Image of Intersection under Injection

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

Then:
 * $\forall A, B \subseteq S: f \left({A \cap B}\right) = f \left({A}\right) \cap f \left({B}\right)$

iff $f$ is an injection.

General Result
Let $\mathcal R \subseteq S \times T$ be a relation.

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

Then:
 * $\displaystyle \forall \mathbb S \subseteq \mathcal P \left({S}\right): f \left({\bigcap \mathbb S}\right) = \bigcap_{X \in \, \mathbb S} f \left({X}\right)$

iff $f$ is an injection.

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

Therefore One-to-Many Image of Intersections applies:


 * $\forall A, B \subseteq S: \mathcal R \left({A \cap B}\right) = \mathcal R \left({A}\right) \cap \mathcal R \left({B}\right)$

iff $\mathcal R$ is a one-to-many relation.

Given that $f$ is a mapping, it follows that:
 * $\forall A, B \subseteq S: f \left({A \cap B}\right) = f \left({A}\right) \cap f \left({B}\right)$

iff $f$ is an injection.

Proof of General Result
Follows directly from the same approach as the above, and from One-to-Many Image of Intersections: General Result.