# Mapping is Bijection iff Composite with Direct Image Mapping with Complementation Commutes

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## Theorem

Let $S$ and $T$ be sets.

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

Then:

- $f$ is a bijection

- $f^\to \circ \complement_S = \complement_T \circ f^\to$

where:

- $f^\to: \powerset S \to \powerset T$ denotes the direct image mapping of $f$
- $\complement_S: \powerset S \to \powerset S$ denotes the complement relative to $S$
- $\complement_T: \powerset T \to \powerset T$ denotes the complement relative to $T$
- $\powerset S$ and $\powerset T$ denote the power sets of $S$ and $T$ respectively.

## Proof

### Sufficient Condition

Let $f$ be a bijection.

Thus a fortiori $f$ is:

From One-to-Many Image of Set Difference: Corollary 2 we have:

- $\forall X \in \powerset S: \map {\paren {f^\to \circ \complement_S} } X = \map {\paren {\complement_{\Img f} \circ f^\to} } X$

By definition of surjection:

- $\Img f = T$

and so:

- $\forall X \in \powerset S: \map {\paren {f^\to \circ \complement_S} } X = \map {\paren {\complement_T \circ f^\to} } X$

That is:

- $f^\to \circ \complement_S = \complement_T \circ f^\to$

$\Box$

### Necessary Condition

Let $f$ be a mapping such that:

- $f^\to \circ \complement_S = \complement_T \circ f^\to$

We have that:

\(\displaystyle \map {f^\to} S\) | \(=\) | \(\displaystyle \map {f^\to} {\relcomp S \O}\) | Definition of Relative Complement | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \relcomp T {\map {f^\to} \O}\) | by hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \relcomp T \O\) | Definition of Direct Image Mapping of Mapping: $\map {f^\to} \O = \O$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle T\) | Definition of Relative Complement |

Hence, by definition, $f$ is a surjection.

Now consider $A, B \subseteq S$:

\(\displaystyle \map {f^\to \circ \complement_S} {A \cap B}\) | \(=\) | \(\displaystyle \map {f^\to} {\relcomp S {A \cap B} }\) | Definition of Composition of Mappings | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map {f^\to} {\relcomp S A \cup \relcomp S B}\) | De Morgan's Laws: Relative Complement of Intersection | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map {f^\to} {\relcomp S A} \cup \map {f^\to} {\relcomp S B}\) | Image of Union under Mapping | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map {f^\to \circ \complement_S} A \cup \map {f^\to \circ \complement_S} B\) | Definition of Composition of Mappings | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map {\complement_T \circ f^\to} A \cup \map {\complement_T \circ f^\to} B\) | by hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \relcomp T {\map {f^\to} A} \cup \relcomp T {\map {f^\to} B}\) | Definition of Composition of Mappings | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \relcomp T {\map {f^\to} A \cap \map {f^\to} B}\) | De Morgan's Laws: Relative Complement of Intersection | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \complement_T \circ \map {f^\to} {A \cap B}\) | \(=\) | \(\displaystyle \relcomp T {\map {f^\to} A \cap \map {f^\to} B}\) | by hypothesis | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map {f^\to} {A \cap B}\) | \(=\) | \(\displaystyle \map {f^\to} A \cap \map {f^\to} B\) | Relative Complement Mapping on Powerset is Bijection, and so Bijection iff Left and Right Cancellable |

It follows from Image of Intersection under Injection that $f^\to$ is an injection.

Hence from Mapping is Injection if its Direct Image Mapping is Injection:

- $f$ is an injection.

We have therefore that $f$ is both an injection and a surjection.

Hence, by definition, $f$ is a bijection.

$\blacksquare$

## Sources

- 1975: T.S. Blyth:
*Set Theory and Abstract Algebra*... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $3$