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 codomain from Codomain of Composite Relation and the definition of Inverse Relation.

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

This completes the proof.