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In the finite-limit convention, a Cartesian category is a category with all finite limits, equivalently a terminal object and all pullbacks in a category. Some authors use the term for finite products alone, so the convention matters in constructions requiring equalizers.

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.

Here Cartesian category means a category with all finite limits. Let $m:Y\hookrightarrow X$ be a monomorphism, and let ${\epsilon }_X\hookrightarrow X\times PX$ be universal membership for a power object $PX$. Pull it back to $Y\times PX$ and then regard the result as a subobject of $X\times PX$. Its unique classifier is an endomorphism $r:PX\to PX$: on a parameterized subobject it intersects with $Y$. Intersecting twice has the same result as intersecting once, so uniqueness of classification gives $r^2=r$.

Take the equalizer $q:Q\hookrightarrow PX$ of $r$ and the identity, and restrict membership to $Y\times Q$. For any subobject $S\hookrightarrow Y\times Z$, its composite with $m\times 1_Z$ is a subobject of $X\times Z$, hence is uniquely classified by $s:Z\to PX$. Since it is contained in $Y\times Z$, its intersection with that subobject changes nothing, so $rs=s$ and $s$ factors uniquely through $Q$. Conversely every map $Z\to Q$ gives the original kind of family by pulling back the restricted membership. These inverse operations givenaturally in $Z$. Thus $Q$ is a power object of a subobject, even if no power objects other than $PX$ were assumed.

Now consider the slice category $\mathcal{E}/A$, which has finite limits formed by ambient pullbacks and has terminal object $1_A:A\to A$. The constant family $A\times X\to A$ has slice power object $A\times PX\to A$: for $Z\to A$,and a classifier $Z\to PX$ is equivalently the map $Z\to A\times PX$ over $A$. Every family $p:X\to A$ is a subobject of this constant family through its monic graph $(p,1_X):X\hookrightarrow A\times X$. Apply the first construction inside the slice. Consequently every slice of a Cartesian category with power objects again has power objects.

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 $\mathcal{E}$.

For a monomorphism $Y\hookrightarrow X$ in a Cartesian category, intersect the universal membership relation on $X\times PX$ with $Y\times PX$. The resulting subobject is classified by $r:PX\to PX$, and $r^2=r$. The equalizer of $r$ and the identity represents precisely families of subobjects of $X$ contained in $Y$, hence is a power object for $Y$. This construction only requires the power object of $X$ and finite limits.

If a Cartesian category has power objects, its slice categories do too. In $\mathcal{E}/A$, the constant family $A\times X\to A$ has power object $A\times PX\to A$, since its product with a family $Y\to A$ is $X\times Y$. An arbitrary family $p:X\to A$ is a subobject of the constant family through its graph $(p,1_X)$, so the power object of a subobject construction applies in the slice.