One-to-Many Relation Composite with Inverse is Coreflexive

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

Then the composite of $\mathcal R$ with its inverse is a subset of the diagonal relation:
 * $\mathcal R^{-1} \circ \mathcal R \subseteq \Delta_X$

That is, by this result, $\mathcal R^{-1} \circ \mathcal R$ is both symmetric and antisymmetric.

Proof
As $\mathcal R$ is one-to-many, it follows from Inverse of Many-to-One Relation is One-to-Many that $\mathcal R^{-1}$ is many-to-one.

Let $\left({x, z}\right) \in \mathcal R^{-1} \circ \mathcal R$.

Then $\exists y \in S: \left({x, y}\right) \in \mathcal R, \left({y, z}\right) \in \mathcal R^{-1}$.

But as $\left({x, y}\right) \in \mathcal R$, it follows that $\left({y, x}\right) \in \mathcal R^{-1}$ from the definition of inverse relation.

But as $\mathcal R^{-1}$ is many-to-one, it follows that $\left({y, z}\right) = \left({y, x}\right)$.

Thus if $\left({x, z}\right) \in \mathcal R^{-1} \circ \mathcal R$, it follows that $\left({x, z}\right) = \left({x, x}\right)$.

Thus $\mathcal R^{-1} \circ \mathcal R \subseteq \Delta_X$.

The result follows.