Many-to-One Relation Composite with Inverse is Transitive

Theorem
Let $\RR \subseteq S \times T$ be a relation which is many-to-one.

Then the composites (both ways) of $\RR$ and its inverse $\RR^{-1}$, that is, both $\RR^{-1} \circ \RR$ and $\RR \circ \RR^{-1}$, are transitive.

Proof
Let $\RR \subseteq S \times T$ be many-to-one.

Then, from the definition of many-to-one:


 * $\RR \subseteq S \times T: \forall x \in S: \tuple {x, y_1} \in \RR \land \tuple {x, y_2} \in \RR \implies y_1 = y_2$

Also, note that from Inverse of Many-to-One Relation is One-to-Many, $\RR^{-1}$ is one-to-many.

Let $\tuple {a, b}, \tuple {b, c} \in \RR^{-1} \circ \RR$.

Thus $\RR^{-1} \circ \RR$ is transitive.

Now let $\tuple {p, q}, \tuple {q, r} \in \RR \circ \RR^{-1}$.

But $\RR$ is many-to-one.

This means that:
 * $\forall x \in S: \tuple {x, y_1} \in \RR \land \tuple {x, y_2} \in \RR \implies y_1 = y_2$

So:

Thus (trivially) $\RR \circ \RR^{-1}$ is transitive.