Order-continuous Boolean homomorphisms extend uniquely to completions

AC is assumed as F5. F6 gives BPI, and F3 makes $A=\widehat{B}$ a complete Boolean algebra.

Set $D=i[B]$ and $h_0=\pi \circ i^{-1}:D\to C$. F4 makes $i:B\to D$ a Boolean isomorphism under the BPI obtained in step 1.1. An order isomorphism and its inverse preserve every existing least and greatest bound: applying the inverse to a competing bound gives exactly the required inequality in the original order. Thus $h_0$ is order-continuous in the sense of F1. F2 makes $D$ order dense in $A$.

Consider all Boolean homomorphisms $h:E\to C$ with $D\subseteq E\subseteq A$ a subalgebra and $h$ extending $h_0$, ordered by graph inclusion. They form a set of subsets of $A\times C$ and include $h_0$. A nonempty chain has a union graph which is a function because any two chain members agree on common arguments. Its domain is a subalgebra and it preserves bounds, complements and binary joins, since any finite list of arguments belongs to one chain member. The union is consequently an upper bound in this poset. The empty chain has upper bound $h_0$. All hypotheses of F7 are now verified; using the assumed AC, it gives a maximal member $h:E\to C$.

Suppose $x\in A\setminus E$. Completeness of $C$ gives $v=\bigvee \{h(a):a\in E,a\le x\}$. For $a,b\in E$ with $a\le x\le b$, every element of this join is at most $h(b)$, so $h(a)\le v\le h(b)$. For $a,b\in E$ put $t(a,b)=(a\wedge x)\vee (b\wedge \mathrm{\neg }x)$. Then $t(a,a)=a$, $t(1,0)=x$, $\mathrm{\neg }t(a,b)=t(\mathrm{\neg }a,\mathrm{\neg }b)$, and $t(a,b)\vee t(a^{\mathrm{\prime }},b^{\mathrm{\prime }})=t(a\vee a^{\mathrm{\prime }},b\vee b^{\mathrm{\prime }})$, by distributivity and the partition $x,\mathrm{\neg }x$ of one. These identities show that $E_1=\{t(a,b):a,b\in E\}$ is a subalgebra containing $E\cup \{x\}$, and any such subalgebra contains every displayed expression. Thus it is exactly the subalgebra generated by that set.

Define a prospective extension by $h_1(t(a,b))=(h(a)\wedge v)\vee (h(b)\wedge \mathrm{\neg }v)$. If $t(a,b)=t(a^{\mathrm{\prime }},b^{\mathrm{\prime }})$, intersecting the equality with $x$ shows $(a\mathrm{△}{a}^{\mathrm{\prime }})\wedge x=0$. Hence $x\le \mathrm{\neg }(a\mathrm{△}{a}^{\mathrm{\prime }})$, and step 4.1 gives $v\le h(\mathrm{\neg }(a\mathrm{△}{a}^{\mathrm{\prime }}))=\mathrm{\neg }(h(a)\mathrm{△}h(a^{\mathrm{\prime }}))$. Therefore $h(a)\wedge v=h(a^{\mathrm{\prime }})\wedge v$. Intersecting instead with $\mathrm{\neg }x$ gives $b\mathrm{△}{b}^{\mathrm{\prime }}\le x$, so $h(b)\mathrm{△}h(b^{\mathrm{\prime }})\le v$ and $h(b)\wedge \mathrm{\neg }v=h(b^{\mathrm{\prime }})\wedge \mathrm{\neg }v$. Combining the two identities proves $h_1$ is independent of the representation $t(a,b)$.

The formulas in step 4.1 and the same partition identities with $v$ in place of $x$ give $h_1(\mathrm{\neg }t(a,b))=\mathrm{\neg }{h}_1(t(a,b))$ and $h_1(t(a,b)\vee t(a^{\mathrm{\prime }},b^{\mathrm{\prime }}))=h_1(t(a,b))\vee h_1(t(a^{\mathrm{\prime }},b^{\mathrm{\prime }}))$: in each case apply $h$ to the componentwise expression, then distribute the meets with $v,\mathrm{\neg }v$. Also $h_1(t(0,0))=0$, and complement preservation gives the unit. Thus $h_1$ is a Boolean homomorphism. Finally $h_1(t(a,a))=(h(a)\wedge v)\vee (h(a)\wedge \mathrm{\neg }v)=h(a)$ and $h_1(t(1,0))=v$. It strictly extends $h$ to the domain containing $x$, contradicting maximality. Consequently $E=A$ and a total Boolean extension $h:A\to C$ exists.

We now prove order continuity of this extension. If $H\subseteq A$ has supremum $1$, let $J=\{d\in D:d\le y\text{ for some }y\in H\}$. F2 gives $y=\bigvee \{d\in D:d\le y\}$ for every $y\in A$. Thus every upper bound of $J$ in $A$ bounds every $y\in H$, and ${\bigvee }_{A}J=1$. Since $1\in D$, it is also the supremum of $J$ in $D$. Order continuity of $h_0$ now gives ${\bigvee }_{C}h[J]=1$. Every $d\in J$ lies below a member of $H$, and $h$ is monotone, so ${\bigvee }_{C}h[H]=1$ as well. This argument includes an empty $H$ if $A$ is trivial: then a bound-preserving homomorphism forces $C$ trivial, so both empty suprema equal one as required.

For arbitrary $T\subseteq A$ put $u={\bigvee }_{A}T$. The family $T\cup \{\mathrm{\neg }u\}$ has supremum one: an upper bound contains both $u$ and $\mathrm{\neg }u$. By step 7.1 its image has supremum one, so, writing $w={\bigvee }_{C}h[T]$, we get $w\vee \mathrm{\neg }h(u)=1$. Monotonicity gives $w\le h(u)$. Intersect the displayed equality with $h(u)$ and distribute to obtain $h(u)=w$. Thus $h$ preserves arbitrary suprema. Complementation reverses the order and turns infima into suprema; since $h$ preserves complements, it preserves arbitrary infima too. For $T=\mathrm{\varnothing }$ these identities give $h(0)=0$ and $h(1)=1$, agreeing with its Boolean bounds.

Set $\widehat{\pi }=h$. It extends $h_0$, so $\widehat{\pi }\circ i=\pi$, and step 8.1 proves order continuity. For every $u\in A$, F2 and supremum preservation give $h(u)=\bigvee \{h_0(d):d\in D,d\le u\}=\bigvee \{\pi (b):i(b)\le u\}$. Any other order-continuous extension satisfies exactly this formula, proving uniqueness. If $B$ is trivial, so are $A$ and, by the existence of the given bound-preserving $\pi$, $C$; if only $C$ is trivial, the unique constant map obeys all the same formulas. The two explicit uses of AC were its implication to BPI and the maximal-partial-map application of F7 in step 3.1. QED.