# On the bijectivity of $f:\mathcal P(E)\to\mathcal P(A)\times \mathcal P(B); X\mapsto (X\cap A, X\cap B)$ with $A,B\subset E$

Can anyone clarify this question, I don't get it?

Let $E$ be a set, $A$ and $B$ be non-empty subsets of $E$. Consider the map $f: \mathcal P(E) \to\mathcal P(A) \times\mathcal P(B)$ defined by

$$f(X) := (X \cap A, X \cap B)$$

1. (a) Prove that if there exists a subset $X\subset E$ such that $X \cap A = A$ and $X \cap B = \emptyset$, then $A \cap B = \emptyset$.
(b) Deduce that if $f$ is surjective, then $A \cap B = \emptyset$.
(c) Prove that if $A \cap B = \emptyset$, then $f$ is surjective.
2. Prove that $f$ is injective if and only if $A \cup B = E$.
3. When $f$ is bijective, define its inverse $f^{−1}$.
• $A\cap B = B \Rightarrow B\subset A$ and $A\cap B = \emptyset \Rightarrow B\subset A^C$.
• If $X\cap A = A, X\cap B = \emptyset$, then $f(X) = (A,\emptyset)$.
• If $A\cup B \subsetneq E$ and $C\subset E\setminus (A\cup B) \neq \emptyset$, then $f(X) = f(X\cup C)$.
• You should be able to guess $f^{-1}$. There is only one sensible candiate, just verify it.