Re: Question on exact sequence

[Michael Barr <barr@math.mcgill.ca>]

I think I have now come to understand this, at least in part.  A couple
things you said triggered this realization.  First the point that this
was taking place in the arrow category.  Second that maybe sometimes you
needed a map to get the connecting homomorphism and sometimes not.

The first point made me think maybe rather than the arrow category I
should perhaps be thinking about the category of chain complexes.  The
second somehow made me think about the difference between homology and
homotopy equivalences.

Perhaps this explanation amounts to shooting flies with elephant guns,
but it satsifies me that it has fully explained things.  Let me start by
discribing an important difference between the two situations.  Both
diagrams:

0 --> A --> B --> C --> 0.......A --> A --> B
......|.....|.....|.............|.....|.....|
......|.....|.....|.............|.....|.....|
......|.....|.....|.............|.....|.....|
......v.....v.....v.............v.....v.....v
0 --> A'--> B'--> C'--> 0.......B --> C --> C

with maps as in earlier posts, give 6 term exact sequences, but the
second continues continues to do so when you apply a homfunctor Hom(D,-)
or Hom(-,D), which is not generally true of the first.  Now this, had I
noticed it earlier, would have immediately put me in mind of the
difference between homotopy and homology.  It is a fact that a homology
equivalence between two chain complexes is a homotopy equivalence iff it
remains an equivalence if you apply any covariant (or any contravariant)
homfunctor.  This fact, along with a couple other things I will mention
below, is proved somewhere in "Acyclic Models".  Next, when I drew the
diagram

0 ---> A ---> A + B ---> B ---> 0
.......|........|........|.......
.......|........|........|.......
.......|........|........|.......
.......|........|........|.......
.......v........v........v.......
0 ---> B ---> B + C ---> C ---> 0

which was marked by a peculiar appearance of - signs, I recognized that
it looked like a mapping cone of something and, had I only worked out of
what, I might have realized sooner what was going on.  It is actually
the mapping cone of the map

0 ---> A
|......|
|......|
|......|
v......v
B ---> B
|......|
|......|
|......|
v......v
C ---> 0

Finally, the observation that the chain complex 0 ---> A ---> C ---> 0
is homotopic to 0 ---> A + B ---> B + C ---> 0 is a complete triviality.

So what is the general situation.  Let me raise the question in this
form.  Suppose f: C' ---> C and g: C ---> C'' is a map of chain
complexes.  When can we expect an exact triangle

  H(C') ------> H(C)
...^............./
....\.........../
.....\........./
......\......./
.......\...../
........\.../
.........\.v
.........H(C'')

Clearly we need something to induce the map H(C'') ---> H(C').  The
obvious thing would be a map S(C'') ---> C' (S is the suspension
functor).  The example of two-length sequences shows that this is too
much to hope for.  It looks like what you need is a relation from S(C'')
--> C' that induces, somehow a map on homology.  This assumption holds
for the case 0 --> C' --> C --> C' --> 0 and also holds for the original
curious sequence (the suspension is what saves the day here, as seen
above).  Is there a better way to express this?  I don't see one.
Perhaps the last word on this hasn't been said yet.

Another question I haven't answered but should be doable is whether the
existence of a relation S(C'') - - - -> C' that you assume induces a
morphism on homology is sufficient to make the triangle exact.  Beyond
this, I have some inchoate ideas that I will explore.  Clearly there has
to be some compatibility between R and f and g.

As I said, this explanation is perhaps a little heavy but I know no
simpler one.

Michael




[For admin and other information see: http://www.mta.ca/~cat-dist/ ]