constructions With Sets, Item 1 of Proposition 4.3.10.1.2 again, we have $U\subset V$.

• •

  Item 2f$\implies $Item 2d: Suppose that the following condition is true:

  • (★)
  For each $U,V\in \mathcal{P}\webleft (A\webright )$, if $R_{*}\webleft (U\webright )\subset R_{*}\webleft (V\webright )$, then $U\subset V$.

  We need to show that the condition

  • (★)
  For each $U,V\in \mathcal{P}\webleft (A\webright )$, if $R_{!}\webleft (U\webright )\subset R_{!}\webleft (V\webright )$, then $U\subset V$.

  is also true. We proceed step by step:

  • •

    Suppose we have $U,V\in \mathcal{P}\webleft (A\webright )$ with $R_{!}\webleft (U\webright )\subset R_{!}\webleft (V\webright )$.

  • •

    By Chapter 9: Constructions With Relations, of , we have

    \begin{align*} R_{!}\webleft (U\webright ) & = B\setminus R_{*}\webleft (A\setminus U\webright ),\\ R_{!}\webleft (V\webright ) & = B\setminus R_{*}\webleft (A\setminus V\webright ). \end{align*}
  • •

    By Chapter 4: Constructions With Sets, Item 1 of Proposition 4.3.10.1.2 we have $R_{*}\webleft (A\setminus V\webright )\subset R_{*}\webleft (A\setminus U\webright )$.

  • •

    By assumption, we then have $A\setminus V\subset A\setminus U$.

  • •

    By Chapter 4: Constructions With Sets, Item 1 of Proposition 4.3.10.1.2 again, we have $U\subset V$.

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  Item 2b$\implies $Item 2d: Consider the diagram

  
  and suppose that $R\mathbin {\diamond }S\subset R\mathbin {\diamond }T$. Note that, by assumption, given a diagram of the form
  
  if $R_{!}\webleft (U\webright )=R\mathbin {\diamond }U\subset R\mathbin {\diamond }V=R_{!}\webleft (V\webright )$, then $U\subset V$. In particular, for each $x\in X$, we may consider the diagram
  
  for which we have $R\mathbin {\diamond }S\mathbin {\diamond }\webleft [x\webright ]\subset R\mathbin {\diamond }T\mathbin {\diamond }\webleft [x\webright ]$, implying that we have
  \[ S\webleft (x\webright )=S\mathbin {\diamond }\webleft [x\webright ]\subset T\mathbin {\diamond }\webleft [x\webright ]=T\webleft (x\webright ) \]

  for each $x\in X$, implying $S\subset T$.

• •

  Item 2d$\implies $Item 2b: Let $U,V\in \mathcal{P}\webleft (A\webright )$ and consider the diagram

  
  By Remark 8.5.1.1.2, we have
  \begin{align*} R_{!}\webleft (U\webright ) & = R\mathbin {\diamond }U,\\ R_{!}\webleft (V\webright ) & = R\mathbin {\diamond }V. \end{align*}

  Now, if $R_{!}\webleft (U\webright )\subset R_{!}\webleft (V\webright )$, i.e. $R\mathbin {\diamond }U\subset R\mathbin {\diamond }V$, then $U\subset V$ by assumption.