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.