Preimage of Intersection under Mapping

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

Theorem
Let $$T_1$$ and $$T_2$$ be subsets of $$T$$.

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

Generalized Result
Let $$T_i \subseteq T: i \in \N^*_n$$.

Then:
 * $$f^{-1} \left({\bigcap_{i = 1}^n T_i}\right) = \bigcap_{i = 1}^n f^{-1} \left({T_i}\right)$$.

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:


 * $$\mathcal{R} \left({\bigcap_{i = 1}^n T_i}\right) = \bigcap_{i = 1}^n \mathcal{R} \left({T_i}\right)$$

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