Preimage of Union under Mapping

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Theorem

Let $S$ and $T$ be sets.

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

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


Then:

$f^{-1} \sqbrk {T_1 \cup T_2} = f^{-1} \sqbrk {T_1} \cup f^{-1} \sqbrk {T_2}$


This can be expressed in the language and notation of inverse image mappings as:

$\forall T_1, T_2 \in \powerset T: \map {f^\gets} {T_1 \cup T_2} = \map {f^\gets} {T_1} \cup \map {f^\gets} {T_2}$


General Result

Let $S$ and $T$ be sets.

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

Let $\powerset T$ be the power set of $T$.

Let $\mathbb T \subseteq \powerset T$.


Then:

$\ds f^{-1} \sqbrk {\bigcup \mathbb T} = \bigcup_{X \mathop \in \mathbb T} f^{-1} \sqbrk X$


Family of Sets

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

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

$\RR^{-1} \sqbrk {T_1 \cup T_2} = \RR^{-1} \sqbrk {T_1} \cup \RR^{-1} \sqbrk {T_2}$

$\blacksquare$


Also see


Sources

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