The Clowder Project

Firstly note that Item 2 and Item 3 are equivalent by Chapter 9: Constructions With Relations, of . 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 9: Constructions With Relations, , 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.

We can also prove this in a more abstract way, following [Yuan, Mono's and epi's in the category Rel?]:

• •

  Item 1$\implies $Item 2: Assume that $R$ is a monomorphism.

  • •

    We first notice that the functor $\mathrm{Rel}\webleft (\mathrm{pt},-\webright )\colon \mathrm{Rel}\to \mathsf{Sets}$ maps $R$ to $R_{!}$ by Chapter 9: Constructions With Relations, .

  • •

    Since $\mathrm{Rel}\webleft (\mathrm{pt},-\webright )$ preserves all limits by , of , it follows by , of that $\mathrm{Rel}\webleft (\mathrm{pt},-\webright )$ also preserves monomorphisms.

  • •

    Since $R$ is a monomorphism and $\mathrm{Rel}\webleft (\mathrm{pt},-\webright )$ 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 $\mathrm{Rel}\webleft (\mathrm{pt},-\webright )\colon \mathrm{Rel}\to \mathsf{Sets}$ maps $R$ to $R_{!}$ by Chapter 9: Constructions With Relations, .

  • •

    Since the monomorphisms in $\mathsf{Sets}$ are precisely the injections (, of ), it follows that $R_{!}$ is a monomorphism.

  • •

    Since $\mathrm{Rel}\webleft (\mathrm{pt},-\webright )$ is faithful, it follows by , of that $\mathrm{Rel}\webleft (\mathrm{pt},-\webright )$ reflects monomorphisms.

  • •

    Since $R_{!}$ is a monomorphism and $\mathrm{Rel}\webleft (\mathrm{pt},-\webright )$ maps $R$ to $R_{!}$, it follows that $R$ is also a monomorphism.

Finally, we prove the second part of the statement. Assume that $R$ is a monomorphism, let $a,a'\in A$ such that $a\sim _{R}b$ and $a'\sim _{R}b$ for some $b\in B$, and consider the diagram

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 $a=a'$.

consider the diagram

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.