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$⟹$Item 2: Let $U,V\in \mathcal{P}\left(A\right)$ and consider the diagram
  

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

  $$\begin{array}{rl}{R}_{\ast }\left(U\right)& =R\diamond U,\\ R_{\ast }\left(V\right)& =R\diamond V.\end{array}$$

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

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

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

  

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

  

  for which we have $R\diamond S\diamond \left[x\right]=R\diamond T\diamond \left[x\right]$, implying that we have

  $$S\left(x\right)=S\diamond \left[x\right]=T\diamond \left[x\right]=T\left(x\right)$$

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

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

Since $\star {\sim }_{\left[a\right]}a$ and $a{\sim }_{R}b$, we have $\star {\sim }_{R\diamond \left[a\right]}b$. Similarly, $\star {\sim }_{R\diamond \left[a^{\prime }\right]}b$. Thus $R\diamond \left[a\right]=R\diamond \left[a^{\prime }\right]$, and since $R$ is a monomorphism, we have $\left[a\right]=\left[a^{\prime }\right]$, i.e. $a=a^{\prime }$.

Conversely, assume the condition

• For each $a,a^{\prime }\in A$, if there exists some $b\in B$ such that
  $$\begin{array}{r}a{\sim }_{R}b,\\ a^{\prime }{\sim }_{R}b,\end{array}$$

  then $a=a^{\prime }$.

and let $(x,a)\in S$. 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 $x{\sim }_{R\diamond S}b$, and since $R\diamond S=R\diamond T$, we have also $x{\sim }_{R\diamond T}b$. Thus there must exist some $a^{\prime }\in A$ such that $x{\sim }_T{a}^{\prime }$ and $a^{\prime }{\sim }_{R}b$. However, since $a,a^{\prime }{\sim }_{R}b$, we must have $a=a^{\prime }$, and thus $(x,a)\in T$ as well.

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