Homology question
Suppose you have an exact sequence of chain complexes 0 --> C' --> C' + C --> C -->0 in which the middle term has boundary operator given by the matrix [ d -f ] [ 0 d ] for some f: C --> C'. This is more or less the mapping cone of f. Suppose in addition that f(Z(C)) is included in B(C'). It then follows immediately that 0 --> H(C') --> H(C' + C) --> H(C) --> 0 is exact. Is it split? The answer is yes if this is a sequence of abelian groups (or modules over a hereditary ring) and C is projective. Anyone know of a counter-example? Michael 29-Apr-2002 16:12:19 -0300,2233;000000000000-00000000
I know a topological example. Let X be a connected Hausdorf compact space and let a_1, , a_2 be two points from X. Let C,C' be complexes of cochains of Alexander-Spenier of X and a point respectively. Let f:C --> C' be a difference f_1-f_2 , where f_1,f_2 are induced by the inclusions i_1,i_2 of the points a_1,a_2 to X. Then a result is: 0 --> H(C') --> H(C' + C) --> H(C) --> 0 splits if and only if i_1,i_2 induce equal homomorphisms in Steenrod homology of X. This is in my paper Batanin M.A., Mappings of spectral sequences and the generalized homotopy axiom, Siberian Math. Journal, 5, (1987), 22-31. An example when i_1,i_2 induce different homomorphisms was constructed by U.Karimov in "On generalised homotopy axiom" , Reports of Tadjikistan Academie of Science, v.XXII,9, 521-524.,1979. In this example X is a 2-adic solenoid i.e. inverse limit of S^1 <--- S^1 <--- .... where all the maps are two-fold coverings and a_1, a_2 are in different linear connected components. Hope this will help. Michael Batanin. on 26/4/02 11:51 PM, Michael Barr at barr@barrs.org wrote:
Suppose you have an exact sequence of chain complexes 0 --> C' --> C' + C --> C -->0 in which the middle term has boundary operator given by the matrix [ d -f ] [ 0 d ] for some f: C --> C'. This is more or less the mapping cone of f. Suppose in addition that f(Z(C)) is included in B(C'). It then follows immediately that 0 --> H(C') --> H(C' + C) --> H(C) --> 0 is exact. Is it split?
The answer is yes if this is a sequence of abelian groups (or modules over a hereditary ring) and C is projective. Anyone know of a counter-example?
Michael
29-Apr-2002 16:12:19 -0300,1369;000000000000-00000000
Michael Barr <barr@barrs.org> writes:
Suppose you have an exact sequence of chain complexes 0 --> C' --> C' + C --> C -->0 in which the middle term has boundary operator given by the matrix [ d -f ] [ 0 d ] for some f: C --> C'. This is more or less the mapping cone of f. Suppose in addition that f(Z(C)) is included in B(C'). It then follows immediately that 0 --> H(C') --> H(C' + C) --> H(C) --> 0 is exact. Is it split?
Unless I'm misunderstanding something, this is false, even for complexes of free abelian groups. Let C = Z -----> Z, where d is multiplication by 2, and let C' = \Sigma C. Let f : C --> C' be the map representing the non-zero element of Ext_Z(Z/2,Z/2). Then H(Cone(f)) = Z/4. Some additional notes that may (or may not) be of interest. 1) If the sequence splits when C is projective, then it splits for all C, since we can take a cofibrant replacement D ---> C for C. (In particular, D is a complex of projectives and D ---> C is a quasi-isomorphism.) 2) Over a PID (or any hereditary ring, I think), the sequence of homology complexes is split iff the map f is zero in the derived category. Sketch: If f is zero in the derived category, then the triangle C' --> Cone(f) --> C splits in the derived category, so the sequence of homology complexes splits. For the converse, note that over a PID every complex D is quasi-isomorphic to the graded module H(D). So our original triangle is quasi-isomorphic to a triangle H(C') --> H(Cone(f)) --> H(C) (*) where the maps here may only exist in the derived category (ie. there may be some Ext components). If the sequence (*) splits just as graded modules (ignoring any Ext components), then the composite of the splitting H(C) --> H(Cone(f)) with the map H(Cone(f)) --> H(C) induces the identity in homology, and so is an isomorphism. Thus H(Cone(f)) --> H(C) is an epimorphism in the derived category, and so f is zero. Dan 30-Apr-2002 18:47:57 -0300,2198;000000000000-00000000
People interested in splitting in homology etc might be interested in 3. ``Cohomology with chains as coefficients'', {\em Proc. London Math. Soc.} (3) 14 (1964), 545-565. which gives examples of non naturality of various splittings such as in universal coefficient theorems. Ronnie Brown Dan Christensen wrote:
Michael Barr <barr@barrs.org> writes:
Suppose you have an exact sequence of chain complexes 0 --> C' --> C' + C --> C -->0 in which the middle term has boundary operator given by the matrix [ d -f ] [ 0 d ] for some f: C --> C'. This is more or less the mapping cone of f. Suppose in addition that f(Z(C)) is included in B(C'). It then follows immediately that 0 --> H(C') --> H(C' + C) --> H(C) --> 0 is exact. Is it split?
Unless I'm misunderstanding something, this is false, even for complexes of free abelian groups. Let C = Z -----> Z, where d is multiplication by 2, and let C' = \Sigma C. Let f : C --> C' be the map representing the non-zero element of Ext_Z(Z/2,Z/2). Then H(Cone(f)) = Z/4.
Some additional notes that may (or may not) be of interest.
1) If the sequence splits when C is projective, then it splits for all C, since we can take a cofibrant replacement D ---> C for C. (In particular, D is a complex of projectives and D ---> C is a quasi-isomorphism.)
2) Over a PID (or any hereditary ring, I think), the sequence of homology complexes is split iff the map f is zero in the derived category.
Sketch: If f is zero in the derived category, then the triangle C' --> Cone(f) --> C splits in the derived category, so the sequence of homology complexes splits.
For the converse, note that over a PID every complex D is quasi-isomorphic to the graded module H(D). So our original triangle is quasi-isomorphic to a triangle
H(C') --> H(Cone(f)) --> H(C) (*)
where the maps here may only exist in the derived category (ie. there may be some Ext components). If the sequence (*) splits just as graded modules (ignoring any Ext components), then the composite of the splitting H(C) --> H(Cone(f)) with the map H(Cone(f)) --> H(C) induces the identity in homology, and so is an isomorphism. Thus H(Cone(f)) --> H(C) is an epimorphism in the derived category, and so f is zero.
Dan
-- Professor Emeritus R. Brown, School of Informatics, Mathematics Division, University of Wales, Bangor Dean St., Bangor, Gwynedd LL57 1UT, United Kingdom Tel. direct:+44 1248 382474|office: 382681 fax: +44 1248 361429 World Wide Web: home page: http://www.bangor.ac.uk/~mas010/ (Links to survey articles: Higher dimensional group theory Groupoids and crossed objects in algebraic topology) Raising Public Awareness of Mathematics CDRom Version 1.1 http://www.bangor.ac.uk/~mas010/CDadvert.html Symbolic Sculpture and Mathematics: http://www.cpm.informatics.bangor.ac.uk/sculmath/ Centre for the Popularisation of Mathematics http://www.cpm.informatics.bangor.ac.uk/
participants (4)
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Dan Christensen -
Michael Barr -
Michael Batanin -
Ronnie Brown