Preimage of Set Difference under Mapping/Corollary 1

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

Let $T_1 \subseteq T_2 \subseteq T$.

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

where:
 * $\complement$ (in this context) denotes relative complement
 * $f^{-1}$ denotes preimage.

Proof
From One-to-Many Image of Set Difference: Corollary 1 we have:
 * $\complement_{\mathcal R \left({T_2}\right)} \left({\mathcal R \left({T_1}\right)}\right) = \mathcal R \left({\complement_{T_2} \left({T_1}\right)}\right)$

where $\mathcal R \subseteq T \times S$ is a one-to-many relation on $T \times S$.

Hence as $f^{-1}: T \to S$ is a one-to-many relation:
 * $\complement_{f^{-1} \left({T_2}\right)} \left({f^{-1} \left({T_1}\right)}\right) = f^{-1} \left({\complement_{T_2} \left({T_1}\right)}\right)$