When I started in on adjoint functors, I gave the definition in terms of a bijection of hom-sets. Then I showed that we can also specify it in terms of its unit and counit. Both approaches (and their relationship) generalize to the enriched setting.
Given a functor and another , an adjunction is given by natural transformations and . These transformations must satisfy the equations and . By the weak Yoneda Lemma, this is equivalent to giving a -natural isomorphism .
Indeed, a -natural transformation in this direction must be of the form , and one in the other direction must be of the form . The equations and are equivalent, by the weak Yoneda Lemma, to the equations satisfied by the unit and counit of an adjunction.
The -functor that sends an enriched category to its underlying ordinary category sends an enriched adjunction to an ordinary adjunction. The function underlying the -natural isomorphism is the bijection of this underlying adjunction.
As we saw before, a -functor has a left adjoint if and only if is representable for each . Also, an enriched equivalence is an enriched adjunction whose unit and counit are both -natural isomorphisms. Just as for ordinary adjunctions, we have transformations between enriched adjunctions, a category of enriched adjunctions between two enriched categories, enriched adjunctions with parameters, and so on.
Lacking an open thread, a commenter asks…
I am studying inner product spaces. I have noticed that the inner product along with the norm and the concept of angle are defined without any reference to basis. So, I would think that in an inner product space the inner product and magnitude and direction of a vector are independent of basis.
However, I have noticed that when you introduce a basis the values of the components of vectors may change and their inner products will change.
So, is the inner product, norm and angle independent of basis?
I’ve been waiting to get back around to linear algebra, but I’d rather answer questions than ignore them, so:
Actually, inner products are not basis independent. In fact, in a certain sense, an inner product (let’s assume it’s positive definite, which is probably what you’re considering anyhow) is equivalent to a choice of a basis, up to a certain kind of equivalence. Basically, if we pick a basis we get an inner product in which each basis vector has length one and is perpendicular to every other. On the other hand, if we have a positive-definite inner product we can find such an “orthonormal” basis for it. So the two go hand in hand, and there’s generally many different inner products to put on a given vector space.
Angle is pretty much identified with the inverse cosine of an inner product, so there’s nothing new there.
Norm, however, is even more general than inner product. Every inner product gives rise to a norm (as you’ve probably seen), but there exist norms that are not given by any inner product. This shows up a lot in infinite-dimensional linear algebra, which mathematicians like to call “functional analysis”. In particular, a vector space equipped with a norm (that satisfies a technical condition called “completeness” under this norm) is called a Banach space. If the norm comes from an inner product it’s called a Hilbert space. That there are separate terms speaks to the fact that there are Banach spaces which are not Hilbert spaces. And thus there are normed vector spaces which are not inner product spaces.