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

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


Then:

$\displaystyle f^{-1} \left[{\bigcup \mathbb T}\right] = \bigcup_{X \mathop \in \mathbb T} f^{-1} \left[{X}\right]$


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 relation.


Then:

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

where:

$\displaystyle \bigcup_{i \mathop \in I} T_i$ 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:

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

$\blacksquare$


Also see


Sources