11

In preparation for an introductory talk on category theory, I recently spent some time thinking about natural transformations. The first example, or maybe the second, that everyone gives to motivate the concept of a natural transformation is the double dual: a vector space is naturally isomorphic to its double dual, and category theory makes this notion precise by saying that there is a natural isomorphism between the identity functor and the double dual functor $\text{Vec}_k\to\text{Vec}_k$. At this point, whoever is giving the example cautions that the dual functor $\text{Vec}_k^{\text{op}}\to\text{Vec}_k$ is not naturally isomorphic to the identity functor, and this is because making an isomorphism between a vector space and its dual requires choosing a basis.

But no one ever proves it! Implicitly, there is a "theorem" here to the following effect:

"Theorem": There is no natural isomorphism between the identity functor $\text{Vec}_k\to\text{Vec}_k$ and the dual functor $\text{Vec}_k^{\text{op}}\to\text{Vec}_k$.

The problem with this "theorem" is that, to my knowledge, it doesn't make sense to talk about a natural transformation between a covariant and a contravariant functor. The obvious commutative diagram to write down, something like $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\la}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ll} V & \ra{f} & W \\ \da{\eta_V} & & \da{\eta_W} \\ V^* & \la{f^*} & W^* \\ \end{array}, $$ will almost certainly not commute -- take, for instance, $f=0$. This failure is caused by the contravariance of the dual functor, not its unnaturality.

My question, then, is this: make precise the claim that there does not exist a natural isomorphism between the identity functor and the dual functor, and prove it.

user134824
  • 12,694

1 Answers1

4

There is the notion of a dinatural transformation (see Mac Lane's book or the nlab), with which you can compare, in particular, covariant with contravariant functors. This essentially comes down to the diagram which you have already found, and your proof shows that every dinatural transformation $\mathrm{id} \to D$ is zero, where $D$ is the dual-vector-space functor.

By the way, since you are interested in examples of (non)-naturality, I think the following one is quite nice and elementary: If $X$ has more than three elements, let $X^2 \to X$ be the projection to the second coordinate. If $X$ has at most three elements, let $X^2 \to X$ be the projection to the first coordinate. This is not a natural transformation from "$\mathrm{id}_{\mathsf{Set}}^2$" to $\mathrm{id}_{\mathsf{Set}}$.

Martin Brandenburg
  • 181,922
  • I'm not seeing why that transformation is not natural. Am I correct in thinking that $\text{id}^2_\text{Set}$ takes $X$ to $X^2$ and a map $f:X\to Y$ to the map $f^2:X^2\to Y^2$ given by $(x,x)\mapsto(f(x),f(x))$? If so, then it seems like when you write down the pertinent commutative diagram, running in either direction gives $f(x)$. – user134824 Apr 12 '15 at 00:12
  • @user134824: Recall the definition of $X^2$, it consists of pairs $(x,y)$. And $f^2$ is defined by $(x,y) \mapsto (f(x),f(y))$. I won't say more, because you should really solve this on your own. :) – Martin Brandenburg Apr 12 '15 at 01:19
  • Oh, of course! I was thinking of $\Delta_X$, not $X^2$. This really isn't that hard! Thanks – user134824 Apr 12 '15 at 12:45