Subject: categories: squares and cubes

From: Peter Freyd <[email-masked]>
Date: Fri, 14 Jan 2000 16:32:21 -0500 (EST)

About an associative bifunctor I wrote

  There's a general theorem that says that a final coalgebra for 
  X v X  is automatically a final coalgebra for  X v X v X v X,
  indeed, for any such ordered-wedge where the number of copies of  X
  is a power of two. I haven't been able to find the proof for a 
  general theorem that specializes to  X v X v X.

Well, the proof is just a modification of an old one (of mine). And
once in hand, it makes clear the nonsensical nature of the
construction for the thirding map that appeared in my "Derivatives"
posting. (I'll replace it with a better construction in a later
posting.)

The general theorem is that if  T  is an endo-functor on a category 
with coproducts and if  f:F --> TF  is a final coalgebra then

   f     Tfn
F --> TF --> TTF  is a final  TT-coalgebra. (Citation below, albeit
for the dual case -- it appears that not everyone has noticed that
just by replacing the category in question with its dual one can
transfer any theorem about initial algebras to one about final
coalgebras.)

Let  X*Y  denote the values of a bifunctor (not necessarily 
associative) on a category with binary coproducts.

    If  f:F --> F*F  is a final "square coalgebra" then

        f       f*1
     F --> F*F ----> (F*F)*F  is a final "cubical coalgebra".

Given a cubical coalgebra  a:A --> (A*A)*A  consider the square 
coalgebra  A + A*A  --> (A + A*A)*(A + A*A)   whose left coordinate 

            a           r*l
map  is  A --> (A*A)*A ----> (A + A*A)*(A + A*A)  and whose right

                        l*l
coordinate map is  A*A ----> (A + A*A)*(A + A*A)  where  l  and  r  
denote the left and right coprojections for the coproduct.  Let  
A + A*A  --> F  be the induced map to the final square coalgebra and
let  a'  and  a'' be its coordinate maps. Then

           a'                            a''
     A --------> F        and      A*A ------> F
                                                       (add downward
   a |           | f              1 |          | f      arrowheads)

 (A*A)*A -----> F*F                A*A -----> F*F          
         a''*a'                        a'*a'

Consequently:
                             a'
                    A  --------------> F

                  a |                  | f
                            a''*a'
                (A*A)*A ------------> F*F

                  1 |                  | f*1

                (A*A)*A ---------> (F*F)*F
                        (a'*a')*a'

For the uniqueness, suppose that  a'  is as in the last diagram with
the middle arrow removed. Define  a'' so that the penultimate diagram
commutes (f, by Lambek, is an isomorphism). Then the map from  A + A*A
to  F  whose coordinate maps are  a'  and  a''  is a map of square 
coalgebras, hence  a'  is as already described.

For a counterexample for arbitrary categories, consider the discrete
category with two objects and bifunctor that turns it into the
two-element group. On discrete categories coalgebras are, of course,
the same as fixed points, and the only way a functor can have a final
coalgebra is if it has a unique fixed point. The object that serves
as the identity element for the group structure is thus the final
square coalgebra. But both objects are cubical coalgebras, hence 
there is no final cubical coalgebra.

There's nothing special about the number three. Given any iteration
obtainable from the bifunctor, the constructions above can be 
modified to provide proofs and counterexamples. Indeed, we needn't
restrict to bifunctors. For example, if  TXYZ  is a trifunctor on a 
category with binary coproducts, then a final coalgebra for  TXXX
gives rise to a final coalgebra for  TX(T(TXXX)X)(TXX(TXXX)).

But what I can't find a proof for is the analogue of this theorem,
good for any category (same citation):

  If  T  is an endo-functor and if  TT  has a final coalgebra
  then so does  T.

The analogue would be: if  X*Y  is a bifunctor and if  (X*X)*X  has
a final coalgebra then so does  X*X.  All my counterexamples --  here
and in the citation below -- are on discrete categories. For what it's 
worth, if an associative bifunctor on a discrete category is such that
X*X*X  has a final coalgebra, then so does  X*X.

Algebraically Complete Categories, Como Conference, Category Theory, 
     Proceedings of the International Conference held in Como, Italy,
     1990, Lecture Notes in Mathematics, 1488, Springer-Verlag, 1991