Preimage of Union under Mapping/Family of Sets

Theorem
Let $S$ and $T$ be sets.

Let $\left\langle{T_i}\right\rangle_{i \mathop \in I}$ be a family of subsets of $T$.

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

Then:
 * $\displaystyle f^\gets \left({\bigcup_{i \mathop \in I} T_i}\right) = \bigcup_{i \mathop \in I} f^\gets \left({T_i}\right)$

where:
 * $\displaystyle \bigcup_{i \mathop \in I} T_i$ denotes the union of $\left\langle{T_i}\right\rangle_{i \mathop \in I}$
 * $f^\gets$ denotes the mapping induced on the power set $\mathcal P \left({T}\right)$ of $T$ by the inverse $f^{-1}$.

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


 * $\displaystyle \mathcal R^\gets \left({\bigcup_{i \mathop \in I} T_i}\right) = \bigcup_{i \mathop \in I} \mathcal R^\gets \left({T_i}\right)$

where $\mathcal R^\gets$ denotes the mapping induced on the power set $\mathcal P \left({T}\right)$ of $T$ by the inverse $\mathcal R^{-1}$.