# Preimage of Union under Mapping/Family of Sets

## Theorem

Let $S$ and $T$ be sets.

Let $\family {T_i}_{i \mathop \in I}$ be a family of subsets of $T$.

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

Then:

$\ds f^{-1} \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} f^{-1} \sqbrk {T_i}$

where:

$\ds \bigcup_{i \mathop \in I} T_i$ denotes the union of $\family {T_i}_{i \mathop \in I}$
$f^{-1} \sqbrk {T_i}$ denotes the preimage of $T_i$ under $f$.

## Proof 1

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

$\ds \RR^{-1} \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} \RR^{-1} \sqbrk {T_i}$

where $\RR^{-1} \sqbrk {T_i}$ denotes the preimage of $T_i$ under $\RR^{-1}$.

$\blacksquare$

## Proof 2

We have that $f$ is a mapping, and so also a relation.

Thus its inverse $f^{-1}$ is also a relation.

Hence we can apply Image of Union under Relation: Family of Sets:

$\ds \RR \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} \RR \sqbrk {T_i}$

where $\RR \sqbrk {T_i}$ denotes the image of $T_i$ under $\RR$.

$\blacksquare$