Preimage of Union under Mapping

Theorem

Let $S$ and $T$ be sets.

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

Let $T_1$ and $T_2$ be subsets of $T$.

Then:

$f^{-1} \sqbrk {T_1 \cup T_2} = f^{-1} \sqbrk {T_1} \cup f^{-1} \sqbrk {T_2}$

This can be expressed in the language and notation of inverse image mappings as:

$\forall T_1, T_2 \in \powerset T: \map {f^\gets} {T_1 \cup T_2} = \map {f^\gets} {T_1} \cup \map {f^\gets} {T_2}$

Let $S$ and $T$ be sets.

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

Let $\powerset T$ be the power set of $T$.

Let $\mathbb T \subseteq \powerset T$.

Then:

$\ds f^{-1} \sqbrk {\bigcup \mathbb T} = \bigcup_{X \mathop \in \mathbb T} f^{-1} \sqbrk X$

Let $S$ and $T$ be sets.

Let $\family {T_i}_{i \mathop \in I}$ be a family of subsets of $T$.

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

Then:

$\ds f^{-1} \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} f^{-1} \sqbrk {T_i}$

where:

$\ds \bigcup_{i \mathop \in I} T_i$ denotes the union of $\family {T_i}_{i \mathop \in I}$

$f^{-1} \sqbrk {T_i}$ denotes the preimage of $T_i$ under $f$.

As $f$, being a mapping, is also a relation, we can apply Preimage of Union under Relation:

$\RR^{-1} \sqbrk {T_1 \cup T_2} = \RR^{-1} \sqbrk {T_1} \cup \RR^{-1} \sqbrk {T_2}$

$\blacksquare$

1955: John L. Kelley: General Topology ... (previous) ... (next): Chapter $0$: Functions: Theorem $7 \ \text{(b)}$
1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.13 \ \text{(b)}$
1968: A.N. Kolmogorov and S.V. Fomin: Introductory Real Analysis ... (previous) ... (next): $\S 1.3$: Functions and mappings. Images and preimages: Theorem $1$
1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: Mappings: $\S 12 \alpha$
1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.1: \ \text{(jj)}$
2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.2 \ \text{(b)}$