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categories: Barr question



  Mike Barr asked:

    Suppose  X  is an abelian category and  T:X --> Ab  is a left- 
    exact functor. Say an object  E  is  T-effaceable if for every 
    exact sequence  0 --> E --> B --> A --> 0, the induced  TB --> TA  
    is surjective. Now suppose that  0 --> E' --> E --> E'' --> 0  is 
    an exact sequence in which  E  and  E' are  T-effaceable. Does it 
    follow that  E'' is?

  (Well, actually he asked the dual question. But they're equivalent.)

  The answer is no.

  Suppose we have an abelian category with a map  x:E --> E  where

     1)  E is injective;
     2)  the image of  x  is not injective; and
     3)  x  is not a zero-divisor in the ring of endomorphisms of E.

  Let  E'  denote the kernel of  x, E'' the image and  F  the 
cokernel. Let  T  be the representable functor  (F,-).

In order of difficulty:

   E  is  T-effaceable because being condition 1 says that it's 
T-effaceable for any old  T.

   E'' is not  T-effaceable because if the exact sequence
0 --> E'' --> E --> F --> 0  were to be carried into an exact sequence
0 --> (F,E'') --> (F,E) --> (F,F) --> 0  then the identity element in
(F,F)  would have a pre-image and the  0 --> E'' --> E --> F --> 0  
would split, forcing  E'' to be injective (violating condition 2).

   E' is  T-effaceable because given  0 --> E' --> B --> A --> 0  
exact we show that  0 --> (F,E') --> (F,B) --> (F,A) --> 0  exact by 
showing that any  F --> A  has a pre-image in  (F,B); we do that by 
first constructing the exact diagram (add your own downwards 
arrowheads):
                      0 --> E'--> C --> F --> 0     

                           1|     |     |

                      0 --> E'--> B --> A --> 0  

by taking the right-hand square as a pullback; we need to show that
the top sequence splits; construct:

                      0 --> E'--> C --> F --> 0     

                           1|     |     |

                      0 --> E'--> E --> E''--> 0     

by using the injectivity of  E  to obtain the middle vertical and
the exactness of the rows to obtain the right-hand vertical; note
         x
that  E --> E --> F --> E'' --> E  is necessarily the zero-map, 
hence  E --> F --> E'' --> E  is a zero-divisor for  x, and condition
3 together with the epiness of  E --> F  and the mononess of  E'' -> E
forces  F --> E'' to be the zero-map; that forces  C --> E --> E''
to be the zero-map; and that says that  C --> E  can be factored 
through  E' --> E; but that implies that  E' --> C  is a split mono;
so  C --> F  is a split epi.

   How do we know that the three numbered conditions can be met? 
There must be an easier way, but the one that occurs to me is to use
the category Joel Cohen named in his book the Freyd Category (named 
not after me, he says, but my daughter). It's the abelian category in
which the stable-homotopy category of finite complexes appears as
the full subcategory of injective-projectives.  Take  E  to be 
the sphere. Its ring of endomorphisms is the ring of integers and
we can take  x = 2.  Using that the endomorphisms form a connected
ring (no non-trivial idempotents) we know that  x  is not epi (else
it would be a split epi) and its image, being a proper non-trivial
subobject, can't be injective.

  (If you try for a dual example by replacing "injective" with
"projective", avoid the temptation of taking  E  to be free -- can't
work.)

  When Mike and I started, category theory was classified (by MR and 
just about everybody) as homological algebra. Mike was right, in fact,
when he said, "I thought this would be a simple application of the
snake lemma", albeit in the other direction. Assuming that  T  could
be taken as a functor of the form  (F,-)  a little homological algebra
says that what's needed is an object  E'  where  Ext(F,E')  is trivial
but  Ext^2(F,E')  is not. I thought it might be best -- since category
theory is no longer a subset of homological algebra -- to get rid of
the Exts.