Proposition 6.122.
For every \(0\)-localic topos \(L\), there exists a complete Boolean algebra \(\Lambda\) and a surjection \(\Shv(\Lambda) \twoheadrightarrow L\).
Proof
Write \(L=\Shv(\Uu)\) for a frame \(\Uu\). For every strict inequality \(u<v\) in \(\Uu\), consider the interval frame
\[[u,v]=\{x\in\Uu\mid u\leq x\leq v\}.\]
The map \(q_{u,v}\colon\Uu\to[u,v]\) given by \(x\mapsto (x\lor u)\land v\) preserves arbitrary joins and finite meets, and sends \(u\) and \(v\) to the bottom and top elements of the interval, respectively.For any frame \(A\), let \[\operatorname{Reg}(A)=\{a\in A\mid \neg\neg a=a\}.\]
This is a complete Boolean algebra: finite meets are inherited from \(A\), joins are given by \(\neg\neg(\bigvee_i a_i)\), and complementation is given by \(a\mapsto\neg a\). The double-negation map \(A\to\operatorname{Reg}(A)\), \(a\mapsto\neg\neg a\), is a frame homomorphism. Applied to the nontrivial frame \([u,v]\), it still distinguishes its bottom and top elements. Hence the composite \[h_{u,v}\colon\Uu\xrightarrow{q_{u,v}}[u,v]\xrightarrow{\neg\neg}\operatorname{Reg}([u,v])\]
distinguishes \(u\) from \(v\).Now set \[\Lambda:=\prod_{u<v}\operatorname{Reg}([u,v])\]
and let \(h\colon\Uu\to\Lambda\) have components \(h_{u,v}\). Products of complete Boolean algebras are complete Boolean algebras, and \(h\) is a frame homomorphism. It is injective, since the component indexed by \((u,v)\) distinguishes any prescribed strict inequality \(u<v\). The corresponding map of locales therefore has conservative inverse-image functor, which is to say that the induced morphism \(\Shv(\Lambda)\to\Shv(\Uu)\) is surjective.