Re: an early exercise in Mac Lane

Re: an early exercise in Mac Lane

[Fred E.J. Linton]

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>

Re: an early exercise in Mac Lane

[Michael Barr]

I am afraid that does not work.  You have to take an identity to an
identity.  The suggested conjugation by choosing an automorphism for each
object does that.  It may be the answer Mac Lane had in mind, although
that works for any category, as noted.  I cannot think of any other
example.


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On Fri, 9 Jul 1999, john baez wrote:

> Lyle Ramshaw writes:
> 
> >   5. Find two different functors T: Grp --> Grp with object function
> >   T(G) = G the identity for every group G.
> > 
> > One such functor, of course, is the identity on every arrow as well.
> > So the challenge is to find a functor that leaves all objects
> > unchanged, but changes around at least some arrows.
> 
> > I've spent some time trying to construct a more interesting solution
> > to the exercise: a functor from Grp to Grp that leaves objects alone
> > and transforms arrows in some way that clearly changes the structure.
> > In particular, I started out hoping to take some non-null arrows to
> > null arrows.  
> 
> I assume that by "null arrow" you mean what some folks call "the trivial
> homomorphism".  
> 
> Why not go all the way and consider the functor that leaves objects
> alone and maps all arrows to null arrows?
> 
> Best,
> John Baez
> 
> 
> 

an early exercise in Mac Lane

[john baez]

Lyle Ramshaw writes:

>   5. Find two different functors T: Grp --> Grp with object function
>   T(G) = G the identity for every group G.
> 
> One such functor, of course, is the identity on every arrow as well.
> So the challenge is to find a functor that leaves all objects
> unchanged, but changes around at least some arrows.

> I've spent some time trying to construct a more interesting solution
> to the exercise: a functor from Grp to Grp that leaves objects alone
> and transforms arrows in some way that clearly changes the structure.
> In particular, I started out hoping to take some non-null arrows to
> null arrows.  

I assume that by "null arrow" you mean what some folks call "the trivial
homomorphism".  

Why not go all the way and consider the functor that leaves objects
alone and maps all arrows to null arrows?

Best,
John Baez

an early exercise in Mac Lane

[ramshaw]

I am puzzled by an exercise on page 15 of Mac Lane's classic text
"Categories for the Working Mathematician": exercise 5 for section 3
of Chapter I:

  5. Find two different functors T: Grp --> Grp with object function
  T(G) = G the identity for every group G.

One such functor, of course, is the identity on every arrow as well.
So the challenge is to find a functor that leaves all objects
unchanged, but changes around at least some arrows.

There is one easy way to do that, which may well be what Mac Lane had
in mind as the solution to this exercise.  For each group G, we choose
some automorphism of G, say a_G.  We then define our functor T to take
the arrow f: G --> H to the arrow

         (a_H)^(-1) f a_G : G --> H.

This preserves identities and compositions, so it does define a
functor.  But it seems like cheating, in a way.  For one thing, this
new functor is structurally the same as the identity functor.  All
that we've done is to change the scheme by which we name the elements
of the various groups, using the automorphism a_G as our rule for
renaming the elements of G.

Note also that we haven't used any properties of the particular
category Grp.  This renaming technique would work equally well for Ab
or Set.  If renaming was what Mac Lane had in mind, why did he specify
Grp in the exercise?

I've spent some time trying to construct a more interesting solution
to the exercise: a functor from Grp to Grp that leaves objects alone
and transforms arrows in some way that clearly changes the structure.
In particular, I started out hoping to take some non-null arrows to
null arrows.  For example, consider the homomorphism

         h : C_2 --> A_5

from the cyclic group C_2 to the alternating group A_5 that takes the
lone non-identity element of C_2 to the permutation (12)(34).  That
permutation is even, and hence belongs to A_5; and it has order two,
so the map h is a homomorphism.

Initially, I thought that it would be pretty safe for my functor to
map h to the null homomorphism.  Note that A_5 is a simple group, so
any homomorphism from A_5 to any other group has to be either
injective or null -- there are no intermediate possibilities.

But then I thought about situations like the following:

           C_2  --->  C_2 x G

            |            |
            | h          | (h, id)
            V            V

           A_5  <---  A_5 x G

The map on the top line takes t |--> (t, e) where e is the identity of
G, while the map on the bottom line is the projection (s, u) |--> s.

This diagram commutes before I apply my functor, and my functor
presumably leaves the top and bottom arrows alone.  If my functor
takes the h on the left to the null homomorphism, it must also take
the h component of the map (h, id) on the right to null; but it can't
alter the id component of that map.  So my functor has to take some
arrows only partway to the null arrow.  Given an arbitrary arrow, how
far towards the null arrow should my functor take it?  Finding a rule
that works seems hard, even for direct products, not to mention
semi-direct products and non-split extensions.

Is there some interesting solution to this exercise that I am missing?
Or did Mac Lane have the boring, renaming solution in mind?

Thanks,
Lyle Ramshaw