BooleanAlgebraHomomorphism 2008-04-29 01:08:27 2008-04-29 02:29:25 Definition Boolean lattice kernel complete Boolean algebra homomorphism $\kappa$-complete Boolean algbra homomorphism \usepackage{amssymb,amscd} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{mathrsfs} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics \usepackage{xypic} \usepackage{pst-plot} % define commands here \newcommand*{\abs}[1]{\left\lvert #1\right\rvert} \newtheorem{prop}{Proposition} \newtheorem{thm}{Theorem} \newtheorem{ex}{Example} \newcommand{\real}{\mathbb{R}} \newcommand{\pdiff}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\mpdiff}[3]{\frac{\partial^#1 #2}{\partial #3^#1}} Let $A$ and $B$ be Boolean algebras. A function $f:A\to B$ is called a \emph{Boolean algebra homomorphism}, or homomorphism for short, if $f$ is a $\lbrace 0,1\rbrace$-\PMlinkname{lattice homomorphism}{LatticeHomomorphism} such that $f$ respects $'$: $f(a')=f(a)'$. Typically, to show that a function between two Boolean algebras is a Boolean algebra homomorphism, it is not necessary to check every defining condition. In fact, we have the following: \begin{enumerate} \item if $f$ respects $'$, then $f$ respects $\vee$ iff it respects $\wedge$; \item if $f$ is a lattice homomorphism, then $f$ respects $0$ and $1$ iff it respects $'$. \end{enumerate} The first assertion can be shown by de Morgan's laws. For example, to see the LHS implies RHS, $f(a\wedge b)= f((a'\vee b')')= f(a'\vee b')'=((f(a')\vee f(b'))'= f(a')'\wedge f(b')'= f(a)'' \wedge f(b)'' = f(a)\wedge f(b)$. The second assertion can also be easily proved. For example, to see that the LHS implies RHS, we have that $f(a')\vee f(a)= f(a'\vee a)=f(1)=1$ and $f(a')\wedge f(a)=f(a'\wedge a)=f(0)=0$. Together, this implies that $f(a')$ is the complement of $f(a)$, which is $f(a)'$. If a function satisfies one, and hence all, of the above conditions also satisfies the property that $f(0)=0$, for $f(0)=f(a\wedge a')=f(a)\wedge f(a')= f(a)\wedge f(a)'=0$. Dually, $f(1)=1$. As a Boolean algebra is an algebraic system, the definition of a Boolean algebra homormphism is just a special case of an algebra homomorphism between two algebraic systems. Therefore, one may similarly define a Boolean algebra monomorphism, epimorphism, endormophism, automorphism, and isomorphism. Let $f:A\to B$ be a Boolean algebra homomorphism. Then the \emph{kernel} of $f$ is the set $\lbrace a\in A\mid f(a)=0\rbrace$, and is written $\ker(f)$. Observe that $\ker(f)$ is a Boolean ideal of $A$. Let $\kappa$ be a cardinal. A Boolean algebra homomorphism $f:A\to B$ is said to be $\kappa$-complete if for any subset $C\subseteq A$ such that \begin{enumerate} \item $|C|\le \kappa$, and \item $\bigvee C$ exists, \end{enumerate} then $\bigvee f(C)$ exists and is equal to $f(\bigvee C)$. Here, $f(C)$ is the set $\lbrace f(c)\mid c\in C\rbrace$. Note that again, by de Morgan's laws, if $\bigwedge C$ exists, then $\bigwedge f(C)$ exists and is equal to $f(\bigwedge C)$. If we place no restrictions on the cardinality of $C$ (i.e., drop condition 1), then $f:A\to B$ is said to be a \emph{complete Boolean algebra homomorphism}. In the categories of $\kappa$-complete Boolean algebras and complete Boolean algebras, the morphisms are $\kappa$-complete homomorphisms and complete homomorphisms respectively.