Subsection 8.4.7 (00LD): Monomorphisms

In this section we characterise the epimorphisms in the category $\mathsf{Rel}$, following

Let $R\colon A\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}B$ be a relation. The following conditions are equivalent:

1.
The relation $R$ is a monomorphism in $\mathsf{Rel}$.
2.
The direct image function

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

associated to $R$ is injective.
3.
The codirect image function

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

associated to $R$ is injective.

Moreover, if $R$ is a monomorphism, then it satisfies the following condition, and the converse holds if $R$ is total:

(★)

\begin{align*} a\sim _{R}b,\\ a’\sim _{R}b,\end{align*}

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

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

(★)

\begin{align*} a\sim _{R}b,\\ a’\sim _{R}b,\end{align*}

consider the diagram

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.

The Clowder Project

8.4.7 Monomorphisms