Let $V$ be a complete and cocomplete [cartesian closed category](http://ncatlab.org/nlab/show/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](http://ncatlab.org/nlab/show/enriched+category) with powers and copowers (a.k.a. cotensors and tensors), hence all $V$-enriched [weighted limits](http://ncatlab.org/nlab/show/weighted+limit). 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](https://mathoverflow.net/q/259829/49)) 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 $Z^{X\times Y} \cong (Z^X)^Y$.



Enriched cartesian closed categories
I believe I managed to cook up an actual counterexample where both $C$ and $V$ are presheaf toposes. I'm going to leave my original attempt below since I still think it is instructive.


Let $V$ be presheaves over the category of finite sets $\mathsf{Fin}$ and let $C$ be preasheaves over the poset $\mathbb{N}$ (i.e., towers of sets). Then $C$ carries a $V$-enrichment given as $C(X, Y)\_A = C\_0(X, m \mapsto Y\_{m |A|})$ for $A \in \mathsf{Fin}$ and $X, Y \in C$. This makes $C$ into a (co)tensored $V$-category, but internal homs in $C$ are not enriched.


This does look quite criptic. Let me explain this as an instance of a general construction.


Let's take two small categories $P$ and $J$ with $P$ symmetric monoidal and acting on $J$. Then we set $V$ to be presheaves on $P$ with Day convolution and $C$ presheaves on $J$. We construct a $V$-enrichment as follows. Let $F\_p \colon J \to J$ be the action of $p \in P$ on $J$. This extends via a left Kan extension to a functor $L\_p \colon C \to C$ which has a right adjoint $R\_p \colon C \to C$. We set $C(X, Y)\_p = C\_0(X, R\_p Y)$. This makes $C(X, Y)$ into a $P$-presheaf and the composition operations are induced by the $P$-action of $J$. We also have (co)tensors. The tensor of $X \in C$ by a representable presheaf $P(-,p)$ is $L\_p X$ and the cotensor is $R\_p X$ (and they extend by (co)limits to all presheaves).


$C$ is clearly cartesian closed and I believe there are many cases when the internal homs are not $V$-enriched. My old example below has a variant with $P = \mathbb{Z}/2$ (as a discrete monoidal category) and $J = \Delta$ with action by opposites.


In order to make $V$ cartesian, we can just take $P$ cartesian, e.g., $P = \mathsf{Fin}$. Then a natural candidate for $J$ is some category with finite coproducts so that $P$ acts by tensors. I took $J = \mathbb{N}$ to keep things easy, but I don't know if this is really the simplest choice. If we compute the functors $R\_A$ for $A \in \mathsf{Fin}$, we indeed get $(R\_A X)\_m = X\_{m |A|}$.


Now, let $A = \mathbf{2}$ be a two element set. Then $C(X \times Y, Z)\_\mathbf{2} = C\_0(X \times Y, R\_\mathbf{2} Z) = C\_0(X, (R\_\mathbf{2} Z)^Y)$ and $C(X, Z^Y)\_\mathbf{2} = C\_0(X, R\_\mathbf{2} (Z^Y))$, but $(R\_\mathbf{2} Z)^Y$ and $R\_\mathbf{2} (Z^Y)$ are clearly different in general.




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### Original answer


This is not an answer, but it is a bit too long for a comment. And perhaps it provides some hint towards an actual solution. Basically, I have a counterexample where the monoidal structure of $V$ is not cartesian, but I don't see if it can be modified to make it cartesian.


Let $V$ be the category of $\mathbb{Z}/2$-[graded sets](https://ncatlab.org/nlab/show/graded+set). The category $C = \mathsf{Cat}$ has a nice $V$-enrichment where even morphisms are covariant functors and odd morphisms are contravariant functors. It also has (co)tensors, the tensor of a category $X$ by a graded set $A$ is $X \times A\_0 \sqcup X^{\mathrm{op}} \times A\_1$ and the cotensor is $X^{A\_0} \times (X^{\mathrm{op}})^{A\_1}$. However, the internal homs in $C$ (with respect to the cartesian monoidal structure) are not enriched. Indeed, the odd elements of $C(X \times Y, Z)$ are functors that are contravariant in both $X$ and $Y$, but the odd elements of $C(X, Z^Y)$ are functors that are contravariant only in $X$.


