Prove comnplement function on [n] is bijective

Please critique the following proof.

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

Prove f is one-to-one: Given [n] where n is odd. Assume $\displaystyle \mathcal{O}_{1}$ and $\displaystyle \mathcal{O}_{2}$ are two unequal sets of odd sized subsets on [n]. We have defined the set complement function $\displaystyle f(\mathcal{E}_{1})=\mathcal{\bar{E}}_{1}=\mathcal{ O}_{1}$ and $\displaystyle f(\mathcal{E}_{2})=\mathcal{\bar{E}}_{2}=\mathcal{ O}_{2}$.

If $\displaystyle \mathcal{O}_{1}$ is the complement of $\displaystyle \mathcal{E}_{1}$ on [n] and $\displaystyle \mathcal{O}_{2}$ is the complement of $\displaystyle \mathcal{E}_{2} $ on [n] then $\displaystyle \mathcal{E}_{1} \neq\mathcal{E}_{2}$. But $\displaystyle \mathcal{E}_{1}$ is the set of even subsets on [n] and $\displaystyle \mathcal{E}_{2}$ is also the the set of even subsets on [n], so $\displaystyle \mathcal{E}_{1}$ must = $\displaystyle \mathcal{E}_{2}$ since the number of even subsets on [n] is fixed. Therefore $\displaystyle \mathcal{\bar{E}}_{1}=\mathcal{\bar{E}}_{2}$ on [n] so $\displaystyle \mathcal{O}_{1}=\mathcal{O}_{2}$ and therefore $\displaystyle f(\mathcal{E}_{1})=f(\mathcal{E}_{2})$ 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$\displaystyle \subseteq$[n], we need to find another set A such that f(A)=B. Choose A=$\displaystyle \bar{\mathcal{E}}$. Then f($\displaystyle \bar{\mathcal{E}}$) = $\displaystyle \mathcal{E}$. Therefore f is onto. Since f is one-to-one and onto, f is bijective.