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# 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 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. $|C|\leq\kappa$, 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.

## Mathematics Subject Classification

06E05*no label found*03G05

*no label found*06B20

*no label found*03G10

*no label found*

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