Preimage of Set Difference under Mapping

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.

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.

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.

From Inverse of Mapping is One-to-Many Relation, we have that $f^{-1}: T \to S$ is one-to-many.

$\RR \sqbrk {T_1 \setminus T_2} = \RR \sqbrk {T_1} \setminus \RR \sqbrk {T_2}$

where in this context $\RR = f^{-1}$.

$\blacksquare$

1955: John L. Kelley: General Topology ... (previous) ... (next): Chapter $0$: Functions: Theorem $7 \ \text{(a)}$
1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: Mappings: $\S 12 \alpha$
1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.12$: Set Inclusions for Image and Inverse Image Sets: Exercise $4 \ \text{(b)}$
1993: Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (2nd ed.) ... (previous) ... (next): $\S 1$: Naive Set Theory: $\S 1.6$: Functions: Exercise $1.6.3 \ \text{(iii)}$
2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.2 \ \text{(d)}$
2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 2$