Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

Let us call a (co)functor $F:\mathcal{C}\to\mathcal{C}$ on a category $\mathcal{C}$ involutive, if $F^2$ is naturally isomorphic to $1_\mathcal{C}$.

For example, if $\mathcal{C}=\operatorname{Vect}_\mathbb{K}^0$ is the category of finite-dimensional $\mathbb{K}$-vector spaces with $\mathbb{K}$-linear maps, then the cofunctor $F$ of taking duals is idempotent via the natural transformation $\xi_X:X\ni x\mapsto(\alpha\mapsto\alpha(X))\in X^{**}$. One then easily computes that $(\xi_X)^*=(\xi_{X^*})^{-1}$ or equivalently $$F(\xi_X)=(\xi_{F(X)})^{-1}.$$

Is this just by coincidence, or is it true for arbitrary involutive cofunctors? The corresponding statement for involutive functors should be $F(\xi_X)=\xi_{F(X)}$, what can be said about that?

Cheers, Robert

share|cite|improve this question
A generalised version of this was asked on MO just a few days ago. – Zhen Lin Sep 5 '12 at 1:33
Do you say that my questions is implicitly answered there as well? Actually I don't really understand what they are talking about :) – Robert Rauch Sep 5 '12 at 7:45
A involutary functor is, in particular, an example of a self-adjoint functor. – Zhen Lin Sep 5 '12 at 7:50

Your Answer


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

Browse other questions tagged or ask your own question.