Preimage of Union under Relation

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

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

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

Generalized Result
Let $$T_i \subseteq T: i \in \mathbb{N}^*_n$$.

Then $$\mathcal R^{-1} \left({\bigcup_{i = 1}^n T_i}\right) = \bigcup_{i = 1}^n \mathcal R^{-1} \left({T_i}\right)$$.

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