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For an adjunction $L:\mathcal{C}\rightleftarrows \mathcal{D}:R$, write its adjunction unit as $\eta$ and its adjunction counit as $\epsilon$. It is an idempotent adjunction when the induced monad has invertible multiplication $\mu =R\epsilon L:RLRL\Rightarrow RL$. The dual condition is invertibility of the induced comonad comultiplication $\delta =L\eta R:LR\Rightarrow LRLR$.

To prove equivalence of these conditions, suppose $\mu$ is invertible. The two unit identities imply $T\eta =\eta T={\mu }^{-1}$, where $T=RL$. For any algebra for a monad $a:TA\to A$, naturality giveswhile $a{\eta }_A=1_A$. Thus ${\eta }_A$ is invertible. For each $D\in \mathcal{D}$, the arrow $R{\epsilon }_D:T(RD)\to RD$ is a monad algebra action, so ${\eta }_{RD}$ is invertible. Applying $L$ proves ${\delta }_D=L{\eta }_{RD}$ invertible. Apply this implication to the opposite adjunction, interchanging the roles of the monad and comonad, to obtain the converse. Therefore idempotence is self-dual.

The open-set frame $O(X)$ is ordered by inclusion. Its arbitrary joins are unions; arbitrary meets are interiors of intersections. Finite intersections are already open, so finite meets are ordinary intersections, including the empty meet $X$. Distributivity of intersection over arbitrary unions proves the frame law. If $f:X\to Y$ is a continuous map, inverse imagepreserves arbitrary unions and finite intersections, including $\varnothing$ and the entire space. It is therefore a frame homomorphism. Since $(gf{)}^{-1}=f^{-1}{g}^{-1}$ and $1_X^{-1}=1_{O(X)}$, this defines $O:\mathbf{Top}\to {\mathbf{Frm}}^{\mathrm{op}}$.

Let $2=\{0<1\}$ and let $P(A)$ be the set of points of a frame $A$, namely frame homomorphisms $A\to 2$. For $a\in A$, put $U_a=\{p:p(a)=1\}$. Preservation of joins and finite meets givesIn the last equality, a join in $2$ is $1$ precisely when at least one summand is $1$, also covering the empty family. Thus the sets $U_a$ are all the opens of a topology, rather than merely a proposed basis.

For a frame homomorphism $h:A\to B$, define $P(h):P(B)\to P(A)$ by $p\mapsto p\circ h$. Since $P(h{)}^{-1}(U_a)=U_{h(a)}$, it is continuous. Composition and identities follow from composition of functions, giving $P:{\mathbf{Frm}}^{\mathrm{op}}\to \mathbf{Top}$.

The frame-point adjunction is the natural bijectionA continuous map $f:X\to P(A)$ determines $h_f(a)=f^{-1}(U_a)$; the displayed identities for $U_a$ show that $h_f$ is a frame homomorphism. Conversely a frame homomorphism $h:A\to O(X)$ determines $f_h(x)(a)=1$ exactly when $x\in h(a)$. Membership in unions and finite intersections proves that $f_h(x)$ is a frame homomorphism to $2$, and $f_h^{-1}(U_a)=h(a)$ proves continuity. These constructions are inverse. Precomposing in $X$ becomes inverse image of open sets; precomposing a frame homomorphism becomes composition of frame points. Hence the bijection is natural in both arguments and proves $O⊣P$.

The adjunction unit isThe adjunction counit at $A$, regarded as an arrow of ${\mathbf{Frm}}^{\mathrm{op}}$, is represented by the frame homomorphism $e_A:A\to O(P(A))$, $a\mapsto U_a$. It is surjective because every open of $P(A)$ is of this form. We can compute $P(e_A):P(O(P(A)))\to P(A)$ explicitly: it sends $q$ to $p=q\circ e_A$. For every open $U_a$ of $P(A)$,Thus ${\eta }_{P(A)}P(e_A)=1$. Conversely, evaluation at $p$ sends $U_a$ to $p(a)$, so $P(e_A){\eta }_{P(A)}=1$. Both maps are continuous by their definitions. Therefore $P(e_A)$ is a homeomorphism. The multiplication of the induced monad $PO$ at $X$ is $P(e_{O(X)})$, which is invertible by this calculation. Consequently the frame-point adjunction is idempotent. This does not assert that every ${\eta }_X$ is a homeomorphism: it says that applying the point-space construction a second time makes no further change.

A point is a frame homomorphism to the two-element frame $2=\{0<1\}$. Its inverse image of $1$ is a completely prime filter: membership of an arbitrary join forces membership of one term. The point space $P(A)$ has opens $U_a=\{p:p(a)=1\}$. Indeed $U_{a\wedge b}=U_a\cap U_b$ and $U_{{\bigvee }_i{a}_i}={\bigcup }_i{U}_{a_i}$, so these sets already form a topology.

For a frame homomorphism $h:B\to A$, nuclei are equivalent when their restrictions to $h(B)$ agree. The least representative of the class of $j$ isThe least representatives form a frame generated by $c(A)$ and $o(h(B))$. To see minimality, each summand is below $j$ and its value at $hb$ is $j(hb)$; any equivalent nucleus therefore bounds every summand. This construction gives the pushout of $B\to N(B)$ along $h$.

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.