Relative Complement Mapping on Powerset is Bijection/Proof 2

Proof
Let $f: \powerset S \to \powerset S$ be a mapping defined as:
 * $\forall T \in \powerset S: \map f T = \relcomp S T$

It is to be demonstrated that $f$ is a bijection.

By definition of relative complement:
 * $\relcomp S T = S \setminus T = \set {x \in S: x \notin T}$

and so it can be seen that $f$ is well-defined.

Let $T_1, T_2 \in \powerset S: \map f {T_1} = \map f {T_2}$.

Then:

That is, $f$ is an injection.

Also:

That is, for all $T \in \powerset S$ there exists an $X$ such that $T = \map f X$.

This demonstrates that $f$ is a surjection.

The result follows by definition of bijection.