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The double-negation nucleus on a frame gives a dense sublocale whose frame is a complete Boolean algebra. Its elements are regular open elements, with Boolean joins $\neg \neg {\bigvee }_i{a}_i$.

This nucleus on a frame determines the closed sublocale complementary to the open determined by $a$. It is the least nucleus on a frame taking $0$ to $a$. In the frame of nuclei on a frame, $c(a)$ and the open nucleus $o(a)$ are complements.

A sublocale is flat when its nucleus on a frame preserves finite joins, including $0$. Equivalently its fixed elements are closed under ambient finite joins. In particular every flat sublocale is a dense sublocale.

The Heyting implication defines a nucleus on a frame whose sublocale is the open determined by $a$. It is the least nucleus on a frame sending $a$ to $1$.

Work in the frame $A$, writing $a\to x$ for Heyting implication. A nucleus on a frame is an inflationary, idempotent map preserving finite meets. The open nucleus and closed nucleus areTheir meet in the frame of nuclei on a frame is pointwise. The distributivity of a Heyting algebra and the defining adjunction for Heyting implication giveIf a nucleus on a frame $l$ bounds both, then $l(a)=1$ and $a\le l(0)$. Consequently $1=l(a)\le l(l(0))=l(0)$, so $l$ is the top nucleus. Thus $o(a)$ and $c(a)$ are complements in $N(A)$.

For any nucleus on a frame $j$, put $t_a=o(a)\wedge c(j(a))$. At an arbitrary $x$,since $j(a)\wedge (a\to x)\le j(a)\wedge j(a\to x)=j(a\wedge (a\to x))\le j(x)$. Moreover $t_a(a)=j(a)$. ThereforeIndeed the join is at most $j$, and its value at each $a$ is at least $j(a)$. This proves generation by the open nuclei and closed nuclei, without assuming that joins of nuclei on a frame are computed pointwise.

For the given frame homomorphism $h$, restrict the displayed join to $a=h(b)$ and denote the resulting nucleus on a frame by $r(j)$. The same inequalities show $r(j)\le j$ and $r(j)(h(b))=j(h(b))$. If $k$ has the same values on $h(B)$, its restricted formula is identical, so $r(j)\le k$. Hence $r(j)$ is the least representative of the equivalence class. We will also use the consequence

The least representatives form a subframe. For a family $j_i=r(j_i)$, let $J={\bigvee }_i{j}_i$ in $N(A)$. Restricted monotonicity gives $j_i\le r(J)$ for every $i$, hence $J=r(J)$. For two least representatives, distribution in $N(A)$ expresses their meet as the join, over $b,d\in B$, ofHere $o(hb)\wedge o(hd)=o(h(b\vee d))$ and $c(u)\wedge c(v)=c(u\wedge v)$. Sinceeach summand is below $r(j\wedge k)$. The opposite inequality always holds, proving closure under binary meets. The top is $c(1)$ and is a least representative. Also $r(c(a))=c(a)$, because the summand at $b=0$ is $c(a)$; and $r(o(hd))=o(hd)$, because the summand at $b=d$ is $o(hd)$. It follows from the restricted normal form that $N_B(A)$ is exactly the subframe generated by $c(A)$ and $o(h(B))$.

The map $c:A\to N_B(A)$ preserves finite meets by the formula above. A nucleus on a frame $l$ bounds every $c(a_i)$ exactly when $a_i\le l(0)$ for every $i$, equivalently ${\bigvee }_i{a}_i\le l(0)$. The least such nucleus is $c({\bigvee }_i{a}_i)$, so $c$ preserves arbitrary joins too, including $0$. Evaluation at $0$ recovers $a$, so this frame homomorphism is injective.

For the pushout, we prove the universal property explicitly. Let $u:A\to C$ be a frame homomorphism for which every $u(hb)$ has a complement $d_b$ in $C$. Define, for $j\in N_B(A)$,To prove preservation of arbitrary joins, let $u_{\ast }$ be the right adjoint of $u$ and set $k_z(a)=u_{\ast }(z\vee u(a))$ for $z\in C$. This is a nucleus on a frame: inflationarity follows from the adjunction, finite-meet preservation follows from distribution and meet preservation by $u_{\ast }$, and $u{u}_{\ast }\le 1$ gives $k_z{k}_z\le k_z$; inflationarity gives the reverse inequality. Complementation and the adjunction now giveThe last equivalence uses leastness of $j$. Thus $F$ is a left adjoint, so preserves arbitrary joins. For binary meets, expand $F(j)\wedge F(k)$. A term indexed by $b,d$ isThis proves $F(j)\wedge F(k)\le F(j\wedge k)$; monotonicity proves the reverse inequality. The term at $b=0$ proves $F(1)=1$. Directly,For the second identity the term at $b=d$ gives $d_d$, while every term is disjoint from $u(hd)$, because $hd\wedge (hd\to hb)\le hb$.

