This is a little exercise I've been fiddling with for a while now.
Let $f\colon X\to Y$ and $g\colon Y\to X$ be functions. I want to show that there are subsets $A\subseteq X$ and $B\subseteq Y$ such that $f(A)=B$ and $g(Y-B)=X-A$.
Of course, if $f$ is surjective, then taking $A=X$, one has $f(A)=B=Y$ and $g(Y-Y)=g(\emptyset)=\emptyset=X-X=X-A$, and you're done. So I suppose $f$ is not surjective.
I tried to approach it by contradiction. It seems that for any $A\subseteq X$, there is obviously a $B\subseteq Y$ such that $f(A)=B$, so if the result is not the case, for each pair of subsets $A$ and $f(A)$, we must have $g(Y-f(A))\neq X-A$. Then for every $f(A)\subseteq Y$, there exists a $y\in Y-f(A)$ such that $g(y)\notin X-A$. This would imply $g(y)\notin X \lor g(y)\in A$, but since $g(y)\in X$, we must have $g(y)\in A$.
The only thing I can glean from this is that $g$ is surjective, since for any singleton $\{x\}\subseteq X$, we could then find a $y\in Y-\{f(x)\}$ such that $g(y)=x$. But I don't quite see how to get a contradiction. What direction should I head? I attempted to apply the fact that a monotone function on power sets has a fixed point, but that only seems to apply then the function maps from a power set into itself. Thanks!