# Boolean algebra homomorphism

Let $A$ and $B$ be Boolean algebras. A function $f:A\to B$ is called a Boolean algebra homomorphism, or homomorphism for short, if $f$ is a $\{0,1\}$-lattice homomorphism (http://planetmath.org/LatticeHomomorphism) such that $f$ respects ${}^{\prime}$: $f(a^{\prime})=f(a)^{\prime}$.

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:

1. 1.

if $f$ respects ${}^{\prime}$, then $f$ respects $\vee$ iff it respects $\wedge$;

2. 2.

if $f$ is a lattice homomorphism, then $f$ respects $0$ and $1$ iff it respects ${}^{\prime}$.

The first assertion can be shown by de Morgan’s laws. For example, to see the LHS implies RHS, $f(a\wedge b)=f((a^{\prime}\vee b^{\prime})^{\prime})=f(a^{\prime}\vee b^{% \prime})^{\prime}=((f(a^{\prime})\vee f(b^{\prime}))^{\prime}=f(a^{\prime})^{% \prime}\wedge f(b^{\prime})^{\prime}=f(a)^{\prime\prime}\wedge f(b)^{\prime% \prime}=f(a)\wedge f(b)$. The second assertion can also be easily proved. For example, to see that the LHS implies RHS, we have that $f(a^{\prime})\vee f(a)=f(a^{\prime}\vee a)=f(1)=1$ and $f(a^{\prime})\wedge f(a)=f(a^{\prime}\wedge a)=f(0)=0$. Together, this implies that $f(a^{\prime})$ is the complement of $f(a)$, which is $f(a)^{\prime}$.

If a function satisfies one, and hence all, of the above conditions also satisfies the property that $f(0)=0$, for $f(0)=f(a\wedge a^{\prime})=f(a)\wedge f(a^{\prime})=f(a)\wedge f(a)^{\prime}=0$. 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 kernel of $f$ is the set $\{a\in A\mid f(a)=0\}$, 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

1. 1.

$|C|\leq\kappa$, and

2. 2.

$\bigvee C$ exists,

then $\bigvee f(C)$ exists and is equal to $f(\bigvee C)$. Here, $f(C)$ is the set $\{f(c)\mid c\in C\}$. 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 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.

Title Boolean algebra homomorphism BooleanAlgebraHomomorphism 2013-03-22 18:02:05 2013-03-22 18:02:05 CWoo (3771) CWoo (3771) 5 CWoo (3771) Definition msc 06E05 msc 03G05 msc 06B20 msc 03G10 Boolean homomorphism kernel complete Boolean algebra homomorphism $\kappa$-complete Boolean algbra homomorphism