Showing if a function is injective or surjective problem $F : \Bbb{P}(X) \rightarrow \Bbb{P}(X) ; U \rightarrow (U-A) \cup (A-U)$
My intuition has been telling me that this function is bijective but I having the most difficult time trying to show this. Any help would be appreciated, thank you!!
edit: So far, I introduced sets $U_1$ and $U_2$, I am assuming $f(U_1)=f(U_2)$ and trying show $U_1=U_2$ in order to show injection. So I have $(U_1-A)\cup(A-U_1)=(U_2-A)\cup(U_2-A)$ and I'm unfamiliar with how to manipulate this into $U_1=U_2$. I haven't worked much with set notation before.
As for showing that f is surjective, I've tried drawing some Venn diagrams to help me process the question but I'm not sure where to go. Thank you for the quick replies!
 A: If $U\ne V$ and WLOG let $x\in U$ such that $x\not\in V$ so


*

*if $x\in A$ then $x\not\in F(U)$ and $x\in F(V)$ so $f(U)\ne F(V)$

*if $x\not\in A$ then $x\in F(U)$ and $x\not\in F(V)$ so $f(U)\ne F(V)$
hence $F$ is injective.


Now let $W\in \Bbb P(X)$ so we verify that $F(W\Delta A)=W$ so $F$ is surjective.
A: Trying it directly seems rough, especially since it's hard to picture if $U_1 = U_2$ using a Venn Diagram. I'm not sure how familiar you are with this idea, but $(U - A) \cup (A-U)$ is also called the symmetric difference of $U$ and $A$. It is denoted $U \triangle A$. Using the associativity of $\triangle$ with your function $F(U) = U \triangle A$, you can show 
$$(F \circ F)(U) = F(F(U)) = F(U \triangle A) = (U \triangle A) \triangle A = U \triangle (A \triangle A) = U \triangle \emptyset = U.$$ 
You can probably show associativity (the fourth equal sign) using Venn diagrams, comparing the diagram of each side. This entire equation tell us that your function has a left and right inverse. It's a bijection. 
A: Hint $\Delta$ is associative, hence
$$A \Delta  (A \Delta W)= (A \Delta A) \Delta W=W$$
