Can exponentials be distinct from hom-functors in enriched categories?

I'm looking at translations of statements involving functions into the language of category theory, and it seems that there are at least two category-theoretic notions collecting functions between objects $A$ and $B$: exponential objects $B^A$, and $\mathrm{Hom}(A,B)$.

In $\mathbf{Set}$ these notions coincide, which leads me to wonder if the same happens in other categories. Since exponential objects remain in the same category, this seems only to be a sensible question in categories enriched over themselves. I know of two other such categories: $\mathbf{Cat}$, small categories, in which I believe the notions still coincide (although I haven't checked the details, I assume the notation $\mathcal D^\mathcal C$ for functor categories is no accident), and $\mathbf{Ab}$, abelian groups.

Now I'm a little stuck because hom-sets in $\mathbf{Ab}$ do have an abelian-group structure, but they aren't exponential objects, and indeed $\mathbf{Ab}$ has almost no exponentials. My guess is that this is essentially because the evaluation map $(f,x) \mapsto f(x)$ isn't a homomorphism of groups.

My question is: are there examples in which hom-objects and exponentials both exist, but are non-isomorphic objects of the same category?

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For $\bf{Ab}$ instead of the direct product, one usually considers the tensor product for the 'monoidal structure'. Then $f\otimes x \mapsto f(x)$ will be a homomorphism, and the 'exponentials' will be present as right adjoint. –  Berci Jan 27 '13 at 0:03

Sure. Let $G$ be a non-trivial abelian group. Then the category of sets with a left $G$-action, $\mathbf{B} G$, is a symmetric monoidal closed category in two different ways. First, $\mathbf{B} G$ is a topos, so it is cartesian closed. The exponential in $\mathbf{B} G$ turns out to be essentially the same as in $\textbf{Set}$, in the sense that the forgetful functor $\mathbf{B} G \to \textbf{Set}$ preserves exponentials. In particular, the underlying set of the exponential $Y^X$ in $\mathbf{B} G$ is in general not the hom-set $\textrm{Hom}_G(X, Y)$!
Nonetheless, the hom-set $\textrm{Hom}_G(X, Y)$ does have a left $G$-action, namely the pointwise action inherited from $Y$. (This is where we use the fact that $G$ is abelian.) Moreover:
• The left $G$-action on $\textrm{Hom}_G(X, Y)$ is natural in $X$ and $Y$.
• $\textrm{Hom}_G(X, -)$ has a left adjoint, $- \otimes_G X$.
• $\otimes_G$ and $\textrm{Hom}_G$ together make $\mathbf{B} G$ into a symmetric monoidal closed category.
Hmm.. My intuition would lead towards the $\otimes$ version for finding the 'right internal hom'.. –  Berci Jan 27 '13 at 0:10