Complement of Preimage equals Preimage of Complement

Theorem
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.

That is:
 * $S \setminus f^{-1} \sqbrk {T_1} = f^{-1} \sqbrk {T \setminus T_1}$

This can be expressed in the language and notation of inverse image mappings as:
 * $\forall T_1 \in \powerset T: \relcomp S {\map {f^\gets} {T_1} } = \map {f^\gets} {\relcomp T {T_1} }$

Proof
From One-to-Many Image of Set Difference: Corollary 2 we have:
 * $\relcomp {\Img {\mathcal R} } {\mathcal R \sqbrk {S_1} } = \mathcal R \sqbrk {\relcomp S {S_1} }$

where:
 * $S_1 \subseteq S$
 * $\mathcal R \subseteq T \times S$ is a one-to-many relation on $T \times S$.

Hence as $f^{-1}: T \to S$ is a one-to-many relation:
 * $\relcomp {\Preimg f} {f^{-1} \sqbrk {T_1} } = f^{-1} \sqbrk {\relcomp T {T_1} }$

But from Preimage of Mapping equals Domain, we have that:
 * $\Preimg f = S$

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