Bijections from $A$ to $B$, where $A$ is the set of subsets of $[n]$ that have even size and $B$ is the set of subsets of $[n]$ that have odd size Let $A$ be the set of subsets of $[n]$ that have even size, and let $B$ be the set of subsets of $[n]$ that have odd size.  Establish a bijection from $A$ to $B$.  The following bijection is suggested for $n=3$:
$$\matrix{A: & \{1,2\} & \{1,3\} & \{2,3\} & \varnothing\\

B: & \{1,2,3\} & \{1\} & \{2\} & \{3\}}$$
I know that first we have to establish a function that is both surjective and injective so that it is bijective. I don't know where to take a step from here in the right direction. So I need a bit of guidance. 
Something suggested is let f be the piecewise function:
$$f(x) = x \setminus \{n\}\text{ if }n \in x\text{ and }f(x) = x\cup\{n\}\text{ if }n \notin x$$
 A: The function says to map a subset either to itself minus $\{n\}$ (if the subset contains $n$) or to itself union $\{n\}$ (if the subset does not contain $n$). Clearly a subset cannot both contain $n$ and not contain $n$, so there is no problem deciding which case we are in.
To prove the $f$ is injective, let $A_1$ and $A_2$ be two subsets and consider $f(A_1) = f(A_2)$. Either we have
$$
A_1 - \{n\} = f(A_1) = f(A_2) = A_2 - \{n\}
$$
or
$$
A_1 \cup \{n\} = f(A_1) = f(A_2) = A_2 \cup \{n\}.
$$
In either case, we can conclude that $A_1 = A_2$, and so $f$ is injective.
To prove that $f$ is surjective, let $B$ be an element of the range of $f$ (i.e. a subset of odd size). If $B$ contains $n$, then $B - \{n\}$ is a subset of even size that maps to $B$ under $f$. If $B$ does not contain $n$, then $B \cup \{n\}$ is a subset of even size that maps to $B$ under $f$. Since everything in the image has something in the domain that maps to it, $f$ is surjective.
A: Number the members of the set $1,2,3,4,\ldots,n$.
For every subset with an even number of elements, there is a corresponding set with an odd number of elements, that corresponds in this way:


*

*If $1$ is a member of the set with an even number of elements, then delete $1$ from the set to get a set with an odd number of elements.

*If $1$ is not a member of the set with an even number of elements, then add $1$ to the set to get a set with an odd number of elements.


For example, suppose the set is $\{1,2,3,4\}$.  Then we have this correspondence between sets with an even number of elements and sets with an odd number of elements:
$$
\begin{array}{rcl}
\text{even} & & \text{odd} \\
\hline
\varnothing & \leftrightarrow & \{1\} \\
\{1,2\} & \leftrightarrow & \{2\} \\
\{1,3\} & \leftrightarrow & \{3\} \\
\{1,4\} & \leftrightarrow & \{4\} \\
\{2,3\} & \leftrightarrow & \{1,2,3\} \\
\{2,4\} & \leftrightarrow & \{1,2,4\} \\
\{3,4\} & \leftrightarrow & \{1,3,4\} \\
\{1,2,3,4\} & \leftrightarrow & \{2,3,4\}
\end{array}
$$
This won't work with the empty set because we don't have an element to which we can assign the number $1$.
A: The example shown for $n=3$ illustrates the suggested bijection (which I corrected from the version in your post): the two sets in each column differ only by the presence or absence of the number $3$. Note that $x$ and $f(x)$ always differ in size by exactly $1$, so one of them must have even size, and the other must have odd size. This shows that the suggested $f$ really does take even-sized sets to odd-sized sets. Now you have to show that $f:A\to B$ is injective and surjective.


*

*To show that $f$ is injective, assume that $x,y\in A$ and $f(x) = f(y)$, and try to show that this forces $x=y$.

*To show that $f$ is surjective, start with a set $y\in B$ and try to find an $x\in A$ such that $f(x)=y$. This isn’t hard if you think about how $f$ works. (HINT: Consider $f$ as a function from $\wp([n])\to\wp([n])$; what does $f$ do to members of $B$?)
A: For $n$ an odd positive integer, there is a natural procedure.  The mapping that takes any subset $E$ of $[n]$ with an even number of elements to its complement $[n]\setminus E$ is a bijection from $A$ to $B$.
Dealing with even positive $n$ is messier.  Here is one way. The subsets of $[n]$ can be divided into two types: (i) subsets that do not contain $n$ and (ii) subsets that contain $n$.  
(i) If $E$ is a subset of $[n]$  with an even number of elements, and does not contain $n$, map $E$ to $[n-1]\setminus E$.  Call this map $\phi$.
(ii) If $E$ is a subset of $n$ with an even number of elements, and contains $n$, map $E$ to $\{n\} \cup \phi^{-1}(E\setminus\{n\})$.     
