Preimage of Set Difference under Mapping

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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.


Corollary 1

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.


Corollary 2

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

Let $T_1$ be a subset of $T$.


Then:

$\relcomp S {f^{-1} \sqbrk {T_1} } = f^{-1} \sqbrk {\relcomp T {T_1} }$

where:

$\complement_S$ (in this context) denotes relative complement
$f^{-1} \sqbrk {T_1}$ 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 $\mathcal R = f^{-1}$.

$\blacksquare$


Sources