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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.