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.

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.

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

So $f$ is a one-to-many relation iff $f$ is also an injection.

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.