# 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:

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

### 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:

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

### 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 mapping.

Then:

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

where:

$\displaystyle \bigcap_{i \mathop \in I} T_i$ denotes the intersection 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 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 $\mathcal R = f^{-1}$.

$\blacksquare$