Labeled and unlabeled categories

When one talks about the category $V_K$ of vector spaces over a field $K$ and considers the dual functor $D$ which maps a vector space $V$ to its dual $V^{*}$ I believe to have in mind something like a labeled category, the labels letting me know which object is the dual of another object. (Or can I see this by carefully looking at the morphisms?)

What I want to know:

Is there - analogously to graphs - a distinction between labelled and unlabeled categories?

Side remark: I see something like a predominance of unlabeled graphs over labeled ones, the former being the more "genuine" graphs (as abstract structures). What's the situation in category theory?

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Hans, the objects in the category of vector spaces are sets endowed with a certain structure. You can probably say that this is a labeling... but then the concept becomes pretty useless as every category is labeled in that sense: each object is labelled by itself! (Regarding your edit to your last comment: I have no idea what "should" means in this context: the deontics of category theory simply escape me!) – Mariano Suárez-Alvarez Mar 11 '11 at 16:20
You keep saying talking about dots... But the vertices in a graph, even if they are "unlabelled", are distinct, just as if you pick two objects in the category of vector spaces (even if they are isomorphic!) then you have two distinct objects. You simply cannot fail to see that the vertices of a complete graph on 3 vertices have an identity which makes them three vertices and not just one. – Mariano Suárez-Alvarez Mar 11 '11 at 16:31
(Actually, to recognize the dual vector space it is not enough to look at the underlying set but one need also check that the operations are the correct ones, of course, but that does not change my point above) – Mariano Suárez-Alvarez Mar 11 '11 at 16:38
@Hans: if $C$ is a small category, and $O$ is the set of objects, the identity function $\mathrm{id}:O\to O$ is injective. I don't know if that means I am "looking into the objects" or not... – Mariano Suárez-Alvarez Mar 11 '11 at 21:29
@Hans: the forgetful functor from $k$-vector spaces to sets is represented by $k$. There is no sense in which you are not allowed to use this functor to make sense of vector spaces. – Qiaochu Yuan Mar 11 '11 at 21:53

$\text{Vect}_k$ is enriched over itself: for any two vector spaces $V, W$, the set $\text{Hom}_k(V, W)$ naturally acquires the structure of a vector space (since linear combinations of linear operators are linear operators), hence defines a functor $\text{Vect}_k \times \text{Vect}_k \to \text{Vect}_k$ contravariant in the first argument.
In particular, for every $V$ the set $\text{Hom}_k(V, k) \equiv V^{\ast}$ naturally acquires the structure of a vector space, and this is the abstract origin of the dual vector space functor $\text{Vect}_k \to \text{Vect}_k$.
By introducing enough extra data, there is no need for "labelings" (and I don't understand what this means anyway). In particular, if you aren't comfortable with singling out $k$ as a $1$-dimensional vector space, you can introduce the monoidal structure on $\text{Vect}_k$, in which the identity object $k$ is singled out as part of the extra data.