AB4.5

AB4.5

[Michael Barr]

The file ab45.zip, that includes both the dvi and ps version, is now
available at ftp.math.mcgill.ca/pub/barr.  It is a preliminary version
because I want to investigate if the condition is really stronger than
AB4.  The following consideration raises the possibility that it might not
be.  One place to look to distinguish AB4* from AB4.5* might be in the
abelian groups of a topos.  But in order for the abelian groups of a topos
to satisfy AB4*, you need AC (in the topos).  But AC implies dependent
choice and then it looks like you have AB4.5*, which really is just like
dependent choice.

Michael

AB4.5

[Michael Barr]

A day after I sent my previous note, I realized the (or an) answer.  In
fact, I seem to have discovered a new infinite exactness condition that
is between AB4 and AB5 and, inevitably, I call AB4.5.  So far I know
only that it is implied by AB5, implies AB4 and, since module categories
satisfy both AB5 and AB4.5*, it is definitely weaker than AB5.  I do
not, at this time, know that it is stronger than AB4.

Here is how it works.  Call a poset superdirected if it is directed and
all down segments are finite.  The natural numbers is one example and
the set of finite subsets of any set is another.  If I is superdirected,
A an abelian category and F,G:  I --> A are two diagrams, say that a
natural transformation alpha:  F --> G is supernatural (OK, suggestions
for better names welcome) if for any i in I, the following:  Look at the
diagram consisting of all the Fj for j =< i and all the Gj for j < i.
The arrows in the diagram are all the alpha j, for j =< i, as well as
all the arrows from the restrictions of F and G. The diagram has a
cocone to Gi and that cocone is monic (meaning the natural map from the
colimit to Gi is monic) for each i in I, call alpha supernatural.  Now
the condition is that whenever alpha is supernatural, the induced colim
F --> colim G is monic.

To see that this implies AB4, suppose for all x in X, A_x >--> B_x.  Let
I consist of all the finite subsets of X.  For i in I, let Fi be the
direct sum of all the A_x, x in i and Gi be the direct sum of the
correspnding B_x.  The supernaturality is satisfied and hence the sum of
the A_x maps monically to the sum of the B_x.

The application to my question comes as follows.  To recall, I am
supposing that
       del            del
  ... -----> (C_n,d) -----> (C_{n-1},d) ---> ... ---> (C_0,d) --> 0
 is a chain complex of differential abelian groups, except that we
assume d.del = -del.d.  Suppose (C_n,d) is exact for each n. Let C be
the direct sum of the C_n with boundary given by the matrix
          (   d     0     0   ...    )
          (  del    d     0   ...    )
          (   0    del    d   ...    )
          (   .     .     .   .      )
          (   .     .     .    .     )
          (   .     .     .     .    )
 The finite truncations F^n = C_0 +...+ C_n are readily
shown to be exact:  F^0 = C_0 is by hypothesis and there is an exact
sequence     0 --> F^{n-1} --> F^n --> C_n --> 0.  Thus each F^n is
exact.  Now let Z^n and B^n be the kernel and image, resp. of the
boundary operator on F^n. Then the rows of
       0 ----> Z^n -------> F^n -------> B^n ----> 0
                |            |            |
                |            |            |
                |            |            |
                v            v            v
       0 --> Z^{n+1} ---> F^{n+1} ---> B^{n+1} --> 0
 Since the right hand vertical arrow is monic, it follows that the left
hand square is a pullback, which in turn implies that the induced map
F^n +_{Z^n} Z^{n+1} ---> F^{n+1} is monic.  Taking colimits, and using
this Ab4.5 condition together with the right exactness of
colimits implies that 0 --> colim Z^n --> C --> B --> 0 is exact, so
that Z = colim Z^n and then B^n = Z^n for all n implies B = Z.

As for AB4.5* on module categories (for which it is sufficient to prove
it for Ab), it is a bit complicated notationally for this medium, but
not hard.  The point is that when you try to construct an element in the
inverse limit, the crucial thing is that you can do it one index at a
time and no choice you make at an earlier stage ever has to be changed
at a later one, so you can just continue.  Of course, AC is involved.
This might a way to find a category that satisfies AB4*, but not AB4.5*;
the trouble is that AC is also used to show AB4*.

Michael