Complement function: how to prove surjective?

Question

Given some set A and a complement function C(K) = A - K from the power set of A onto the power set of A, how can I formally prove that it is surjective?

I think I get it, but can't get it on paper. (For context, I have already proven that it is injective.) Every element of the power set of A will be both in the domain and range because the power set of A minus one set yields the rest of its power set, and

P(A) - empty set = P(A)

and

P(A) - P(A) = empty set

... but that's not equivalent to saying there's a function f(x) = P(x) - {x} , right?

Answer 1

To prove that a function is surjective, it's enough to prove that it's bijective; and so it's enough to prove that it has an inverse.

What is the inverse of $C: K \mapsto A \setminus K$?

Answer 2

Not sure I got your question.

Let $T,S \subset A$.

$C: P(A) \rightarrow P(A).$

$T \mapsto A$ \ $T.$

Surjective:

Let $S \in P(A)$, range, then

$A$ \ $S \in P(A).$

Choose $A$ \ $S \in P(A)$, domain: Then

$C(A$ \ $S) = A$ \ $(A$ \ $S) =S,$ hence surjective.