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:
 * $$f^{-1} \left({T_1 \cap T_2}\right) = f^{-1} \left({T_1}\right) \cap f^{-1} \left({T_2}\right)$$.

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

Let $$\mathbb T \subseteq \mathcal P \left({T}\right)$$.

Then:
 * $$f \left({\bigcap \mathbb T}\right) = \bigcap_{X \in \, \mathbb T} f \left({X}\right)$$

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 we can apply One-to-Many Image of Intersections:
 * $$\mathcal R \left({T_1 \cap T_2}\right) = \mathcal R \left({T_1}\right) \cap \mathcal R \left({T_2}\right)$$

and:
 * $$\mathcal R \left({\bigcap \mathbb T}\right) = \bigcap_{X \in \, \mathbb T} \mathcal R \left({X}\right)$$

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