Preimage of Intersection under Mapping

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

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

Then:
 * $f^{-1} \sqbrk {T_1 \cap T_2} = f^{-1} \sqbrk {T_1} \cap f^{-1} \sqbrk {T_2}$

This can be expressed in the language and notation of inverse image mappings as:
 * $\forall T_1, T_2 \in \powerset T: \map {f^\gets} {T_1 \cap T_2} = \map {f^\gets} {T_1} \cap \map {f^\gets} {T_2}$

Proof
As $f$, being a mapping, is also a many-to-one relation, it follows from Inverse of Many-to-One Relation is One-to-Many that its inverse $f^{-1}$ is a one-to-many relation.

Thus Image of Intersection under One-to-Many Relation applies:
 * $\mathcal R \sqbrk {T_1 \cap T_2} = \mathcal R \sqbrk {T_1} \cap \mathcal R \sqbrk {T_2}$

where here $\mathcal R = f^{-1}$.

Also see

 * Image of Intersection under Mapping


 * Preimage of Intersection under Relation


 * Image of Union under Mapping
 * Preimage of Union under Mapping