TITLE: Enriched cartesian closed categories
QUESTION [14 upvotes]: Let $V$ be a complete and cocomplete cartesian closed category.  Feel free to assume more about $V$ if necessary; in my application $V$ is simplicial sets, so it is a presheaf topos and hence has all sorts of nice properties you might want (except that its underlying-set functor $V(1,-) : V \to \rm Set$ is not faithful or conservative).
Let $C$ be a complete and cocomplete $V$-enriched category with powers and copowers (a.k.a. cotensors and tensors), hence all $V$-enriched weighted limits.  And suppose that the underlying ordinary category $C_0$ is cartesian closed, i.e. we have natural isomorphisms of hom-sets
$$ C_0(X\times Y, Z) \cong C_0(X, Z^Y). $$
Is it necessarily the case that $C$ is also cartesian closed in the $V$-enriched sense, i.e. we have $V$-natural isomorphisms of $V$-valued hom-objects
$$ C(X\times Y, Z) \cong C(X, Z^Y)? $$
I can't decide whether I think this is likely to be true or not.  I used to believe that it was true, and I believe I have implicitly used it (or more precisely its generalization to the locally cartesian closed case, which should follow from the simple one applied to enriched slice categories) in a published paper or two.  However, right now I can't see how to prove it, and when stated clearly it sounds unlikely, since usually it is an extra condition on an adjunction between enriched categories to be an enriched adjunction (even when one of the functors is known to be an enriched functor, as is the case here for the cartesian product).  On the other hand, it is true when $C=V$ itself: a cartesian closed category, defined by an isomorphism of hom-sets, automatically enhances to an isomorphism of hom-objects $Z^{X\times Y} \cong (Z^X)^Y$.

REPLY [5 votes]: $\require{AMScd}$As Theo noted, the question amounts to whether the action $A \odot Y$ is given by the formula $(A \odot 1) \times Y$. The algebraic analogue would be to ask whether a module $C$ over a ring $V$ which also happens to have a ring structure is automatically a $V$-algebra. Of course there is no reason that this should be true, and if not for the condition in the question that $V$ and $C$ be cartesian monoidal categories, it ought to be easy to pick a simple counterexample in algebra and cook up a corresponding enriched category. However, I don't know how to extend the analogy to incorporate the cartesian condition, so instead let's just try simple cartesian closed categories $V$ and see what happens.
Obviously $V = \mathrm{Set}$ is too simple. Perhaps the next simplest example is $V = \mathcal{P}(\cdot \to \cdot) = \mathrm{Set}^{\cdot \leftarrow \cdot}$, where $\mathcal{P}(A)$ denotes the category of presheaves on $A$. Let's label the morphism of $\cdot \to \cdot$ as $e : u \to i$ and write $\mathbf{y}$ for the Yoneda embedding. Then the category $V$ is freely generated under colimits by the morphism $\mathbf{y}e : \mathbf{y}u \to \mathbf{y}i$. The object $\mathbf{y}i$ is also the terminal object of $V$ so let's denote it by $1_V$. We also see that the diagonal map $\mathbf{y}u \to \mathbf{y}u \times \mathbf{y}u$ is an isomorphism. Then, informally, it appears that $V$ with its cartesian closed structure as an "algebra" (~ closed symmetric monoidal locally presentable category) is generated by the object $\mathbf{y}u$ together with a morphism $\mathbf{y}e : \mathbf{y}u \to 1_V$ and an isomorphism $\mathbf{y}u \times \mathbf{y}u \to \mathbf{y}u$ inverting the diagonal map of $\mathbf{y}u$.
Now suppose $C$ is a category enriched, tensored and cotensored over $V$. Then the tensor $- \odot - : V \times C \to C$ is an adjunction of two variables, so it is determined by specifying the left adjoints $U(-) = \mathbf{y}u \odot - : C \to C$ and $1_V \odot - : C \to C$, which we can identify with $\mathrm{id}_C$, together with a natural transformation $\varepsilon : U \to \mathrm{id}_C$. Furthermore the diagonal map $\mathbf{y}u \to \mathbf{y}u \times \mathbf{y}u$ induces a natural transformation
$$
\delta : U = \mathbf{y}u \odot - \to (\mathbf{y}u \times \mathbf{y}u) \odot - \cong \mathbf{y}u \odot (\mathbf{y}u \odot -) = U \circ U
$$
which is easily seen to (together with $\varepsilon$) make $U$ into a comonad on $C$. Moreover, it is an idempotent comonad because the diagonal map of $\mathbf{y}u$ is an isomorphism.
To construct an example we need the converse direction, so suppose that $C$ is a complete and cocomplete category equipped with an idempotent comonad $U$ which is also a left adjoint. I claim that $C$ then has the structure of a category enriched, tensored and cotensored over $V$ for which $\mathbf{y}u \odot - = U(-)$. We could write down the enrichment (the Hom-objects are given by $C(X, Y) = (\varepsilon_X^* : \mathrm{Hom}(UX, Y) \leftarrow \mathrm{Hom}(X, Y))$) and cotensor and check a lot of conditions but I'll instead assume that $C$ is a locally presentable category and equip it with the structure of a (pseudo)module over $V$. The construction is just the above argument in reverse. A functor $\alpha : V \otimes C \to C$ (recall $V = \mathcal{P}(\cdot \to \cdot)$) is specified by a functor from $\cdot \to \cdot$ to the category of left adjoints $C \to C$, that is, a natural transformation between left adjoints, for which we take $\varepsilon : U \to \mathrm{id}_C$. We must give an associator which is a natural transformation between the two functors $V \otimes V \otimes C \to C$ built from $\alpha$ and the multiplication $\times$ on $V$; since $\times$ factors through the construction $\mathcal{P}$, this amounts to giving a natural isomorphism between two squares
\begin{CD}
    U @>1>> U\\
    @V 1 V V @VV \varepsilon V\\
    U @>>\varepsilon> \mathrm{id}_C
\end{CD}
and
\begin{CD}
    U \circ U @>U\varepsilon>> U\\
    @V \varepsilon U V V @VV \varepsilon V\\
    U @>>\varepsilon> \mathrm{id}_C
\end{CD}
for which we take the comultiplication of $U$ (which is an isomorphism) in the top-left corner and the identity elsewhere. The unitor between $C \to V \otimes C \to C$ and the identity $C \to C$ is an equality. Finally, we have to check two commutative diagrams. The pentagon identity is an equality between two natural transformations between left adjoints $V \otimes V \otimes V \otimes C \to C$. We may check it object-wise on the generators $\mathbf{y}u$ and $\mathbf{y}i$ of $V$. When all the $V$ generators are $\mathbf{y}u$, this is the coassociativity of $U$. In the remaining cases it should be a formal identity (though I haven't checked). Similarly the relation between the unitor and the associator should reduce to the counitality of $U$. Thus the action $\alpha : V \otimes C \to C$ amounts to giving $C$ the desired structure. (There may well be a more high-brow way of describing this whole construction, but this way is just about manageable.)
Now, for a counterexample to the original question, it suffices to find a locally presentable cartesian closed category $C$ together with an idempotent comonad $U$ which is a left adjoint but for which the natural transformation $UY \to U1_C \times Y$ is not an isomorphism. For example, we could take $C$ to be simplicial sets and $U$ to be the 0-skeleton functor, sending a simplicial set $Y$ to $Y_0$ viewed as a discrete simplicial set. Then $UY$ is 0-skeletal, but $U1 \times Y = Y$ is not.