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In the finite-limit setting, a Cartesian comonad is a comonad whose underlying endofunctor preserves finite limits. Its category of coalgebras for a comonad has finite limits created by the forgetful functor. On an elementary topos, equalizer constructions of exponentials in a coalgebra topos and the subobject classifier of a coalgebra topos make that coalgebra category an elementary topos.

For a comonad, a coalgebra is a map $a:A\to GA$ satisfying ${ϵ}_{A}a=1_A$ and ${\delta }_{A}a=Gaa$. A morphism $f:(A,a)\to (B,b)$ obeys $bf=Gfa$. This is distinct from the tensor-based notion of a coalgebra in algebra.

For a comonad, the cofree coalgebra on $X$ is $(GX,{\delta }_X)$. A map $UA\to X$ transposes to the coalgebra morphism $Gfa:A\to GX$. This gives the adjunction $U⊣R$ with the forgetful functor.

A full subcategory is coreflective when its inclusion $I$ has a right adjoint $q$. The counit $IqX\to X$ is universal for maps to $X$ from objects of the subcategory. The comonad $Iq$ is idempotent. For a finite-limit-closed coreflective subcategory of a topos, its coalgebra construction can prove that the subcategory is itself a topos.

Discretization $D$ and indiscretization $J$ retain underlying sets. Every function $DX\to Y$ is continuous, and every function $X\to JY$ is continuous, giving the adjunction. Its induced monad and comonad are idempotent, since repeated discrete topology or indiscrete topology construction changes nothing.

For a finite-limit-preserving comonad on a topos, start with the cofree coalgebra on the ambient exponential $B^A$. Its underlying evaluation is $e=\mathrm{ev}(ϵ\times 1)$. Transpose the two expressions $be$ and $Ge(\delta \times a)$ first using the ambient exponential and then the cofree adjunction. Their coalgebra equalizer represents exactly the maps whose evaluations respect coalgebra structure. Thus coalgebra exponentials need not be the underlying ambient exponentials.

An adjunction $L⊣R$ is idempotent when its induced monad has invertible multiplication. This is equivalent to its induced comonad having invertible comultiplication. For an idempotent monad, the two maps ${\eta }_{TA}$ and $T{\eta }_A$ agree, since both invert the multiplication. If $a:TA\to A$ is an algebra, naturality then gives ${\eta }_{A}a=Ta{\eta }_{TA}=T(a{\eta }_A)=1$. Thus every algebra unit is invertible. Apply this to $R{\epsilon }_B$ to obtain the self-dual adjunction condition.

A comonad is idempotent when its comultiplication $G\to G^2$ is invertible. Its coalgebra category identifies with the coreflective subcategory of objects on which the counit is invertible. A coreflective inclusion produces such a comonad.

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.

For an adjunction $L:\mathcal{C}\rightleftarrows \mathcal{D}:R$, with unit $\eta$ and counit $\epsilon$, call the adjunction idempotent when the multiplicationof its induced monad is invertible. We show that this idempotent adjunction condition is equivalent to invertibility of $\delta =L\eta R:LR\to (LR{)}^2$, the comultiplication of the induced comonad.

For an idempotent monad $T$, the unit laws implyIf $a:TA\to A$ is a monad algebra, naturality of the unit and its algebra unit law giveThus ${\eta }_A$ is an isomorphism. Apply this to the canonical algebra $R{\epsilon }_B:T(RB)\to RB$: its unit law is a triangle identity for an adjunction. It follows that ${\eta }_{RB}$, and therefore $L{\eta }_{RB}$, is invertible for every $B$. Hence $\delta$ is invertible.

Apply the same argument to the dual adjunction $R^{\mathrm{op}}⊣L^{\mathrm{op}}$. It gives the converse. Idempotence is self-dual: the induced monad is idempotent exactly when the induced comonad is idempotent. In particular, idempotence requires the unit to be invertible on every object in the image of the right adjoint.

Write a coalgebra for a comonad as $(A,a)$ with $a:A\to GA$, ${ϵ}_{A}a=1_A$ and ${\delta }_{A}a=Gaa$. Since the Cartesian comonad $G$ preserves finite limits, the forgetful functor $U:{\mathcal{E}}_G\to \mathcal{E}$ creates them: for an underlying limit $K$ of coalgebras, the maps $K\to A_i\to G{A}_i$ induce a unique $K\to GK$ using $GK\cong {\lim }_{i}G{A}_i$. Its counit and coassociativity equations follow after every limiting projection, which jointly detect equality. The same argument makes the limiting cone universal among coalgebra cones. Thus ${\mathcal{E}}_G$ is a Cartesian category; in particular, the product structure on $A\times B$ is $a\times b$ followed by the comparison $GA\times GB\cong G(A\times B)$.

An exponential object needs more than the underlying ambient $B^A$. The cofree coalgebra functor is $R(D)=(GD,{\delta }_D)$, with adjunctionThe inverse sends $\tilde{h}$ to ${ϵ}_D\tilde{h}$; the comonad laws verify that these are inverse and that $\tilde{h}$ is a coalgebra map.

