Prove comnplement function on [n] is bijective

Please critique the following proof.

Let and be the sets of even- and odd-sized subsets of [n], respectively. If n is odd then the set complement function maps sets in to sets in . Is this a bijection? Prove or disprove.

Prove f is one-to-one: Given [n] where n is odd. Assume and are two unequal sets of odd sized subsets on [n]. We have defined the set complement function and .

If is the complement of on [n] and is the complement of on [n] then . But is the set of even subsets on [n] and is also the the set of even subsets on [n], so must = since the number of even subsets on [n] is fixed. Therefore on [n] so and therefore proving f is one-to-one.

Now prove f in onto. We are given that f is the complement function on [n]. Given a set B [n], we need to find another set A such that f(A)=B. Choose A= . Then f( ) = . Therefore f is onto. Since f is one-to-one and onto, f is bijective.