Preimage of Set Difference under Mapping

Theorem
Let $$f: S \to T$$ be a mapping. Let $$C$$ and $$D$$ be subsets of $$T$$.

Then:
 * $$f^{-1} \left({C \setminus D}\right) = f^{-1} \left({C}\right) \setminus f^{-1} \left({D}\right)$$

where $$\setminus$$ denotes set difference.

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 one-to-many.

Thus we can apply One-to-Many Image of Set Difference:


 * $$\mathcal{R} \left({A \setminus B}\right) = \mathcal{R} \left({A}\right) \setminus \mathcal{R} \left({B}\right)$$