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:

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

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.