Order-reversing bijection on partially ordered sets Let $(X, \preceq)$ and $(Y, \ll)$ be partially ordered sets. Suppose there are functions $f : X \to Y$ and $g : Y \to X$ such that $$x \preceq g(y) \iff y \ll f(x)$$ for all $x \in X$, $y \in Y$. 
Define the closed elements of $X$ to be all elements of the form $g(f(x))$ and the closed elements of $Y$ to be all elements of the form $f(g(y))$. I have to prove that there is an order-reversing bijection between the closed elements of $X$ and the closed elements of $Y$. 
I'm really not sure how to do this. I'm thinking the map would have to be something simple but I can't see what it would be. If you just map $g(f(x)) \mapsto f(x)$ then that isn't necessarily injective and also maps onto $f(X)$ which might be larger than the set of closed elements of $Y$.
I feel like I have to use the property of $f$ and $g$ listed above to derive some other relationship between $f$ and $g$ ... maybe something like $f(g(f(x))) = f(x)$, but I'm not sure.
 A: You can prove that $fgf = f$ and $gfg = g$. $f(x) ≤ f(x)$ implies $x ≤ gf(x)$, the same for $fg$. Also $x ≤ x' ≤ gf(x')$ implies $f(x) ≥ f(x')$ so in fact the assumption gives that both $f$ and $g$ are order-reversing and they form a Galois connection or adjoint pair on poset categories. $x ≤ gfgf(x)$ gives $f(x) ≥ fgf(x)$ and $gf(x) ≤ gf(x)$ gives $fgf(x) ≥ f(x)$ so $fgf = f$ and $gfg = g$. Now it follows that $f$ and $g$ restricted to closed elements are desired mutually inverse order-reversing bijections ($fgf = f$ implies $gfgf = gf$ so closed elements are exactly fix-points of $gf$ and the same for $fg$. Again by $fgf = f$ we have that closed elements are exactly images of $f$, $g$.)
Also note that the only asymetric result of the construction is $x ≤ gf(x)$ and $y ≤ fg(y)$. If the assumption was $y ≥ f(x) \iff x ≥ g(y)$ then you would have the same result but $x ≥ gf(x)$ and $y ≥ fg(y)$ would hold. This has to do with what is called “contravariant adjoint situation”. Since both $f$, $g$ are order-reversing, neither of them is right nor left adjoint but both are “adjoints on the right”. For the reversed assumption they would be “adjoints on the left”.
