Preimage of Union under Relation

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

Let $T_1$ and $T_2$ be subsets of $T$.

Then:
 * $\mathcal R^{-1} \left({T_1 \cup T_2}\right) = \mathcal R^{-1} \left({T_1}\right) \cup \mathcal R^{-1} \left({T_2}\right)$

Proof
This follows from Image of Union, and the fact that $\mathcal R^{-1}$ is itself a relation, and therefore obeys the same rules.