Composition of Relation with Inverse is Symmetric

Theorem
Let $$\mathcal{R} \subseteq S \times T$$ be a relation.

Then the composition of $$\mathcal{R}$$ with its inverse $$\mathcal{R}^{-1}$$ is symmetric:


 * $$\mathcal{R}^{-1} \circ \mathcal{R}$$ is a symmetric relation on $$S$$;
 * $$\mathcal{R} \circ \mathcal{R}^{-1}$$ is a symmetric relation on $$T$$.

Proof
Note that this result holds for any $$\mathcal{R} \subseteq S \times T$$, and does not require that $$\left({S; \mathcal{R}}\right)$$ necessarily be a relational structure.

$$ $$ $$ $$

Thus $$\left({a, b}\right) \in \mathcal{R}^{-1} \circ \mathcal{R} \implies \left({b, a}\right) \in \mathcal{R}^{-1} \circ \mathcal{R}$$ and thus $$\mathcal{R}^{-1} \circ \mathcal{R}$$ is symmetric.

As $$\mathcal{R} = \left({\mathcal{R}^{-1}}\right)^{-1}$$ from Inverse of Inverse Relation, it follows that $$\mathcal{R} \circ \mathcal{R}^{-1} = \left({\mathcal{R}^{-1}}\right)^{-1} \circ \mathcal{R}^{-1}$$ is likewise a symmetric relation.

The domain of $$\mathcal{R}^{-1} \circ \mathcal{R}$$ is $$S$$ from Domain of Composite Relation, as is its range from Range of Composite Relation and the definition of Inverse Relation.

Similarly, the range of $$\mathcal{R} \circ \mathcal{R}^{-1}$$ is $$T$$, as is its domain.

This completes the proof.