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.