Preimage of Union under Mapping

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

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

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

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

Let $$\mathcal P \left({T}\right)$$ be the power set of $$T$$.

Let $$\mathbb T \subseteq \mathcal P \left({T}\right)$$.

Then:
 * $$f^{-1} \left({\bigcup \mathbb T}\right) = \bigcup_{X \in \, \mathbb T} f^{-1} \left({X}\right)$$

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


 * $$\mathcal R^{-1} \left({\bigcup \mathbb T}\right) = \bigcup_{X \in \, \mathbb T} \mathcal R^{-1} \left({X}\right)$$