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 $\mathrm{ker}(f)$. Observe that $\mathrm{ker}(f)$ 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.
Title  Boolean algebra homomorphism 

Canonical name  BooleanAlgebraHomomorphism 
Date of creation  20130322 18:02:05 
Last modified on  20130322 18:02:05 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  5 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 06E05 
Classification  msc 03G05 
Classification  msc 06B20 
Classification  msc 03G10 
Synonym  Boolean homomorphism 
Defines  kernel 
Defines  complete Boolean algebra homomorphism 
Defines  $\kappa $complete Boolean algbra homomorphism 