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