Date: Wed, 8 Sep 1999 10:12:06 -0400 (EDT) From: Michael Barr Subject: categories: nonpareil dinats 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 = for all a in A and all x in X. The nth power of such a 4-tuple is simply . 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