I think I have tracked my question to its lair. I will omit the details since most of you will not care and the remaining two or three will be able to readily fill them in. As George suggested, if you have a map f: C' ---> C of complexes you get an exact sequence 0 ---> C ---> C_f ---> SC' ---> 0 in which SC' is just like C' except the index on each term is increaed by 1 and the boundary operated d' replaced by -d'. As for the mapping cone C_f, the nth term is C_n + C'_{n-1} and the boundary operator is given by the matrix (d f_{n-1};0 -d). So to get back to my original question, given the diagram f_1 g_1 C'_1 --------> C_1 --------> C"_1 .|..............|.............| .|..............|.............| .|..............|.............| .|d'............|d............|d" .|..............|.............| .|..............|.............| .v.....f_0......v.....g_0.....v C'_0 --------> C_0 --------> C"_0 (I have changed notation in order to be compatible with thinking of them as complexes), when is there an exact snake. What I will do will be the case in which the snake starts with a monic and ends with an epic (so 0s to be added at the ends). Clearly a sufficient condition to get a snake (I don't know if it is necessary, but I doubt it) is that C_f be homologous to C". (C_g being homologous to C' would work equally well.) Even stating conditions on which two complexes are homologous is problematic. For example a complex is homologous to its homology sequence with null boundary, but that is certainly not induced by an map. But it might be induced by some relation; I haven't thought on that. See my paper in TAC, Vol. 16, No. 7 to see that I could not give a satisfactory answer to the question: when does an additive functor between abelian categories preserve homology? The mapping cone sequence is the following. .........................................1 ...........................A'_1 -------------------> A'_1 ............................|.........................| ............................|.........................| ............................|(f_1)....................| ............................|(d' )....................|d' ............................|.........................| ............................|.........................| ...........(1)..............|.........................| ...........(0)..............v..........(0 -1).........v A_1 ------------------> A_1 + A'_0 ----------------> A'_0 .|..........................| .|..........................| .|..........................|(d f_0) .|d.........................| .|..........................| .|..........................| .v..........1...............v A_0 ---------------------> A_0 so my original question boils down to whether the chain complex ..............(f_1) ..............(-d')................(d f_0} 0 -----> A'_1 -------> A_1 + A'_0 ----------> A_0 -----> 0 and d" ...........0 -----------> A"_1 -------------> A"_0 ----> 0 are homologous. A trivial computation should convince you that there is no homomorphism of complexes in either direction. But there is a canonical relation R in the following diagram ..............(-d')................(d f_0} 0 -----> A'_1 -------> A_1 + A'_0 ----------> A_0 -----> 0 ...........................|...................| ...........................|...................| ...........................|...................| ...........................|R..................|g_0 ...........................|...................| ...........................|...................v ...........................v........d" ...........0 -----------> A"_1 -------------> A"_0 ----> 0 Where R = {(a_1,a'_0,a'_1,a"_1) \in A_1 x A'_0 x A'_1 x A"_1 | g_1a_1 = g_1f_1a'_1 = a"_1 & d'a'_1 = a_0'}. It can be thought of as the matrix (g_1 g_1f_1(d_')^{-1}). And, mirabile dictu, this relation actually induces a homomorphism of homology groups! So this is the answer to my question. Incidentally in the original case of f : A --> B and g: B ---> C giving A --> A --> B |.....|.....| |.....|.....| |.....|.....| v.....v.....v B --> C --> C The mapping cone sequence of the left hand complex is readily seen to be homomopic to the right hand one. The right hand one is a retract of the mapping cone and the other composite is homomotopic to the identity. I would like to end this thread with the observation that I agree completely with George's perception that the difference between homology and homotopy is just the that between a functor being an equivalence and being full, faithful, and representative. Here is another example of the same phenomenon. If a complex has null homology, then for each cycle z there must by an element c(z) such that dc(z) = z. Jon Beck called this a "cycle operator" and the existence of a cycle operator is equivalent to the complex being exact. But if (and only if) the cycle operator can be chosen as a morphism in the category, the complex is homotopic to the null sequence (that is, is contractible). Michael [For admin and other information see: http://www.mta.ca/~cat-dist/ ]