For any
linear space V, let
V∗ denote its dual
space.
It is a fact that a
space V is
isomorphic to its dual
V∗. The usual way to show this is to pick some basis for
V and transform it into a basis for
V∗. As a consequence of this, a
space V is
isomorphic to its double-dual
V∗∗: repeat the construction twice to produce an
isomorphism V≅V∗ and then another
isomorphism V∗≅V∗∗.
There is another, somewhat more ‘direct’ method of showing that
V≅V∗∗. Namely, note that an element
λ∈V∗∗ is a co-co-
vector; ie, a
function
λ:(V→k)→k
where
k is the field underlying
V. To show that
V≅V∗∗ we must transform any
vector v∈V into a corresponding co-co-
vector λv:(V→k)→k in a way that preserves the
structure of
linear spaces. There is (if you’ll excuse the
language) an “obvious” thing to try here, which is:
λv:(V→k)→kλv(f)=f(v)
As it turns out, this choice does produce a valid
isomorphism V≅V∗∗, but we will skip the details here.
The more important point is that this construction of
V≅V∗∗ differs in some qualitative way from the previous construction
V≅V∗≅V∗∗. Namely, whereas the previous construction required first choosing a basis for
V (and then
V∗) and hence relied on knowing
in particular which
V we were looking at, this construction does not. In some sense, this second construction for
V≅V∗∗ is
defined “independently” of choice of
V.
As such, we want to say that this second construction is ‘natural’, and it does turn out to be a natural transformation (we also skip the proof of this). In particular, we have that the collection
{(v↦λv):(V→k)→k}V a linear space over k
of components forms a natural transformation between the
functors
idVectkand(−)∗∗
where the first
functor is the identity
functor over the
category of
vector spaces on the field
k and the second
functor is over the same
category and sends a
space to its double-dual.
commute for all choices of . This holds pretty easily:
and hence, yes, is natural.
Now we check . Turns out that no, is not natural; we can construct a square that fails to commute. For let . Now consider the square
where is the inclusion map. This square fails to commute because
but
Hence is not natural.
The point is that in some sense does something “different” on different choices of , and the above square for the morphism detects that it is acting “differently” on versus .
For to be natural, this square must commute for all choices of . However, it fails in the following case. Define a function
where . Now consider the square where . This gives:
that is, for every ,
on the specific value this says
which is impossible to satisfy: there is no for which .