Reddy's question

[Peter Freyd <pjf@saul.cis.upenn.edu>]

Uday Reddy poses the following (with a few changes in notation]:

>Consider a monoid  <M,*,e>  in a CCC.  The operations of interest
>are natural transformations  E:[-,M] -> M  that satisfy the
>following equations (in the internal language of the CCC):
>
>               E_A(\x.e)  =  e
>         E_A(\x. a * gx)  =  a * E_A(g)
>         E_A(\x. gx * a)  =  E_A(g) * a
>    E_A(\x. E_B(\y.hxy))  =  E_B(\y. E_A(\x.hxy))

I wonder if naturality is really desired: it would seem to force  M  to
be trivial. By the familiar Yoneda-lemma argument, E  must be constant
as far as the "points" of  [A,M]  are concerned. (Actually one doesn't
need the argument, just the lemma itself; consider the transformation
that  E  induces between set-valued functors  (-,M) -> (1,M); Yoneda
says it must be constant.) The condition

    E_A(\x.e)  =  e

forces just which constant it is. That is, for any  f:A -> M  it will
be the case that  E_A  will send  f  to  e.

But then either condition

    E_A(\x. a * gx) = a * E_A(g)

or

    E_A(\x. gx * a) = E_A(g) * a

will force  M  to be trivial. (It's clear in the \-calculus notation.
But that argument would be implicitly using the fact that  E_A  is
constant and we officially know only that it's constant on points.
So take, say, the second condition. It says in diagrammatic language:


              1 x K                 P            [A,*]
   [A,M] x M  -----> [A,M] x [A,M] ---> [A,MxM]  -----> [A,M]

         |                                                |
         |  E_A x 1_M                                     |  E_A
         |                                                |
                                    *
       M x M   ---------------------------------------->  M


where  K  is the standard "constant-map" operator that's adjoint to
the projection  AxM -> M  and  P  is the standard operator that
defines  MxM  (given products in Set). Specialize to  A = M  and
precompose with  <f,1>: M -> [M,M] x M  where  f  doesn't matter. If
the commutative rectangle above is chased clockwise one obtains the
map constantly valued  e. Chased counterclockwise one obtains the
identity map. And, of course, it didn't matter what  f  is.)