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The category ${\mathcal{E}}^G$ consists of coalgebras for a comonad and their structure-preserving morphisms. Its forgetful functor has the cofree coalgebra as right adjoint. If $\mathcal{E}$ is a topos and $G$ preserves finite limits, those limits are created by the forgetful functor, while exponentials in a coalgebra topos and the subobject classifier of a coalgebra topos can be constructed as equalizers inside cofree objects.

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.

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.

For the categorical presheaf $X$, its category of elements has objects $(C,x)$ with $x\in X(C)$. A morphism $(C,x)\to (D,y)$ is a morphism $f:C\to D$ satisfying $X(f)(y)=x$. Composition in a category is inherited from $\mathcal{C}$: if also $X(g)(z)=y$, then $X(gf)(z)=X(f)X(g)(z)=x$. The identity morphisms are inherited as well. The forgetful functor sends $(C,x)$ to $C$ and $f$ to $f$.

A universal element is a pair $(R,r)$ for which each $x\in X(C)$ is uniquely of the form $X(f)(r)$ for $f:C\to R$. Thus $(R,r)$ is a terminal object of the category of elements, with the variance appropriate to a categorical presheaf.

Given a universal element, defineThe defining uniqueness makes each map a bijection; $X(u){\psi }_C(f)={\psi }_{C^{\prime }}(fu)$ gives naturality for $u:C^{\prime }\to C$. Hence $\psi$ is a natural isomorphism and $X$ is a representable presheaf. Conversely, from a natural isomorphism $\psi :\mathcal{C}(-,R)\cong X$, take $r={\psi }_R(1_R)$. The Yoneda lemma gives ${\psi }_C(f)=X(f)(r)$; its bijectivity makes $(R,r)$ a universal element. Therefore the two descriptions coincide:

Suppose $X\cong \mathcal{C}(-,R)$. As a functor on the opposite category, a representable presheaf preserves every existing categorical limit: a colimit in $\mathcal{C}$ is defined by the bijection between morphisms from its vertex into $R$ and compatible families of morphisms from its diagram objects into $R$. This remains valid for a possibly large diagram in a category whenever that colimit exists and the compatible-family collection is the corresponding set.

The category of elements has a terminal object $(R,r)$, where $r$ is the image of $1_R$ under the representation. For each $(C,x)$ let $u_{C,x}:C\to R$ be its unique morphism to $(R,r)$. These form a cocone for the forgetful functor $G$. Any competing cocone $v_{C,x}:C\to A$ satisfiesThe component at $(R,r)$ uniquely determines the mediating morphism. Thus the representing object is the colimit of the elements projection:

For the monad $\mathbb{T}=(T,\eta ,\mu )$ define the free algebra functor byThe unit and associativity identities of the monad make $(TX,{\mu }_X)$ an algebra for a monad, and naturality of $\mu$ makes $Tf$ a morphism of algebras for a monad. Let $G^{\mathbb{T}}$ be the forgetful functor. For $(A,a)$ in the Eilenberg-Moore category, setThe inverse candidate is a monad algebra morphism, sinceThe two composites are identities:These use respectively the unit law for a monad algebra, the algebra-morphism equation, and a unit identity of the monad. The formulas commute with precomposition in $X$ and postcomposition by monad algebra morphisms, so they form a natural bijection. Therefore the free-algebra functor is left adjoint to forgetting:Its adjunction unit is ${\eta }_X$, and its adjunction counit at $(A,a)$ is $a:(TA,{\mu }_A)\to (A,a)$.

Suppose every algebra action is an isomorphism. Its inverse is ${\eta }_A$, by the unit law for a monad algebra. For any underlying morphism $f:A\to B$ between two algebras for a monad, naturality of $\eta$ givesTherefore every underlying morphism is automatically a monad algebra morphism. The forgetful functor is always faithful, because a monad algebra morphism has no data beyond its underlying morphism, and it is now full as well. Hence invertible algebra actions imply

An opmonoidal monad on a monoidal category is a monad $(T,\eta ,\mu )$ whose endofunctor is an opmonoidal functor and whose unit and multiplication of a monad are opmonoidal natural transformations. Suppress only the canonical parentheses. Explicitly,The composite opmonoidal functor $T^2$ has tensor comparison $T_2(TX,TY)T(T_2(X,Y))$ and unit comparison $T_{0}T(T_0)$, explaining the last two equations.

For two algebras for a monad $(X,a)$ and $(Y,b)$, defineLet $c=(a\otimes b)T_2(X,Y)$. The unit law for a monad algebra follows at once from the opmonoidality of $\eta$:For the multiplication law, naturality of $T_2$, the algebra laws, and the opmonoidality of $\mu$ giveThe two unit-comparison equations above similarly make $(I,T_0)$ a algebra for a monad. If $f,g$ are morphisms of algebras for a monad, naturality of $T_2$ shows that $f\otimes g$ is an algebra morphism.

For a third algebra $(Z,d)$, the base associator is also an algebra morphism: its intertwining equation is precisely the opmonoidal associativity axiom, followed by $a\otimes b\otimes d$. The two base unitors are algebra morphisms by the opmonoidal unit axioms. Their pentagon and triangle commute because they commute after the faithful forgetful functor, and the lifted maps have exactly the same underlying morphisms.

The Eilenberg-Moore category is therefore monoidal, with these lifted constraints. Its forgetful functor $U:{\mathcal{C}}^T\to \mathcal{C}$ preserves the tensor product, unit object and constraints exactly, so it is a strict monoidal functor.