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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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    Can you please not use the `sup` tag for something for which it is not designed? There is no need to use smaller fonts: if you prepend "Side remark" to it, we all know immediately it is a side remark.2011-03-11
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    There is no need to label anything in your example: the dual space to a vector space $V$ is very easily recognizable by the fact that its underlying set is the set of all linear functions of $V$.2011-03-11
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    But isn't mapping an object to its underlying set just labeling it? (Does this mean one cannot/should not try to see a category in isolation?)2011-03-11
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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!)2011-03-11
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    @Mariano: and I used to believe that the objects of a category were **dots**! What is right and what is wrong?2011-03-11
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    ... just like the nodes of a graph are dots, labeled by letters or sets endowed with a certain structure ("explaining the edges") or not at all.2011-03-11
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    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.2011-03-11
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    (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)2011-03-11
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    @Mariano: What you say sums up to: I cannot conceive the category $V_K$ without "looking into" the objects and seeing what they are - sets endowed with a certain structure. It's trivial then to see that V* is the dual of V: "its underlying set is the set of all linear functions of V". But I believed that objects gain their individuality - up to isomorphism - from the surrounding morphisms.2011-03-11
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    @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...2011-03-11
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    @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.2011-03-11

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$\text{Vect}_k$ 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.

Side remark: I do not understand your side remark. There are many categories of graphs, and people are interested in both labeled and unlabeled graphs. What does "genuine" mean?

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    I find this rather advanced. Is "enriched category" a standard concept? By labeling I mean e.g. the labeling of an object of a concrete category "by itself". And I always thought that such labelings are not essential because the objects gain their "inner structure" by the surrounding morphisms.2011-03-11
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    This process of "introducing extra data" or "singling out objects": under which label is this handled usually?2011-03-11
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    @Hans: I don't know what "standard" means. I still don't know what you mean by "labeling." Introducing extra data is the same in category theory as it is in set theory. My advice to you is the same as always: instead of asking unfocused questions, why don't you _learn and do some mathematics_?2011-03-11