Preimage of Set Difference under Mapping

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

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

Then:
 * $f^{-1} \sqbrk {T_1 \setminus T_2} = f^{-1} \sqbrk {T_1} \setminus f^{-1} \sqbrk {T_2}$

where:
 * $\setminus$ denotes set difference
 * $f^{-1} \sqbrk {T_1}$ denotes preimage.

Proof
From Inverse of Mapping is One-to-Many Relation, we have that $f^{-1}: T \to S$ is one-to-many.

Thus we can apply One-to-Many Image of Set Difference:
 * $\mathcal R \sqbrk {T_1 \setminus T_2} = \mathcal R \sqbrk {T_1} \setminus \mathcal R \sqbrk {T_2}$

where in this context $\mathcal R = f^{-1}$.