abelian stuff

abelian stuff

[Peter Freyd]

At the recent Ottawa conference I repeated, in passing, something I've
been saying for over 30 years: given a subcategory of an abelian
category the finite bi-closure is, of course, obtainable by
alternately adjoining limits and colimits countably often; remarkably
enough, one need only do two cycles (assuming direct sums that's
kernels, cokernels, kernels, cokernels). The proof is a consequence of
of the construction of free abelian categories. Susan asked me later
if I had ever actually worked out a direct proof. Nope.

But it became apparent to me, thinking about what I had had to say
about free abelian categories that I didn't know 30 years ago, that it
shouldn't take two cycles, just one:


           In the abelian setting, adjoining finite limits
               commutes with adjoining finite colimits.


(No -- for those old enough to remember -- this is not still 1969 and
Ottawa is not Seattle.)

There are some finicky points one must deal with when talking about
images of functors (in the case at hand, functors such as those that
assign kernels to maps), so let me just cut here to a critical little
lemma that says, in effect, "a kernel of a map between cokernels is a
cokernel of a map between kernels:"

LEMMA: In an abelian category an exact diagram of the form (all
       vertical arrows are downwards):

         c
      B --> A --> F --> O

    b |   a |     |

      B'--> A'--> F'--> O
         c'

     may be enlarged to an exact diagram of the form:


            O --> K'--> A + B'+ B --> A + A'

                  |         |           |

            O --> K ----> A + B'------> A'

                  |
         c
      B --> A --> F --> O

    b |   a |     |

      B'--> A'--> F'--> O
         c'

The exactness of  K'--> K --> F --> F' says, precisely, that the
kernel of  F --> F' is the cokernel of  K'--> K.

The middle vertical  K --> F  is  K --> A + B'--> A  --> F  where
A + B'--> A  is the projection map. (For present purposes a projection
map is one given by a matrix in which each entry is  1  if such fits
else  0.)

The two top-right vertical maps are also projection maps.

The two top-right horizontal maps are given (as luck would have it) by
matrices in which each entry is  1  or a single letter (a,b,c,c') if
such fits, else  0.

It's not hard to verify this lemma in the category of abelian groups,
which -- given the exact representation theorem -- suffices for all
abelian categories. As if often the case, the proof of the dual of the
lemma harder in the concrete case:

CO-LEMMA: In an abelian category an exact diagram of the form (all
          vertical arrows are downwards):

                                         c'
                          O --> K'--> A'--> B'

                                |     | a   | b

                          O --> K --> A --> B
                                         c

          may be enlarged to an exact diagram of the form:

                                         c'
                          O --> K'--> A'--> B'

                                |     | a   | b

                          O --> K --> A --> B
                                         c
                                |

          A'------> A + B'----> F --> O

          |           |         |

        A'+ A --> A + B'+ B --> F'--> O


For the record, note that "a kernel of a map between kernels is a
kernel," that is, the same initial exact diagram may be enlarged to a
diagram:with exact rows and exact left column:

                            O

                            |

                     O ---> K''--> A'--> A + B;

                            |      |       |
                                       c'
                     O ---> K'---> A'----> B'

                            |      | a     | b

                     O ---> K ---> A ----> B
                                       c

The two upper-right verticals are projection maps and the upper right
horizontal map is given by a matrix in which each entry is a single
letter (a,c').

Also for the record: the finite bi-completion of an additive
category is not abelian. Start with the additive closure of a single
map  f:A --> B. In the finite bi-completion the canonical map from
Cok(Ker(f))  to  Ker(Cok(f))  will not be an isomorphism. In the
finite _abelian_ finite bi-completion, of course, it must be. (It is
precisely this isomorphism, invoked at the end of the Lemma in the
case  f  =  K --> F, that delivers the commutativity of finite left
and right completions.)