Prove $f:P(\mathbb{N})^2\rightarrow P(\mathbb{N})^2$ , is defined by $f(\langle A,B\rangle) = \langle A,A \Delta B \rangle$ is an bijection. Prove $f:P(\mathbb{N})^2\rightarrow P(\mathbb{N})^2$ , is defined by $f(\langle A,B\rangle) = \langle A,A \Delta B \rangle$ is an bijection.
($A\Delta B $ is a Symmetric difference of A, B)
Attempt:
I tried to find the inverse function.
I tried to use a Symmetric difference in the outcome as the inverse function.
$(A\Delta B) \Delta A$
I'm having trouble finding a bijection in groups. I am used to finding bijections in regular functions.
 A: To give some intuition: we have two sets, and the function $f$ returns the first set $A$, followed by the set whose elements are the elements that are exclusively in one of the sets ($A \Delta B$). Clearly, we can reconstruct the first set $A$, but can we reconstruct the second set $B$?
The answer is yes! We can notice that the intersection of $A$ and $A \Delta B$ (call it $I$) is exactly the elements which are not in $B$, so if we remove these elements from $A$, then we will be left with exactly $A \cap B$. Furthermore, if we remove $I$ from $A \Delta B$, we recover the elements which are only in $B$, but not in $A$ (why?). So we have both the set of elements that are in $B$ and not in $A$, as well as the set of elements in $B$ and in $A$. Clearly, if we take the union we will recover $B$.
Putting our procedure into symbols, the inverse function will be
$$f^{-1}(\langle A, B\rangle) = \langle A, (A \setminus (A \cap B)) \cup (B \setminus (A \cap B)) \rangle$$
(Here $\setminus$ represents the set difference.) But as lisyarus suggests, this expression can be written more compactly: can you see what it is? Hint: it should look familiar...
Remark: If you are at all interested in boolean logic and bits, this statement is the equivalent of saying for two bits $a$ and $b$, the function $f$ which takes $(a, b)$ to $(a, a \oplus b)$ is invertible (and is its own inverse). Here $\oplus$ represents the XOR operation between the bits.
