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 $\lbrace 0,1\rbrace$ -lattice homomorphism 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:

1. if $f$ respects $'$ , then $f$ respects $\vee$ iff it respects $\wedge$ ;
2. if $f$ is a lattice homomorphism, then $f$ respects $0$ and $1$ iff it respects $'$ .

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 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

1. $|C|\le \kappa$ , and
2. $\bigvee C$ exists,