For $(A,a)$ and $(B,b)$, put $D=B^A$ and $W=(GB{)}^A$ in the ambient Cartesian closed category. Define $p,q:GD\to W$ byand by transposing the compositeto obtain $q$. Their cofree-adjoint transposes are coalgebra mapsTake their coalgebra equalizer $e:H\hookrightarrow R(D)$, which exists by the finite-limit construction.

To verify its universal property, let $h:C\to D$ transpose to $t:C\times A\to B$. Its coalgebra transpose $\tilde{h}=Ghc$ factors through $H$ exactly when $p^{♯}\tilde{h}=q^{♯}\tilde{h}$. By the cofree adjunction, this equality is equivalent to $p\tilde{h}=q\tilde{h}$. The first side is $b^{A}h$, because ${ϵ}_{D}Ghc=h$; transposing the second side gives $Gt$ composed with the product coalgebra structure on $C\times A$. The equality therefore says exactlywith the canonical product comparison understood: precisely the condition that $t$ is a coalgebra map. Thusnaturally in the coalgebra $C$. Its evaluation is ambient evaluation after ${ϵ}_{D}e\times 1_A$, and the preceding condition makes it a coalgebra map. This equalizer is the coalgebra exponential, so the category of coalgebras is cartesian closed. This Cartesian closure for a finite-limit-preserving comonad uses no subobject classifier on $\mathcal{E}$.

Write the comonad as $(G,ϵ,\delta )$ and its category of coalgebras for a comonad as ${\mathcal{E}}^G$. A coalgebra for a comonad is a map $a:A\to GA$ with ${ϵ}_{A}a=1_A$ and ${\delta }_{A}a=Gaa$. A morphism $f:(A,a)\to (B,b)$ satisfies $bf=Gfa$. Let $U:{\mathcal{E}}^G\to \mathcal{E}$ be the forgetful functor and let $R(X)=(GX,{\delta }_X)$ be the cofree coalgebra. The adjunction $U⊣R$ has the explicit correspondenceWe construct the three pieces of the elementary topos structure.

Because $G$ preserves finite limits, each underlying finite limiting cone has a unique coalgebra structure induced by the structures on its vertices. The counit and coassociativity equations can be checked after its jointly monic projections. Thus $U$ creates finite limits and reflects isomorphisms. A morphism of coalgebras is monic exactly when its underlying morphism is monic, by the diagonal criterion using the created pullback.

For exponentials in a coalgebra topos, fix coalgebras $(A,a)$ and $(B,b)$ and put $E=B^A$ in $\mathcal{E}$. On the cofree coalgebra $R(E)$ there is an underlying evaluationThe two mapsuse $G(GE\times A)\cong GGE\times GA$ in the second expression. Transpose them in $\mathcal{E}$ to maps $r,s:GE\rightrightarrows (GB{)}^A$, and then transpose across $U⊣R$ to coalgebra morphisms $\tilde{r},\tilde{s}:R(E)\rightrightarrows R((GB{)}^A)$. Take their equalizer $C$ in ${\mathcal{E}}^G$.

An underlying map $UX\times A\to B$ corresponds to a coalgebra map $h:X\to R(E)$. The equation saying that the original map is a coalgebra morphism is precisely $rh=sh$, since ${\delta }_{E}h=Ghx$ for the structure $x:X\to GX$. By the cofree adjunction, this is equivalent to $\tilde{r}h=\tilde{s}h$, hence to unique factorization through $C$. Thereforenaturally in $X$. This constructs the required exponential object.

For the subobject classifier of a coalgebra topos, let $\top :1\hookrightarrow \Omega$ be the underlying subobject classifier, and let $\kappa :G\Omega \to \Omega$ classify the mono $G\top :G1\cong 1\hookrightarrow G\Omega$. Its cofree transpose is the coalgebra endomorphismDefine ${\Omega }_G$ as the equalizer of $k$ and $1_{R\Omega }$. The transpose of $\top$ factors through this equalizer and gives ${\top }_G:1\to {\Omega }_G$.

Indeed, for a subobject $S\hookrightarrow UX$ classified by $\chi :UX\to \Omega$, the pullback $x^{-1}(GS)$ has characteristic map $\kappa G\chi x$. It is always contained in $S$, by naturality of the counit. Equality holds exactly when $x$ restricts to a coalgebra structure on $S$; its axioms then follow by composing with the monomorphisms $GS\hookrightarrow GX$ and $GGS\hookrightarrow GGX$. Under $U⊣R$, the classifying map becomes $h=G\chi x:X\to R\Omega$. The equality of subobjects is ${ϵ}_{\Omega }h=\kappa h$. Transposing this equality gives $h=kh$, so exactly the coalgebra subobjects correspond to maps $X\to {\Omega }_G$. Their pullback of ${\top }_G$ is the desired subcoalgebra, and uniqueness follows from uniqueness of $\chi$.

Thus we have finite limits, exponentials and a subobject classifier:The construction does not assume that $G$ preserves the underlying exponentials or underlying subobject classifier.