Detection of Lattice Homomorphism.

Question

If $(L,*,+)$ and $(S,\cdot,\vee)$ are two lattices, a mapping $g\colon L\to S$ is called a lattice homomorphism from $L$ to $S$ if for any $a,b \in L$ we have $g(a*b) = g(a) \cdot g(b)$ and $g(a+b) = g(a) \vee g(b)$.

I can't understand what $g(a*b) = g(a) \cdot g(b)$ and $g(a+b) = g(a) \vee g(b)$ means. How can we detect homomorphism between two lattices?

Answer 1

Let $L$ be the lattice $\langle\wp(\{a,b\}),\cap,\cup\rangle$, and let $S$ be the lattice $\langle\{0,1\},\land,\lor\rangle$, where $m\land n=\min\{m,n\}$ and $m\lor n=\max\{m,n\}$. Define $g:\wp(\{a,b\})\to\{0,1\}$ as follows: for each subset $X$ of $\{a,b\}$, $g(X)=1$ if and only if $a\in X$. The four elements of $\wp(\{a,b\})$ and their images under $g$ are shown in the following table.

$$\begin{array}{c|c} X&g(X)\\ \hline \varnothing&0\\ \{a\}&1\\ \{b\}&0\\ \{a,b\}&1 \end{array}$$

Now let’s check that $g$ is a lattice homomorphism. Suppse that $X,Y\subseteq\{a,b\}$; we need to show that

$$g(X\cap Y)=g(X)\land g(Y)\tag{1}$$ and $$g(X\cup Y)=g(X)\lor g(Y)\;.\tag{2}$$

You can verify this by brute force: there are only $4^2=16$ possible pairs of $X$ and $Y$ to check. However, brute force isn’t really needed. I’ll take $(1)$ first.

• Suppose that $g(X\cap Y)=0$; that means that $a\notin X\cap Y$, so either $a\notin X$, or $a\notin Y$ (or both). If $a\notin X$, then $g(X)=0$, and if $a\notin Y$, then $g(Y)=0$, so at least one of $g(X)$ and $g(Y)$ is $0$. This of course implies that $g(X)\land g(Y)=0$, which verifies $(1)$ for this case.
• If instead $g(X\cap Y)=1$, then $a\in X\cap Y$, so $a\in X$ and $a\in Y$. But then $g(X)=1$ and $g(Y)=1$, so $g(X)\land g(Y)=1\cap 1=1$, and we’ve verified $(1)$ for this case as well. There are only these two cases, so we’ve actually proved that $(1)$ holds for all $X,Y\in\wp(\{a,b\})$.

$(2)$ is proved similarly, so I’ll be a little more concise.

• If $g(X\cup Y)=0$, then $a\notin X\cup Y$, so $a\notin X$ and $a\notin Y$. Thus, $g(X)=0=g(Y)$, and $g(X)\lor g(Y)=0\lor 0=0$, and we’ve verified $(2)$ for this case.
• If $g(X\cup Y)=1$, then $a\in X\cup Y)$, so at least one of $X$ and $Y$ contains $a$, and at least one of $g(X)$ and $g(Y)$ is $1$. That implies that $g(X)\lor g(Y)=1$, so $(2)$ holds in this case as well, and we’ve completed the proof that $g$ is a lattice homomorphism.

Note that since $g$ is not a bijection, it certainly isn’t a lattice isomorphism. It ‘collapses’ the power set lattice to the two-element lattice, distinguishing subsets of $\{a,b\}$ only as to whether they contain $a$ or not instead of looking at all of their elements.