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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.

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.

Let $(G,\epsilon ,\delta )$ be a Cartesian comonad, so its endofunctor preserves finite limits. A coalgebra for a comonad is $(A,a)$ with $a:A\to GA$, ${\epsilon }_{A}a=1_A$ and ${\delta }_{A}a=Gaa$. Its morphisms satisfy $bf=Gfa$. Write $U:{\mathcal{E}}^G\to \mathcal{E}$ for the forgetful functor and $R(X)=(GX,{\delta }_X)$ for the cofree coalgebra. Transposition isThe forgetful functor creates finite limits, since $G$ preserves them. We construct both exponential objects and the subobject classifier.

For coalgebras $(A,a),(B,b)$, set $D=B^A$ and $W=(GB{)}^A$ in the ambient topos. Define $p,q:GD\to W$ by transposingwhere $GD\times GA\cong G(D\times A)$. Their cofree transposes are coalgebra morphismsLet $H$ be their coalgebra equalizer. For a coalgebra $(C,c)$, an arrow $C\to R(D)$ corresponds to a map $h:C\times A\to B$. Its evaluation respects coalgebra structure exactly whenUnder the displayed transpositions this is precisely the condition that it equalize $\hat{p},\hat{q}$. Thus ${\mathcal{E}}^G(C,H)\cong {\mathcal{E}}^G(C\times A,B)$ naturally, and $H$ is the required exponential in a coalgebra topos.

Let $\kappa :G\Omega \to \Omega$ classify $G\top :G1\cong 1\hookrightarrow G\Omega$. In $R(\Omega )$ take the equalizerThe truth map into $R(\Omega )$ is $G\top$ and factors through this equalizer. To verify its universal property, take an underlying subobject $m:S\hookrightarrow A$ of a coalgebra. It carries a compatible coalgebra structure exactly when $a$ maps $S$ into $GS$. Such a structure, if it exists, is unique because $Gm$ is monic; its laws follow by cancellation of $m,Gm,G^{2}m$. Moreover $a^{-1}(GS)\subseteq S$ always: applying ${\epsilon }_A$ to a point of $GS$ lands in $S$. Thus stability is equivalent to $S=a^{-1}(GS)$.

If $\chi :A\to \Omega$ classifies $S$, this last equality is exactlyThe cofree transpose $\hat{\chi }=G\chi a:A\to R(\Omega )$ satisfies this equation precisely when it factors through ${\Omega }_G$. Indeed transposing the equality $\hat{\chi }=G\kappa {\delta }_{\Omega }\hat{\chi }$ gives the equation above. Pulling back its truth map recovers $S$, because $\kappa$ classifies $G\top$. Hence ${\Omega }_G$ is the subobject classifier of a coalgebra topos. We have obtained finite limits, exponentials and a classifier, proving ${\mathcal{E}}^G$ is a topos.

Now let $f:\mathcal{H}\to \mathcal{E}$ be a geometric morphism, and put $L=f^{\ast }$, $V=f_{\ast }$, with unit $\eta$. Both functors preserve finite limits, so $G=LV$ is a Cartesian comonad on $\mathcal{H}$. Put $\mathcal{D}={\mathcal{H}}^G$. Its comparison functoris left exact and has right adjointTo check the adjunction, transpose an underlying map $LX\to A$ to $X\to VA$. The coalgebra-morphism equation becomes exactly the equality of the two composites into $VLVA$, so the transposed arrow factors uniquely through the displayed equalizer.

The counit $KJ(A,a)\to (A,a)$ is invertible. Applying $L$ to the defining equalizer gives the equalizer of $Ga,{\delta }_A:GA\rightrightarrows G^{2}A$. The structure map $a:A\to GA$ is exactly that equalizer: it equalizes by coassociativity; if $t:T\to GA$ equalizes, apply ${\epsilon }_{GA}$ to obtain $t=a{\epsilon }_{A}t$, giving its unique factorization through $a$. Thus the underlying comparison counit is an isomorphism, and the forgetful functor reflects isomorphisms. Therefore $J$ is full and faithful.

The adjunctions $U⊣R$ and $K⊣J$ defineHere $p^{\ast }=U$ is faithful, so $p$ is a surjective geometric morphism; $i_{\ast }=J$ is full and faithful, so $i$ is a geometric embedding. Their inverse-image composite is $UK=L$, establishing the surjection-embedding factorization of a geometric morphism.

