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Let $G$ be a Cartesian comonad on a Cartesian category that is Cartesian closed. Its forgetful functor creates finite limits. For coalgebras $(A,a)$ and $(B,b)$, set $D=B^A$, $W=(GB{)}^A$. Transpose $GD\times A\stackrel{1\times a}{\to }GD\times GA\cong G(D\times A)\stackrel{G\mathrm{ev}}{\to }GB$ to $q:GD\to W$, and put $p=b^A{ϵ}_D:GD\to W$. Their cofree coalgebra transposes $Gp{\delta }_D,Gq{\delta }_D:R(D)\rightrightarrows R(W)$ have a coalgebra equalizer. A map into this equalizer is exactly a map $C\times A\to B$ respecting coalgebra structures, so it is the required exponential object. No subobject classifier on the base category is needed.

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 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 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.