The Clowder Project

is faithful, i.e. iff the morphism

is injective for each $S,T\in \text{Obj}\webleft (\mathbf{Rel}\webleft (B,X\webright )\webright )$. However, $\textup{Hom}_{\mathbf{Rel}\webleft (B,X\webright )}\webleft (S,T\webright )$ is either empty or a singleton, in either case of which the map $R^{*}_{S,T}$ is necessarily injective.

• Item (a)$\iff $Item (b): This is simply a matter of unwinding definitions: The relation $R$ is a corepresentably full morphism in $\textbf{Rel}$ iff, for each $X\in \text{Obj}\webleft (\textbf{Rel}\webright )$, the functor
  \[ R^{*}\colon \mathbf{Rel}\webleft (B,X\webright )\to \mathbf{Rel}\webleft (A,X\webright ) \]

  is full, i.e. iff the morphism

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

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

• Item (c)$\iff $Item (d): This is also simply a matter of unwinding definitions: The functor
  \[ R^{-1}\colon \webleft (\mathcal{P}\webleft (B\webright ),\subset \webright )\to \webleft (\mathcal{P}\webleft (A\webright ),\subset \webright ) \]

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

  \[ R^{-1}_{U,V}\colon \textup{Hom}_{\mathcal{P}\webleft (B\webright )}\webleft (U,V\webright )\to \textup{Hom}_{\mathcal{P}\webleft (A\webright )}\webleft (R^{-1}\webleft (U\webright ),R^{-1}\webleft (V\webright )\webright ) \]

  is surjective, i.e. iff whenever $R^{-1}\webleft (U\webright )\subset R^{-1}\webleft (V\webright )$, we also necessarily have $U\subset V$.

• Item (e)$\iff $Item (f): This is once again simply a matter of unwinding definitions, and proceeds exactly in the same way as in the proof of the equivalence between Item (c) and Item (d) given above.
• Item (d)$\implies $Item (f): Suppose that the following condition is true:
  • For each $U,V\in \mathcal{P}\webleft (B\webright )$, if $R^{-1}\webleft (U\webright )\subset R^{-1}\webleft (V\webright )$, then $U\subset V$.
  We need to show that the condition
  • For each $U,V\in \mathcal{P}\webleft (B\webright )$, if $R_{-1}\webleft (U\webright )\subset R_{-1}\webleft (V\webright )$, then $U\subset V$.
  is also true. We proceed step by step:
  1. Suppose we have $U,V\in \mathcal{P}\webleft (B\webright )$ with $R_{-1}\webleft (U\webright )\subset R_{-1}\webleft (V\webright )$.
  2. By Chapter 7: Constructions With Relations, Item 7 of Proposition 7.4.2.1.3, we have
     \begin{align*} R_{-1}\webleft (U\webright ) & = B\setminus R^{-1}\webleft (A\setminus U\webright ),\\ R_{-1}\webleft (V\webright ) & = B\setminus R^{-1}\webleft (A\setminus V\webright ). \end{align*}

  3. By Chapter 2: Constructions With Sets, Item 1 of Proposition 2.3.10.1.2 we have $R^{-1}\webleft (A\setminus V\webright )\subset R^{-1}\webleft (A\setminus U\webright )$.
  4. By assumption, we then have $A\setminus V\subset A\setminus U$.
  5. By Chapter 2: Constructions With Sets, Item 1 of Proposition 2.3.10.1.2 again, we have $U\subset V$.
• Item (f)$\implies $Item (d): Suppose that the following condition is true:
  • For each $U,V\in \mathcal{P}\webleft (B\webright )$, if $R_{-1}\webleft (U\webright )\subset R_{-1}\webleft (V\webright )$, then $U\subset V$.
  We need to show that the condition
  • For each $U,V\in \mathcal{P}\webleft (B\webright )$, if $R^{-1}\webleft (U\webright )\subset R^{-1}\webleft (V\webright )$, then $U\subset V$.
  is also true. We proceed step by step:
  1. Suppose we have $U,V\in \mathcal{P}\webleft (B\webright )$ with $R^{-1}\webleft (U\webright )\subset R^{-1}\webleft (V\webright )$.
  2. By Chapter 7: Constructions With Relations, Item 7 of Proposition 7.4.3.1.3, we have
     \begin{align*} R^{-1}\webleft (U\webright ) & = B\setminus R_{-1}\webleft (A\setminus U\webright ),\\ R^{-1}\webleft (V\webright ) & = B\setminus R_{-1}\webleft (A\setminus V\webright ). \end{align*}

  3. By Chapter 2: Constructions With Sets, Item 1 of Proposition 2.3.10.1.2 we have $R_{-1}\webleft (A\setminus V\webright )\subset R_{-1}\webleft (A\setminus U\webright )$.
  4. By assumption, we then have $A\setminus V\subset A\setminus U$.
  5. By Chapter 2: Constructions With Sets, Item 1 of Proposition 2.3.10.1.2 again, we have $U\subset V$.
• Item (b)$\implies $Item (d): Consider the diagram
  

  and suppose that $S\mathbin {\diamond }R\subset T\mathbin {\diamond }R$. Note that, by assumption, given a diagram of the form

  

  if $R^{-1}\webleft (U\webright )=R\mathbin {\diamond }U\subset R\mathbin {\diamond }V=R^{-1}\webleft (V\webright )$, then $U\subset V$. In particular, for each $x\in X$, we may consider the diagram

  

  for which we have $\webleft [x\webright ]\mathbin {\diamond }S\mathbin {\diamond }R\subset \webleft [x\webright ]\mathbin {\diamond }T\mathbin {\diamond }R$, implying that we have

  \[ S^{-1}\webleft (x\webright )=\webleft [x\webright ]\mathbin {\diamond }S\subset \webleft [x\webright ]\mathbin {\diamond }T=T^{-1}\webleft (x\webright ) \]

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

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

  By , we have

  \begin{align*} R^{-1}\webleft (U\webright ) & = U\mathbin {\diamond }R,\\ R^{-1}\webleft (V\webright ) & = V\mathbin {\diamond }R. \end{align*}

  Now, if $R^{-1}\webleft (U\webright )\subset R^{-1}\webleft (V\webright )$, i.e. $U\mathbin {\diamond }R\subset V\mathbin {\diamond }R$, then $U\subset V$ by assumption.