Proving surjectivity with partitions with a set

Question

I've got the following sentence :

Let \begin{array}{ll} F :& P(A)\times P(B) \longrightarrow P(E) \\ &(X,Y) \longrightarrow X\cup Y \end{array}

where $A$ and $B$ are two subsets of E. I have to say if it's surjective and/or injective but I'm struggling to give an demonstration.

And I have also the question : what happens when $A\cup B = E $ and when $A\cap B = \emptyset $.

How can I prove that ?

Thanks !!

Answer 1

If $A\cap B=\emptyset$, then for any $X\in\mathcal{P}(A)$ and $Y\in\mathcal{P}(B)$, we have $X\cap Y=\emptyset$. Then for a pair of sets $(X,Y)\ne(X',Y')\in\mathcal{P}(A)\times\mathcal{P}(B)$, if $X\cup Y=X'\cup Y'$, then $X\cap Y'=\emptyset\implies X\subseteq X'$. Again, $X'\cap Y=\emptyset\implies X'\subseteq X$. Hence $X=X'$. This implies $Y=Y'$. This establishes 'one-to-one'ness of the map when $A\cap B=\emptyset$.

If $A\cap B\ne\emptyset$, then we see that $B\setminus A\ne B$, but $A\cup B=A\cup(B\setminus A)$. Hence we get that in this case the function no more remains injective.

If $A\cup B=E$, then for any set $X\in\mathcal{P}(E)$, we have $X=(X\cap A)\cup(X\cap B)$. Now $X\cap A\in\mathcal{P}(A)$ and $X\cap B\in\mathcal{P}(B)$. This proves surjectivity of the function when $A\cup B=E$.

When $A\cap B\ne E$, then take $X=E\setminus (A\cup B)$, then $X\ne\emptyset$ and $X\in\mathcal{P}(E)$. However $X\cap A=\emptyset$ and $X\cap B=\emptyset$. Hence $X$ can not be written as the union of subsets of $A$ and $B$. Hence in this case the function is not surjective.

Answer 2

$F$ is surjective iff $A \cup B = E$.

If $A \cup B = E$ then for any $Y \in P(E)$, $F(A \cap Y, B \cap Y)=Y$ and if $F$ is surjective, we can write $E = F(X,Y)$ for some $(X,Y) \in P(A) \times P(B)$ and then $E = X \cup Y \subseteq A \cup B \subseteq E$ so $E=A \cup B$.

$F$ is injective iff $A \cap B = \emptyset$.

If $A \cap B \neq \emptyset$ then pick $x \in A \cap B$, and note that $F(\emptyset,\{x\})=F(\{x\},\emptyset)$ is $F$ is not injective. If $A$ and $B$ are disjoint, and $F(X,X')=Y$ then $X=A \cap Y$ and $X'=Y \cap B$ so the pair $(X,X')$ are uniquely determined by the image.

So $F$ is a bijection iff both of the previous conditions hold.