Does $(f^{0})^{0} = f$ hold for suplattice homomorphisms? Let $X$ and $Y$ be suplattices and $f: X \rightarrow Y$ a suplattice homomorphism (i.e. $f(\bigvee S)=\bigvee \{f(s)|s \in S\}$ for every subset $S \subseteq X$). Denote by $X^0$ and $Y^0$ the opposite poset of $X$ and $Y$ respectively. Define a function $f^{0}: Y^{0}\rightarrow X^{0}$ via $f^{0}(y)= \bigvee \{x \in X | f(x) \leq y \}$. I am trying to prove that $(f^0)^0=f$. Do you have any advice on how to complete the following proof attempt of mine?

Let $x \in X$. We have to prove that $\bigvee \big\{ y \in Y|\bigvee \{z \in X|f(z) \leq y \}\leq x\big\}=f(x).$

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*We first show that $f(x)$ is an upper bound of the set $\big\{ y \in Y|\bigvee \{z \in X|f(z) \leq y \}\leq x\big\}$: Let $y \in Y$ such that $\bigvee \{z \in X|f(z) \leq y \}\leq x$. Since $f$ preserves arbitrary joins (and is therefore monotone) we know that $\bigvee \{f(z)|z \in X \text{ and } f(z) \leq y\} \leq f(x)$. If I could prove that $\bigvee \{f(z)|z \in X \text{ and } f(z) \leq y\}=y$, this would show that $f(x) \geq y$. However, I do not know how to do it. The element $y$ is clearly an upper bound for $\{f(z)|z \in X \text{ and } f(z) \leq y\}$, but why is it the smallest?

*We next show that $f(x)$ is the smallest upper bound for the set $\big\{ y \in Y|\bigvee \{z \in X|f(z) \leq y \}\leq x\big\}$: Let $a$ be an upper bound of $\bigvee \big\{ y \in Y|\bigvee \{z \in X|f(z) \leq y \}\leq x\big\}.$ If I could prove that $\bigvee \{z \in X|f(z) \leq f(x) \}\leq x$, I would be done. Is it true?


 A: Let $f:M \rightarrow N$ be a homomorphism between sup-lattices $M,N$.
1. Preliminaries
Consider $M$ (or $N$) as a (posetal) category with the elements of $M$ as objects and an arrow from an element $x \in M$ to an element $y \in M$ if and only if $x \leq y$.
Then colimits in $M$ are joins and limits are meets.
A functor $g:M \rightarrow N$ is precisely an order-preserving function.
A morphism of sup-lattices is the same as a functor that preserves colimits.
2. Defining $f^0$
Let me give a more conceptual definition of $f^0$. The map you call $f^0$ is essentially the unique right adjoint of $f$.
2.1. Existence
Since $f$ preserves colimits by the adjoint functor theorem for preorders it has a right adjoint. That is, there exists an order-preserving function $f_*:N \rightarrow M$ such that $$f(x) \leq y \text{ iff } x \leq f_*(y) \qquad(\ast)$$
2.2. Uniqueness
The right adjoint is unique. In what follows I give an elementary proof. Alternatively, you could argue that adjoint functors are unique up to natural isomorphism. Since in a posetal category two objects are isomorphic if and only if they are equal this shows the (strict) uniqueness of the right adjoint of $f$.

Let $f_*$ be a right adjoint of $f$ and let $y \in N$. We prove that $f_*(y)=\bigvee \{x \in M; f(x) \leq y\}$.

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*By $(\ast)$ the element $f_*(y)$ is an upper bound of the set $\{x \in M;f(x) \leq y\}$.

*It is the smallest upper bound: From reflexivity $f_*(y) \leq f_*(y)$ and $(\ast)$ it follows that $f(f_*(y))\leq y$. Therefore, for any upper bound $z$ of $\{x \in M;f(x) \leq y\}$ we have $z \leq f_*(y)$.


As a right adjoint, $f_*$ preserves limits (infima). Hence we obtain a morphism of sup-lattices $f^0:N^0 \rightarrow M^0$.
3. Answering your question
Let us now show that $(f^0)^0=f$. By the uniqueness of the right adjoint it suffices to prove that $f:M \rightarrow N$ is a right adjoint of $f^0: N^0 \rightarrow M^0$. That $f$ is a right adjoint is precisely the claim that $f^0(x) \leq_{opp} y \text{ iff } x \leq_{opp} f(y)$. (Here $\leq_{opp}$ denotes the opposite order.) But this statement is nothing but the claim that $f_*(x) \geq y \text{ iff } x \geq f(y)$, i.e. that $f_*$ is a right adjoint.
4. Additional remarks
One can prove the claim $(fg)^0=g^0f^0$ in a similar way. Using categorical language: This shows that $(-)^0$ gives an (involutive) antiisomorphism of the category of sup-lattices.
For more information see An Extension of the Galois Theory of Grothendieck by Joyal and Tierney.
