Let $S$ and $T$ be sets and let $f: S \to T$ and $g : T \to S$ be arbitrary functions. Prove that there is a subset $A \subset S$ and a subset $B \subset T$ such that $f(A) = B$ and $g(T \setminus B)= S \setminus A$.
My approach :=
We know that $A=S \setminus A^c$ and $B= T \setminus B^c$
Now $f(A)=B$ gives $f(S-A^c)=T \setminus B^c$
Next I will approach...?