MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've made the following observation: let V be a vector space over $\mathbb{R}$ with a inner product $\langle , \rangle$. then there is a "natural contravariant" injective map $V \to \hom(V,\mathbb{R})$. if we apply this twice, we get a "natural covariant" injective map $V \to \hom(\hom(V,\mathbb{R}),\mathbb{R}), v \mapsto (\phi \mapsto \phi(v))$. but the same things happen in category theory: let $C$ be a category (which I assume to be locally-small), then $C \to \hom(C,Set), x \mapsto \hom(x,-)$ is a natural contravariant fully-faithful functor and applying this twice yields (up to natural isomorphism) the natural covariant functor $C \to \hom(\hom(C,Set),Set), x \mapsto (F \mapsto F(x))$ (yoneda-lemma).

so how can we unify these two phenomena? perhaps we can make $V$ to a enriched category over $\mathbb{R}$ and hope that the yoneda-lemma for symmetric closed monoidal categories is the common generalization? what is the right structure on $\mathbb{R}$?

share|cite|improve this question
The yoneda embedding is covariant. – Harry Gindi Dec 28 '09 at 0:03
Interesting analogy. Vote up! Now, any real vector space is naturally embedded into its [algebraic] bidual, without be endowed with a scalar product. How do you "apply this twice", since the dual of V [i.e., hom(V,R)] has no natural pre-Hilbert structure ? – Ady Dec 28 '09 at 0:08
@harry: it is a contravariant functor from C to hom(C,Set). @ady: you're right, I have to think about that! in the finite-dimensional case, we transport the inner product with the isomorphism V -> Hom(V,R). – Martin Brandenburg Dec 28 '09 at 0:25
I asked a somewhat related question here:… – Qiaochu Yuan Dec 28 '09 at 1:24
@Harry: please read what Martin wrote more carefully. – Reid Barton Dec 28 '09 at 2:22
up vote 11 down vote accepted

I don't see the need to try to make vector spaces into categories. I would just say that in each case we have a closed symmetric monoidal category (respectively Vect or Cat), a map f : X ⊗ Y → Z for some objects X, Y, Z (respectively $\langle-,-\rangle$ : V ⊗ V → R and Hom : Cop × C → Set) and we are forming the associated map X → hom(Y,Z) (where hom denotes the internal hom functor). The double dual construction is obtained by setting Y = hom(X,Z) and letting f be the evaluation map; it doesn't depend on anything but X and Z.

That said, there is a great analogy between Vect and Cat, where R and Set play parallel roles: but what corresponds to the construction sending C to Hom(Cop, Set) is the free vector space functor from Set to Vect. The analogy goes something like this. (I am omitting some technical conditions for convenience.)

sets                categories
vector spaces       cocomplete categories (and colimit-preserving functors)
additive structure  colimits
free v.s. on S      category of presheaves on C
the ground field    Set
(comm.) algebras    cocomplete closed (symmetric) monoidal categories
A-modules           cocomplete V-enriched categories

I am not claiming there is a way to take an object in one column and get a corresponding object in the other column (although under some circumstances that may be possible): rather that it is fruitful to use the left-hand column as a way of thinking about the right-hand column.

See this nLab page for an introduction to these ideas.

share|cite|improve this answer

There are many different things that Martin may have had in mind in asking this question, but the obvious answer requires less category theory and not more.

The basic operation is ${-}\to S$ for some fixed "dualising" object in some cartesian closed (or more generally monoidal closed) category.

This operation is part of a monad, of which Martin's natural map is the unit.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.