nonpareil dinats

nonpareil dinats

[Michael Barr]

Here is an amusing observation, doubtless of no importance.  In the middle
of a talk at Boulder in, I think, 1987, Bob Pare' came up with a class of
exotic dinatural endotransformations on the homfunctor on Set.  Namely,
for an endomorphism t: X --> X, let t |--> n where n depends only the
cardinality of the fixpoint set of t.  Since, for f: X --> Y and g: Y -->
X, fg and gf have isomorphic fixpoint sets, this turns out to be
dinatural.  The fact that Fix(fg) is isomoprhic to Fix(gf) is perfectly
general in any category that has the equalizer used to define them.  In
particular, this is true in Set\op and so we can get non-Pare' dinats by
using the same formula, but making n depend instead on the cofixpoint set
of t, where that is the coequalizer of t and the identity.  It is easy to
find examples of endomorphisms that have the same fixpoint set, but
different cofixpoint sets and vice versa, so tere are genuinely new.  Are
there any others?  I don't know.

I started thinking about this after a note from Vaughan Pratt who was
interested in Chu(Set,2) (Surprise!).  He had noted that there was a full
subcategory of chusets of the form (A,0) and you could treat their
endomorphisms separately.  That full subcategory is essentially Set and so
you on that subcategory you can use all the Pare' and non-Pare' dinats.
Leaving those aside, you can do dinatural endotransformations of the
internal hom functor in four ways:  If (A,X) is an object, then an
internal endoarrow is a 4-tuple (f,s,a,x) where f: A --> A, s: X --> X, a
in A and x in X subject to <fa,x> = <a,sx> for all a in A and all x in X.
The nth power of such a 4-tuple is simply <f^n,s^n,a,x>.  Then you can
define dinats by letting (f,s,a,x) |--> (f,s,a,x)^n where n depends on
Fix(f) and Fix(s) OR on Cofix(f) and Cofix(s) OR on Fix(f) and Cofix(s) OR
on Cofix(f) and Fix(s).

Qeustion: Are there any dinats on the internal homfunctor on vector
spaces?  I almost have an argument for finite dimensional spaces, but it
depends on writing every endomorphism as a sum of rank one endomorphisms.

Michael