Sikorski’s extension theorem
Theorem 1 (Sikorski’s ).
Let $A$ be a Boolean subalgebra of a Boolean algebra^{} $B$, and $f\mathrm{:}A\mathrm{\to}C$ a Boolean algebra homomorphism from $A$ to a complete Boolean algebra $C$. Then $f$ can be extended to a Boolean algebra homomorphism $g\mathrm{:}B\mathrm{\to}C$.
Remark. In the category^{} of Boolean algebras and Boolean algebra homomorphisms, this theorem says that every complete Boolean algebra is an injective object.
Proof.
We prove this using Zorn’s lemma. Let $M$ be the set of all pairs $(h,D)$ such that $D$ is a subalgebra^{} of $B$ containing $A$, and $h:D\to C$ is an algebra homomorphism extending $f$. Note that $M$ is not empty because $(f,A)\in M$. Also, if we define $({h}_{1},{D}_{1})\le ({h}_{2},{D}_{2})$ by requiring that ${D}_{1}\subseteq {D}_{2}$ and that ${h}_{2}$ extending ${h}_{1}$, then $(M,\le )$ becomes a poset. Notice that for every chain $\mathcal{C}$ in $M$,
$$(\bigcup \{h\mid (h,D)\in \mathcal{C}\},\bigcup \{D\mid (h,D)\in \mathcal{C}\})$$ 
is an upper bound of $\mathcal{C}$ (in fact, the least upper bound). So $M$ has a maximal element^{}, say $(g,E)$, by Zorn’s lemma. We want to show that $E=B$.
If $E\ne B$, pick $a\in BE$. Let $r$ be the join of all elements of the form $g(x)$ where $x\in E$ and $x\le a$, and $t$ the meet of all elements of the form $g(y)$ where $y\in E$ and $a\le y$. $r$ and $t$ exist because $C$ is complete^{}. Since $g$ preserves order, it is evident that $r\le t$. Pick an element $s\in C$ such that $r\le s\le t$.
Let $F=\u27e8E,a\u27e9$. Every element in $F$ has the form $({e}_{1}\wedge a)\vee ({e}_{2}\wedge {a}^{\prime})$, with ${e}_{1},{e}_{2}\in E$. Define $h:F\to C$ by setting $h(b)=(g({e}_{1})\wedge s)\vee (g({e}_{2})\wedge {s}^{\prime})$, where $b=({e}_{1}\wedge a)\vee ({e}_{2}\wedge {a}^{\prime})$. We now want to show that $h$ is a Boolean algebra homomorphism extending $g$. There are three steps to showing this:

1.
$h$ is a function. Suppose $({e}_{1}\wedge a)\vee ({e}_{2}\wedge {a}^{\prime})=({e}_{3}\wedge a)\vee ({e}_{4}\wedge {a}^{\prime})$. Then, by the last remark of this entry (http://planetmath.org/BooleanSubalgebra), ${e}_{2}\mathrm{\Delta}{e}_{4}\le a\le {e}_{1}\leftrightarrow {e}_{3}$, so that $g({e}_{2})\mathrm{\Delta}g({e}_{4})=g({e}_{2}\mathrm{\Delta}{e}_{4})\le s\le g({e}_{1}\leftrightarrow {e}_{3})=g({e}_{1})\leftrightarrow g({e}_{3})$, which in turn implies that $(g({e}_{1})\wedge s)\vee (g({e}_{2})\wedge {s}^{\prime})=(g({e}_{3})\wedge s)\vee (g({e}_{4})\wedge {s}^{\prime})$. Hence $h$ is welldefined.

