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

if and only if:

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

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:

a surjection
a one-to-many relation


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