The Clowder Project

Firstly note that Item 2 and Item 3 are equivalent by Chapter 7: Constructions With Relations, Item 7 of Proposition 7.4.1.1.3. We then claim that Item 1 and Item 2 are also equivalent:

• Item 1$\implies $Item 2: Let $U,V\in \mathcal{P}\webleft (A\webright )$ and consider the diagram
  

  By Chapter 7: Constructions With Relations, Remark 7.4.1.1.2, we have

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

  Now, if $R\mathbin {\diamond }U=R\mathbin {\diamond }V$, i.e. $R_{*}\webleft (U\webright )=R_{*}\webleft (V\webright )$, then $U=V$ since $R$ is assumed to be a monomorphism, showing $R_{*}$ to be injective.

• Item 2$\implies $Item 1: Conversely, suppose that $R_{*}$ is injective, consider the diagram
  

  and suppose that $R\mathbin {\diamond }S=R\mathbin {\diamond }T$. Note that, since $R_{*}$ is injective, given a diagram of the form

  

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

  

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

  \[ S\webleft (x\webright )=S\mathbin {\diamond }\webleft [x\webright ]=T\mathbin {\diamond }\webleft [x\webright ]=T\webleft (x\webright ) \]

  for each $x\in X$, implying $S=T$, and thus $R$ is a monomorphism.

• Item 1$\implies $Item 2: Assume that $R$ is a monomorphism.
  • We first notice that the functor $\mathrm{Rel}\webleft (\text{pt},-\webright )\colon \mathrm{Rel}\to \mathsf{Sets}$ maps $R$ to $R_{*}$ by Chapter 7: Constructions With Relations, Remark 7.4.1.1.2.
  • Since $\mathrm{Rel}\webleft (\text{pt},-\webright )$ preserves all limits by , of , it follows by of that $\mathrm{Rel}\webleft (\text{pt},-\webright )$ also preserves monomorphisms.
  • Since $R$ is a monomorphism and $\mathrm{Rel}\webleft (\text{pt},-\webright )$ maps $R$ to $R_{*}$, it follows that $R_{*}$ is also a monomorphism.
  • Since the monomorphisms in $\mathsf{Sets}$ are precisely the injections ( of ), it follows that $R_{*}$ is injective.
• Item 2$\implies $Item 1: Assume that $R_{*}$ is injective.
  • We first notice that the functor $\mathrm{Rel}\webleft (\text{pt},-\webright )\colon \mathrm{Rel}\to \mathsf{Sets}$ maps $R$ to $R_{*}$ by Chapter 7: Constructions With Relations, Remark 7.4.1.1.2.
  • Since the monomorphisms in $\mathsf{Sets}$ are precisely the injections ( of ), it follows that $R_{*}$ is a monomorphism.
  • Since $\mathrm{Rel}\webleft (\text{pt},-\webright )$ is faithful, it follows by of that $\mathrm{Rel}\webleft (\text{pt},-\webright )$ reflects monomorphisms.
  • Since $R_{*}$ is a monomorphism and $\mathrm{Rel}\webleft (\text{pt},-\webright )$ maps $R$ to $R_{*}$, it follows that $R$ is also a monomorphism.

Since $\star \sim _{\webleft [a\webright ]}a$ and $a\sim _{R}b$, we have $\star \sim _{R\mathbin {\diamond }\webleft [a\webright ]}b$. Similarly, $\star \sim _{R\mathbin {\diamond }\webleft [a'\webright ]}b$. Thus $R\mathbin {\diamond }\webleft [a\webright ]=R\mathbin {\diamond }\webleft [a'\webright ]$, and since $R$ is a monomorphism, we have $\webleft [a\webright ]=\webleft [a'\webright ]$, i.e. $a=a'$.

Conversely, assume the condition

• For each $a,a'\in A$, if there exists some $b\in B$ such that
  \begin{align*} a\sim _{R}b,\\ a’\sim _{R}b,\end{align*}

  then $a=a'$.

and let $\webleft (x,a\webright )\in S$. Since $R$ is total and $a\in A$, there exists some $b\in B$ such that $a\sim _{R}b$. In this case, we have $x\sim _{R\mathbin {\diamond }S}b$, and since $R\mathbin {\diamond }S=R\mathbin {\diamond }T$, we have also $x\sim _{R\mathbin {\diamond }T}b$. Thus there must exist some $a'\in A$ such that $x\sim _{T}a'$ and $a'\sim _{R}b$. However, since $a,a'\sim _{R}b$, we must have $a=a'$, and thus $\webleft (x,a\webright )\in T$ as well.

A similar argument shows that if $\webleft (x,a\webright )\in T$, then $\webleft (x,a\webright )\in S$, and thus $S=T$ and it follows that $R$ is a monomorphism.