Preimage of Set Difference under Mapping/Corollary 1

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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
$f^{-1} \left[{T_1}\right]$ 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:

$\complement_{f^{-1} \left[{T_2}\right]} \left({f^{-1} \left[{T_1}\right]}\right) = f^{-1} \left[{\complement_{T_2} \left({T_1}\right)}\right]$

$\blacksquare$