Preimage of Union under Mapping

Let $$f: S \to T$$ be a mapping.

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

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

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

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

Proof
As $$f$$, being a mapping, is also a relation, we can apply Preimage of Union:

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

and

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