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

where:
 * $\complement$ (in this context) denotes relative complement
 * $f^{-1} \sqbrk {T_1}$ denotes preimage.

Proof
From One-to-Many Image of Set Difference: Corollary 1 we have:
 * $\relcomp {\RR \sqbrk {T_2} } {\RR \sqbrk {T_1} } = \RR \sqbrk {\relcomp {T_2} {T_1} }$

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

Hence as $f^{-1}: T \to S$ is a one-to-many relation:
 * $\relcomp {f^{-1} \sqbrk {T_2} } {f^{-1} \sqbrk {T_1} } = f^{-1} \sqbrk {\relcomp {T_2} {T_1} }$