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!

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. 
