Image of sup/inf under Galois connection between lattices

Question

This is Lemma 2.4 in Pete Clark's notes on commutative algebra.

Given two posets $(X,\le)$ and $(Y,\le)$, an (antitone) Galois connection between $X$ and $Y$ is a pair of maps $F:X\to Y$ and $G:Y\to X$ such that $F$ and $G$ are antitone (= order-reversing) and for all $x\in X$ and $y\in Y$ $$x\le G(y)\iff y\le F(x)\quad\quad(*).$$

Lemma 2.4: Given a Galois connection as above, if $X$ and $Y$ are lattices, then for all $x_1,x_2\in X$

• $F(x_1\lor x_2)=F(x_1)\land F(x_2)$
• $F(x_1\land x_2)=F(x_1)\lor F(x_2)$

I can prove the first formula (proof below), but I don't see how to show the second one. Can anyone show how?

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Proof of $F(x_1\lor x_2)=F(x_1)\land F(x_2)$: Given $x_1, x_2$, for each $y\in Y$ we have $$ \begin{align*} & y\le F(x_1\lor x_2) & \\ \iff\;& x_1\lor x_2\le G(y) &\text{ from (*)} \\ \iff\;& x_1\le G(y)\quad\text{and}\quad x_2\le G(y) & \text{ from definition of }\lor\\ \iff\;& y\le F(x_1)\quad\text{and}\quad y\le F(x_2) & \text{ from (*)} \\ \end{align*} $$ The desired formula follows.

Answer 1

That's not right. Maybe you're overlooking some underlying hypothesis.
To see that that's not true (the second condition), consider the lattice $\mathbf L$ in the picture below.

We will define a Galois connection $\varphi : \mathbf L \to \mathbf L$.
(In this case, it's the same map in both ways, having also the same poset in both sides.)

Notice that in order to show that $$x \leq \varphi(y) \quad\text{iff}\quad y \leq \varphi(x),$$ it is enough to see that $x \leq y$ implies that $\varphi(y) \leq \varphi(x)$ and that $x \leq \varphi(\varphi(x))$, for all $x,y \in L$ (this is a well known alternative definition—or characterization—of Galois connections).
These conditions are immediate if we define $$\varphi(0)=1,\; \varphi(a) = b,\; \varphi(b) = a,\;\varphi(c) = \varphi(1) = 0.$$ However, $$\varphi(a) \vee \varphi(b) = b \vee a = c < 1 = \varphi(0) = \varphi(a \wedge b).$$

Answer 2

In addition to @amrsa's example, here is another simple counterexample.

Take the Boolean algebra with 2 atoms $X=\{0,a,b,1\}$ with $a$ and $b$ not comparable, $a\lor b=1$, and $a\land b=0$. Take $Y=\{0,1\}$ with $0\le 1$.

Define two maps $F:X\to Y$ and $G:Y\to X$ by $$F(0)=1\quad\text{and}\quad F(1)=F(a)=F(b)=0$$ and $$G(0)=1\quad\text{and}\quad G(1)=0\,.$$ The condition $x\le G(y)\iff y\le F(x)$ is easy to check for each value of $y$, so that $F$ and $G$ form a Galois connection between $X$ and $Y$.

We have $F(a\lor b)=F(1)=0=F(a)\land F(b)$ as expected. But $F(a\land b)=F(0)=1$ is not equal to $F(a)\lor F(b)=0\lor 0=0$.

Answer 3

This is false even in the basic motivating example from algebraic geometry. For instance, there is an antitone Galois connection between the lattice of subsets of $\mathbb{C}^n$ and the lattice of subsets of $\mathbb{C}[x_1,\dots,x_n]$, sending a set $S\subseteq\mathbb{C}^n$ to the set $I(S)$ of polynomials that vanish on it and sending a set of polynomials $S\subseteq \mathbb{C}[x_1,\dots,x_n]$ to the set $V(S)$ of points where they vanish. However, $I$ and $V$ do not turn intersections into unions. For instance, $V(\{x_1\}\cap\{x_1+1\})=V(\emptyset)=\mathbb{C}^n$ whereas $V(\{x_1\})\cup V(\{x_1+1\})$ is not all of $\mathbb{C}^n$.