Direct Image of Intersection with Inverse Image
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Theorem
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let:
- $f^\to: \powerset S \to \powerset T$ denote the direct image mapping of $f$
- $f^\gets: \powerset T \to \powerset S$ denote the inverse image mapping of $f$
where $\powerset S$ denotes the power set of $S$.
Then:
- $\forall A \in \powerset S, B \in \powerset T: \map {f^\to} {A \cap \map {f^\gets} B} = \map {f^\to} A \cap B$
Proof
Let $A \in \powerset S, B \in \powerset T$ be arbitrary.
Then:
\(\ds \map {f^\to} {A \cap \map {f^\gets} B}\) | \(\subseteq\) | \(\ds \map {f^\to} A \cap \map {f^\to} {\map {f^\gets} B}\) | Image of Intersection under Mapping | |||||||||||
\(\ds \) | \(\subseteq\) | \(\ds \map {f^\to} A \cap B\) | Subset of Codomain is Superset of Image of Preimage |
Let $x \in \map {f^\to} A \cap B$.
\(\ds x\) | \(\in\) | \(\ds \map {f^\to} A\) | Definition of Set Intersection | ||||||||||||
\(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds \map f a\) | for some $a \in A$ | ||||||||||
and: | |||||||||||||||
\(\ds x\) | \(\in\) | \(\ds B\) | Definition of Set Intersection | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map f a\) | \(\in\) | \(\ds B\) | from $(1)$ above | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds a\) | \(\in\) | \(\ds \map {f^\gets} B\) | Definition of Inverse Image Mapping | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds a\) | \(\in\) | \(\ds A \cap \map {f^\gets} B\) | Definition of Set Intersection | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map f a\) | \(\in\) | \(\ds \map {f^\to} {A \cap \map {f^\gets} B}\) | Definition of Direct Image Mapping | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds \map {f^\to} {A \cap \map {f^\gets} B}\) | as $x = \map f a$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {f^\to} A \cap B\) | \(\subseteq\) | \(\ds \map {f^\to} {A \cap \map {f^\gets} B}\) | Definition of Subset |
Thus we have:
- $\map {f^\to} {A \cap \map {f^\gets} B} \subseteq \map {f^\to} A \cap B$
and:
- $\map {f^\to} A \cap B \subseteq \map {f^\to} {A \cap \map {f^\gets} B}$
and so:
- $\forall A \in \powerset S, B \in \powerset T: \map {f^\to} {A \cap \map {f^\gets} B} = \map {f^\to} A \cap B$
by definition of set equality.
$\blacksquare$
Sources
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $5$