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

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


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

Also see

 * Mapping Image of Union


 * Mapping Image of Intersection
 * Mapping Preimage of Intersection