Re: Question on exact sequence

[Clemens.BERGER@unice.fr]

Marco Grandis wrote:
> Dear Michael,
>
> The following lemma extends both results.
>
> We have a sequence of consecutive morphisms indexed on the integers
>
>     ...  ---->   An  ---->  An+1  ---->  An+2  ---->  ...
>
> (if your sequence is finite, you extend by zero's). Call  f(n,m)  the
> composite
> from  An  to  Am  (n < m).
>
> Then, writing  H/K  a subquotient  H/(H intersection K),
> there is an unbounded exact sequence of induced morphisms:
>
>    ...  ---->   Ker f(n,n+2) / Im f(n-1,n)
>        ---->   Ker f(n+1,n+2) / Im f(n-1,n+1)
>        ---->   Ker f(n+1,n+3) / Im f(n,n+1) ---->  ...
>
> where morphisms are alternatively induced by an 'elementary' morphism
> (say An  -->  An+1) or by an identity.
>
> At each step, one increases of one unit the first index in the numerator
> and the second index in the denominator, or the opposite
> (alternatively); after
> two steps, all indices are increased of one unit, and we go along in
> the same way.
>
> -  Your lemma comes out of a sequence  A ----> B ----> C  (extended
> with zeros).
>
> -  Snake's lemma, with your letters, comes out of a sequence of three
> morphisms
> whose total composite is 0
>
>                           A ----> B ----> B' ----> C
> taking into account that  A' = Ker(B' ----> C')  and  C = Cok(A ---->
> B).
>
> I like your lemma (and the Snake's). The form above does not look
> really nice.
> Perhaps someone else will find a nicer solution?
>
> However, if one looks at the universal model of a sequence of
> consecutive morphisms,
> in my third paper on Distributive Homological Algebra, Cahiers 26,
> 1985, p.186,
> the exact sequence above is obvious. (Much in the same way as for the
> sequence of
> the Snake Lemma, p. 188, diagrams (10) and (11).) This is how I found
> it.
>
> Best wishes
>
> Marco
>
Dear Michael and Marco,

   as addition to Marco's answer, I would propose the following
notation: for any pair of composable arrows f:A-->B and g:B-->C denote
by H(g,f) the cokernel of ker(gf)-->ker(g), or (what amounts to the same
in an abelian category),  the kernel of coker(f)-->coker(gf). Thus the
object H(g,f) is precisely the one which allows one to glue together the
short exact sequences 0-->ker(f)-->ker(gf)-->ker(g) and
coker(f)-->coker(gf)-->coker(g)-->0.

   If gf=0 then H(g,f)=ker(g)/im(f) is precisely the homology object at
B, but as Ross already mentioned, this object is well defined for any
composable pair of arrows. Now, if we have three composable arrows
f:A-->B, g:B-->C, h:C-->D, then there is a  4-term exact sequence
0-->H(g,f)-->H(hg,f)-->H(h,gf)-->H(h,g)-->0. The snake lemma can be
derived from the special case hgf=0 of this 4-term exact sequence, where
g corresponds precisely to the middle vertical arrow of Michael's diagram.

It is interesting to observe that a proof (without elements !) of this
4-term exact sequence uses some nice composition properties of
Hilton-exact squares (a common generalization of pullback and pushout
squares). Some more details can be found at pg. 24 of
http://math.unice.fr/~cberger/structure1.

Several questions arise naturally: this 4-term exact sequence looks like
a "cocycle". Are there generalizations to n composable arrows ? What
about generalizations to non-abelian categories ?

All the best,
                         Clemens.


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