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

Theorem
Let $S$ and $T$ be sets.

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

Then:
 * $f$ is a bijection


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

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.

Sufficient Condition
Let $f$ be a bijection.

From Relative Complement Mapping on Powerset is Bijection, $\complement_S$ and $\complement_T$ are bijections.

Let $\tuple {X, Y} \in f^\to \circ \complement_S$.

Then:

Necessary Condition
Let $f$ be a mapping such that:


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

We have that:

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

Now consider $A, B \subseteq S$:

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.