From mboxrd@z Thu Jan 1 00:00:00 1970 X-Msuck: nntp://news.gmane.io/gmane.science.mathematics.categories/5230 Path: news.gmane.org!not-for-mail From: Michael Barr Newsgroups: gmane.science.mathematics.categories Subject: Re: Question on exact sequence Date: Tue, 10 Nov 2009 11:15:02 -0500 (EST) Message-ID: Reply-To: Michael Barr NNTP-Posting-Host: lo.gmane.org Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII; format=flowed X-Trace: ger.gmane.org 1257953568 29556 80.91.229.12 (11 Nov 2009 15:32:48 GMT) X-Complaints-To: usenet@ger.gmane.org NNTP-Posting-Date: Wed, 11 Nov 2009 15:32:48 +0000 (UTC) To: Marco Grandis , categories@mta.ca Original-X-From: categories@mta.ca Wed Nov 11 16:32:42 2009 Return-path: Envelope-to: gsmc-categories@m.gmane.org Original-Received: from mailserv.mta.ca ([138.73.1.1]) by lo.gmane.org with esmtp (Exim 4.50) id 1N8FBU-0000Rd-GC for gsmc-categories@m.gmane.org; Wed, 11 Nov 2009 16:32:36 +0100 Original-Received: from Majordom by mailserv.mta.ca with local (Exim 4.61) (envelope-from ) id 1N8EZc-0004sm-T7 for categories-list@mta.ca; Wed, 11 Nov 2009 10:53:28 -0400 Original-Sender: categories@mta.ca Precedence: bulk Xref: news.gmane.org gmane.science.mathematics.categories:5230 Archived-At: That is really fascinating. I assume that when you wrote A --> B --> B' --> C, that last should have been C'. Here is how it arose. Call a map A --> B cotorsion if its cokernel is torsion (for a generalized notion of torsion). Assuming that torsion objects are closed under extension (and that a quotient of torsion is torsion) this allows a proof that a composite of cotorsion maps is cotorsion. Anyway, thanks. Michael On Tue, 10 Nov 2009, 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 > > [For admin and other information see: http://www.mta.ca/~cat-dist/ ]