Apply this construction first with $A=B$, $h=1_B$, target $N_B(A)$, $u=c_{A}h$ and complements $d_b=o_A(hb)$. It gives a frame homomorphism $v_0:N(B)\to N_B(A)$ satisfying $v_0{c}_B=c_{A}h$ and $v_0{o}_B=o_{A}h$. Given any compatible pair $u:A\to C$, $v:N(B)\to C$, take $d_b=v(o_B(b))$; compatibility supplies the required complements. The constructed $F$ satisfies $F{c}_A=u$ and $F{v}_0=v$, since the latter equality holds on the open and closed generators of $N(B)$. Uniqueness follows because $c(A)$ and $o(h(B))$ generate $N_B(A)$. Hence the square with edges $h,c_B,c_A,v_0$ is a pushout in $\mathbf{Frm}$.

For a Boolean algebra $B$, write $\downarrow b$ for its principal lattice ideal. The ideal frame of a Boolean algebra has intersections as meets and ideals generated by unions as joins. The pseudocomplement of an ideal $I$ isIn particular $(\downarrow b{)}^{\ast }=\downarrow \neg b$, so every principal ideal is complemented and regular.

If $B$ is complete, put $s={\bigvee }_{B}I$. Then $I^{\ast }=\downarrow \neg s$ and $I^{\ast \ast }=\downarrow s$. Thus all regular ideals are principal and complemented, so the corresponding locale is extremally disconnected.

Conversely, if the ideal frame of a Boolean algebra is extremally disconnected, each $I^{\ast \ast }$ is complemented. Any complemented ideal is principal: if $J\vee K=B$ and $J\cap K=\{0\}$, the fact that their generated join contains $1$ supplies $b\in J,c\in K$ with $b\vee c=1$. Then $b\wedge c=0$, so $c=\neg b$; every $j\in J$ is disjoint from $c$ and therefore $j\le b$. Thus $J=\downarrow b$. Write $I^{\ast \ast }=\downarrow s$. Since $I\subseteq I^{\ast \ast }$, $s$ bounds $I$. If $t$ is another upper bound, $I\subseteq \downarrow t$ implies $I^{\ast \ast }\subseteq (\downarrow t{)}^{\ast \ast }=\downarrow t$, so $s\le t$. An arbitrary subset has the same upper bounds as the lattice ideal it generates, so every subset of $B$ has a supremum. Hence

Now put $A=\mathcal{O}(X)$ and $B=A_{\neg \neg }$. This is a complete Boolean algebra, with arbitrary Boolean joins $\neg \neg {\bigvee }_A{b}_i$, so the preceding result makes $\gamma X$ extremally disconnected. We construct its map to $X$ using the permitted Joyal extension lemma.

The map $\sigma :\mathrm{Idl}(B)\to B$, $I\mapsto {\bigvee }_{B}I$, is a frame homomorphism, with right adjoint $b\mapsto \downarrow b$. Thus $k(I)=\downarrow {\bigvee }_{B}I$ is a nucleus on a frame whose fixed frame is $B$. It preserves finite joins: both $k(I\vee J)$ and $k(I)\vee k(J)$ equal $\downarrow ({\bigvee }_{B}I{\vee }_B{\bigvee }_{B}J)$, and $k(0)=0$. Consequently the locale with frame $B$ is a flat sublocale of $\gamma X$. It is also the Booleanization of a locale $X$, so it has its inclusion map to $X$.

The Joyal extension lemma says that a map from a flat sublocale to a compact locale that is a regular locale extends uniquely to the ambient locale. Apply it to this inclusion. Its frame formula giveswhere $u\prec a$ is the well-inside relation $\neg u\vee a=1$. This also proves directly that the map is surjective. Indeed $u\prec a$ implies $\neg \neg u\le a$: meet $\neg u\vee a=1$ with $\neg \neg u$. Every member of the ideal $g^{\ast }(a)$ is therefore at most $a$ in $A$, since a finite Boolean join of its generators is below $\neg \neg (u_1\vee \cdots \vee u_n)\le a$; the last inequality follows because $u_1\vee \cdots \vee u_n\prec a$. Regularity givesThus $a$ can be recovered from $g^{\ast }(a)$, so $g^{\ast }$ is injective. A locale map with injective inverse-image frame homomorphism is a surjection. We have obtained an extremally disconnected $\gamma X$ and a surjection $\gamma X\to X$.

A subframe is a subset of a frame closed under arbitrary ambient joins and finite ambient meets, including $0$ and $1$. The inclusion is a frame homomorphism. In contrast, the fixed elements of a nucleus on a frame usually need their own regularized joins and are not a subframe.

A sublocale is determined by a nucleus on a frame $j$ on $\mathcal{O}(X)$. Its frame consists of the fixed elements of $j$, with inherited meets and regularized joins. Increasing the nucleus makes the sublocale smaller.