Relating morphisms between two objects in category of finite sets and relations

Question

In category of finite Sets and Relations (let's call it FinRel), suppose $A$ and $B$ are two objects (sets), and $b1,b2,b3$ are some morphisms (relations) from $A$ to $B$.

Interpreting $b1 .. b3$ as relations, we can define Union of them to be $b4$ (i.e. $b4 = b1 \cup b2 \cup b3$ ). Obviously, $b4$ is a morphism from $A$ to $B$ in FinRel category.

My question is:
how I can specify relatedness of $b4$ with other three morphisms (i.e., $b1..b3$) categorically? In other words, what categorical operation(s) would return me $b4$, provided that $b1$,$b2$, and $b3$ are known?

Thanks

Answer 1

I assume you mean union of relations $b_1,b_2$ and not disjoint union of relations $b_1,b_2$, because if you take the disjoint union of a non-empty relation $R: A\to B$ with itself, you had each edge from $R$ twice in $R+R$. So $R+R$ would not be a relation $A\to B$, because $R+R$ is not a subset of $A\times B$ anymore. But everything works out well if you replace disjoint union by simple union.

Short Answer

You can get $u=``b_1\cup b_2\text{''}$ out of the coproduct in Rel : $$ u:= [b_1,b_2] \cdot \Delta \equiv (A \xrightarrow{\Delta} A + A \xrightarrow{[b_1,b_2]} B ) $$ where $\Delta: A\to A+A$ connects each $a\in A$ with both the two disjoint copies of $a$ in $A+A$.

As the coproduct is associative, $b_1\cup b_2\cup b_3$ can be defined incrementally as e.g. $b_1\cup (b_2 \cup b_3)$ or directly as: $$ u:= [b_1,b_2,b_3] \cdot \Delta_3 \equiv (A \xrightarrow{\Delta_3} A + A +A \xrightarrow{[b_1,b_2,b_3]} B ) $$

Notation

In Rel, coproducts exists and are disjoint unions. The coproduct injections are denoted by inl and inr: $X \xrightarrow{\text{inl}} X+Y \xleftarrow{\text{inr}} Y$.

The morphism $[b_1,b_2]$ denotes the unique morphism from the coproduct $A+A$ into $B$ induced by the morphisms $b_1$ and $b_2$.

Here, $\Delta: A\to A+A$ denotes the right-inverse of $[\text{id}_A,\text{id}_A]: A+A \to A$, i.e. $[\text{id}_A,\text{id}_A]\cdot \Delta= \text{id}_A$. This right-inverse can be defined by $\Delta := \{(a,x)\in A\times (A+A) \mid (a,x)\in \text{inl}\text{ or } (a,x) \in \text{inr}\}$, where $\times$ denotes the cartesian product. (Note that for $|A| \ge 2$, $\Delta$ is not a left-inverse.)

Why it works

$$ (a,b) \in b_1\cup b_2 \\ \Leftrightarrow (a,b) \in b_1\text{ or } (a,b) \in b_2 \\ \Leftrightarrow (a,b) \in [b_1,b_2]\cdot\text{inl or } (a,b) \in [b_1,b_2]\cdot\text{inr} \quad\text{(By universal property)} \\ \Leftrightarrow (a,b) \in [b_1,b_2]\cdot \Delta\quad\text{(By definition of $\Delta$)} $$