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

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. $|C|\le \kappa$, and

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.