For uniqueness, let $S$ be the class of arrows in $\mathcal{E}$ inverted by $f^{\ast }$. In any surjection-embedding factorization $f=i^{\prime }{p}^{\prime }$, conservativity of $p^{\prime \ast }$ shows that $S$ is also the class inverted by $i^{\prime \ast }$. The essential image of the fully faithful $i_{\ast }^{\prime }$ consists exactly of the $S$-local objects $Y$, namely those for which $\mathcal{E}(B,Y)\to \mathcal{E}(A,Y)$ is bijective for every $A\to B$ in $S$. One direction follows from the adjunction. For the converse, the reflector unit $Y\to i_{\ast }^{\prime }{i}^{\prime \ast }Y$ belongs to $S$; locality gives a retraction, and locality of the reflected object makes that retraction also a right inverse. Thus the unit is invertible. The image subcategory is consequently determined by $f$, independently of the factorization. The two intermediate toposes are equivalent over $\mathcal{E}$, and their inverse-image composites with the embeddings determine the surjection legs. This proves uniqueness up to equivalence.

Write the Cartesian comonad as $(G,\epsilon ,\delta )$, with $G$ preserving finite limits. A coalgebra for a comonad is $(A,a)$ with $a:A\to GA$, ${\epsilon }_{A}a=1_A$ and ${\delta }_{A}a=Gaa$. Its forgetful functor $U:{\mathcal{E}}^G\to \mathcal{E}$ is faithful, creates finite limits, and has the cofree coalgebra right adjoint $R(X)=(GX,{\delta }_X)$. We construct the two remaining topos structures explicitly.

For exponential objects, take coalgebras $(A,a),(B,b)$ and start with $R(B^A)$. Put $X=G(B^A)$, with structure $\xi ={\delta }_{B^A}$, and letTranspose the following two maps $X\times A\to GB$ into maps $u,v:X\to (GB{)}^A$:using $G(X\times A)\cong GX\times GA$. The cofree adjunction transposes $u,v$ once more into coalgebra morphisms $R(B^A)\rightrightarrows R((GB{)}^A)$. Let $Q$ be their equalizer. A coalgebra map $(C,c)\to R(B^A)$ corresponds to an arbitrary ambient map $h:C\times A\to B$. It factors through $Q$ precisely whenwhich is exactly the condition that $h$ be a coalgebra morphism. Therefore $Q$ represents ${\mathcal{E}}^G(C\times A,B)$ and is the required exponential in a coalgebra topos.

For the subobject classifier, let $\kappa :G\Omega \to \Omega$ classify the mono $G\top :G1\cong 1\hookrightarrow G\Omega$. In the cofree coalgebra $R\Omega$, formBoth arrows are coalgebra morphisms. The cofree transpose of $\top :1\to \Omega$ factors through this equalizer and gives its true arrow. To verify classification, let $S\hookrightarrow X$ have ambient characteristic map $\chi :X\to \Omega$. It supports a subcoalgebra of $(X,x)$ exactly when it is invariant under $x$, equivalentlyThe counit proves the reverse containment in the first equation; the forward containment supplies the restricted structure map, whose coalgebra laws follow through the mono. Under the cofree adjunction, the second equation says exactly that $G\chi x:X\to R\Omega$ factors through ${\Omega }_G$. Pulling back its true arrow recovers $S$, since $\kappa G\chi x=\chi$. This proves the universal property of the subobject classifier of a coalgebra topos. Hence ${\mathcal{E}}^G$ is a topos.

Now let $f:\mathcal{E}\to \mathcal{F}$ be a geometric morphism, with $L=f^{\ast }⊣H=f_{\ast }$. The comonad $G=LH$ is Cartesian: $L$ preserves finite limits and the right adjoint $H$ preserves limits. Put $\mathcal{D}={\mathcal{E}}^G$, already a topos. The forgetful adjunction $U⊣R$ defines $p:\mathcal{E}\to \mathcal{D}$ with $p^{\ast }=U$, so $p^{\ast }$ is faithful.

The comparison functorpreserves finite limits. It has a right adjoint $J$, given on a coalgebra $(A,a)$ byIndeed, the transposed arrow $Y\to HA$ corresponds to a coalgebra map $KY\to (A,a)$ exactly when it equalizes these two maps. Applying the finite-limit-preserving $L$ shows that $LJ(A,a)$ is the equalizer ofThat equalizer is $a:A\to GA$. If $t:Z\to GA$ equalizes the pair, then $a{\epsilon }_{A}t={\epsilon }_{GA}Gat={\epsilon }_{GA}{\delta }_{A}t=t$, proving the claimed universal property. Consequently the counit $KJ(A,a)\to (A,a)$ is invertible. The fully faithful adjoint criterion makes $J$ full and faithful.

Thus $K⊣J$ defines a geometric embedding $i:\mathcal{D}\to \mathcal{F}$, and $UK=L$ identifies the composite with $f$. The requested factorization iswhere $p$ is a surjective geometric morphism and $i_{\ast }=J$ is full and faithful.

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.