Preimage of Union under Mapping

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

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


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

Also see

 * Image of Union under Mapping


 * Image of Intersection under Mapping
 * Preimage of Intersection under Mapping