2.
$h$ is a Boolean homomorphism. All we need to show is that $h$ respects $\vee $ and ${}^{\prime}$. Let $x=({e}_{1}\wedge a)\vee ({e}_{2}\wedge {a}^{\prime})$ and $y=({e}_{3}\wedge a)\vee ({e}_{4}\wedge {a}^{\prime})$. Then $x\vee y=({e}_{5}\wedge a)\vee ({e}_{6}\wedge {a}^{\prime})$, where ${e}_{5}={e}_{1}\vee {e}_{3}$ and ${e}_{6}={e}_{2}\vee {e}_{4}$. So
$h(x\vee y)$ $=$ $(g({e}_{5})\wedge s)\vee (g({e}_{6})\wedge {s}^{\prime})$ $=$ $((g({e}_{1})\vee g({e}_{3}))\wedge s)\vee ((g({e}_{2})\vee g({e}_{4}))\wedge {s}^{\prime})$ $=$ $(g({e}_{1})\wedge s)\vee (g({e}_{2})\wedge {s}^{\prime})\vee (g({e}_{3})\wedge s)\vee (g({e}_{4})\wedge {s}^{\prime})$ $=$ $h(x)\vee h(y),$ so $h$ respects $\vee $. In addition^{}, $h$ respects ${}^{\prime}$, as ${x}^{\prime}=({e}_{2}^{\prime}\wedge a)\vee ({e}_{1}\wedge {a}^{\prime})$, so that
$h({x}^{\prime})$ $=$ $h(({e}_{2}^{\prime}\wedge a)\vee ({e}_{1}\wedge {a}^{\prime}))=(g({e}_{2}^{\prime})\wedge s)\vee (g({e}_{1})\wedge {s}^{\prime})$ $=$ $(g{({e}_{2})}^{\prime}\wedge s)\vee (g({e}_{1})\wedge {s}^{\prime})={((g({e}_{1})\wedge s)\vee (g({e}_{2})\wedge {s}^{\prime}))}^{\prime}$ $=$ $h{(x)}^{\prime}.$ 
3.
$h$ extends $g$. If $x\in E$, write $x=(x\wedge a)\vee (x\wedge {a}^{\prime})$. Then
$$h(x)=(g(x)\wedge s)\vee (g(x)\wedge {s}^{\prime})=g(x).$$
This implies that $$, and with this, we have a contradiction^{} that $(g,E)$ is maximal. This completes the proof. ∎
One of the consequences of this theorem is the following variant of the Boolean prime ideal theorem:
Corollary 1.
Every Boolean ideal of a Boolean algebra is contained in a maximal ideal^{}.
Proof.
Let $I$ be an ideal of a Boolean algebra $A$. Let $B=\u27e8I\u27e9$, the Boolean subalgebra generated by $I$. The function $f:B\to \{0,1\}$ given by $f(a)=0$ iff $a\in I$ is a Boolean homomorphism. First, notice that $f(a)=0$ iff $a\in I$ iff ${a}^{\prime}\notin I$ iff $f({a}^{\prime})=1$. Next, if at least one of $a,b$ is in $I$, $a\wedge b\in I$, so that $f(a\wedge b)=0=f(a)\wedge f(b)$. If neither are in $I$, then ${a}^{\prime},{b}^{\prime}\in I$, so ${(a\wedge b)}^{\prime}={a}^{\prime}\vee {b}^{\prime}\in I$, or $a\wedge b\notin I$. This means that $f(a\wedge b)=1=f(a)\wedge f(b)$.
Now, by Sikorski’s extension theorem, $f$ can be extended to a homomorphism^{} $g:A\to \{0,1\}$. The kernel of $g$ clearly contains $I$, and is in addition maximal (either $a$ or ${a}^{\prime}$ is in the kernel of $g$). ∎
Remarks.

•
As the proof of the theorem shows, ZF+AC (the axiom of choice^{}) implies Sikorski’s extension theorem (SET). It is still an open question whether the ZF+SET implies AC.

•
Next, comparing with the Boolean prime ideal theorem (BPI), the proof of the corollary above shows that ZF+SET implies BPI. However, it was proven by John Bell in 1983 that SET is independent from ZF+BPI: there is a model satisfying all axioms of ZF, as well as BPI (considered as an axiom, not as a consequence of AC), such that SET fails.
References
 1 R. Sikorski, Boolean Algebras, 2nd Edition, SpringerVerlag, New York (1964).
 2 J. L. Bell, http://plato.stanford.edu/entries/axiomchoice/The Axiom of Choice, Stanford Encyclopedia of Philosophy (2008).
Title  Sikorski’s extension theorem 

Canonical name  SikorskisExtensionTheorem 
Date of creation  20130322 18:01:31 
Last modified on  20130322 18:01:31 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  21 
Author  CWoo (3771) 
Entry type  Theorem 
Classification  msc 06E10 
Synonym  Sikorski extension theorem 