# split exact complexes and biproducts

I have an split exact complex in an abelian category, that is, a chain complex which is exact and maps $s_n : C_n \to C_{n+1}$ st. $dsd = d$. I would like to prove that this implies that $C_n \cong B_n \oplus B_{n-1}$.

I would like to do this by showing that $d_{n+1} s_n + s_{n-1}d_n = id_{C_n}$. Is this even true? If it is how do I prove it?

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Yes, a chain complex $C_\bullet$ is split exact if and only if the identity map of the complex is null-homotopic. Take a look here –  Andrea Gagna Feb 13 '13 at 23:39
yes. but can you use $s_n$ for the chain homotopy, rather than defining $t_n$? –  krey Feb 14 '13 at 12:13
Oh, sorry! No, it is false in general. If you can wait till the end of the next week, I promise I will write a (class of) easy conterexample(s) and a sufficient condition for that to work, with a good amount of details. –  Andrea Gagna Feb 14 '13 at 17:25