Image of Intersection under Injection/General Result

Theorem

Let $S$ and $T$ be sets.

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

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

Then:

$\ds \forall \mathbb S \subseteq \powerset S: f \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} f \sqbrk X$

if and only if $f$ is an injection.

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

$\ds \forall \mathbb S \subseteq \powerset S : \map {f^\to} {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} \map {f^\to} X$

An injection is a type of one-to-one relation, and therefore also a one-to-many relation.

$\ds \forall \mathbb S \subseteq \powerset S: \RR \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk {\mathbb S}$

if and only if $\RR$ is a one-to-many relation.

We have that $f$ is a mapping and therefore a many-to-one relation.

So $f$ is a one-to-many relation if and only if $f$ is also an injection.

It follows that:

$\ds \forall \mathbb S \subseteq \powerset S: f \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} f \sqbrk X$

if and only if $f$ is an injection.

$\blacksquare$

1996: Winfried Just and Martin Weese: Discovering Modern Set Theory. I: The Basics ... (previous) ... (next): Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions: Exercise $7 \ \text {(b)}$