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