Many-to-One Relation Composite with Inverse is Transitive

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

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

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

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


 * $\mathcal R \subseteq S \times T: \forall x \in S: \left({x, y_1}\right) \in \mathcal R \land \left({x, y_2}\right) \in \mathcal R \implies y_1 = y_2$

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

Let $\left({a, b}\right), \left({b, c}\right) \in \mathcal R^{-1} \circ \mathcal R$.

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

Now let $\left({p, q}\right), \left({q, r}\right) \in \mathcal R \circ \mathcal R^{-1}$.

But $\mathcal R$ is many-to-one.

This means that
 * $\forall x \in S: \left({x, y_1}\right) \in \mathcal R \land \left({x, y_2}\right) \in \mathcal R \implies y_1 = y_2$

So:

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