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>