Re: an early exercise in Mac Lane

["Fred E.J. Linton" <FLinton@mail.wesleyan.edu>]

At 01:50 PM 7/8/99 -0700, Lyle Ramshaw was 

> ... puzzled by an exercise on page 15 of Mac Lane's classic text ...

I have three remarks to add to this discussion.

The first is more of a confession: for nigh onto thirty years I seem to have
misread this problem, thinking what it sought was a functor  Grp ---> Grp
commuting with the underlying-set functors, like the passage from a group
to its opposite.  I now stand, belatedly, corrected.

Second (and surely others will be chiming in on this point too),
Baez's suggestion to trivialize all homomorphisms can't work -- you can't be
a functor if you trivialize identity maps.

Finally, the sort of example Mac Lane probably had in mind for an exercise
on such an early page as p. 15:  writing   t  for the only non-identity
automorphism (an central involution, actually) of, say, the chosen group  G
= Z 
(or of G = Z/3Z , if you prefer), define  T: Grp ---> Grp as follows:

 T(X) = X ,  whatever the group  X ,  and, for  f: X --> Y  in  Grp ,  set

        + -
        |   f  ,  if either  X = G = Y  or neither  X  nor  Y  is  G ;
 T(f) = |  ft  ,  if only  X = G ;
        |  tf  ,  if only  Y = G .
        + -

(This ASCII graphic works best if you display it in a fixed-width font.)

I'll spare CATEGORIES readers the straightforward details of the checking 
that  T  really is a functor.  But I'll add the aside that there's nothing 
special about the choices of  G  and  t  above, beyond  G  being a group
and  t  being a central, involutive automorphism of  G .  Even "involutive"
isn't really needed, except for the typographical convenience, here in 
ASCII-land, of not having to compose a  " t-inverse "  with  f  in *one*
(but, please, *not* in the other) of the last two lines of the definition 
of  T(f) :-) .

Cheers,

-- Fred [E.J. Linton,  aka <FLinton@Wesleyan.edu>