I am reading "an introduction to homological algebra by Charles A.Weibel" and the author deifnes split exact complex to be exact complex $\{C_n,d_n:C_n\to C_{n-1}\}$ such that there exists a sequence of maps $s_n:C_n\to C_{n+1}$ such that $d_{n}s_{n-1}d_{n}=d_{n}.$ He then claims that this is equivalent to the condition $ds+sd=id$. I can see why $ds+sd=id$ implies that the complex is split exact. I tried to prove the other direction but I dont see why it is true. So I Tried to write $C_n=Ker(sd)+Ker(sd-Id)$ which holds because $(sd)^2=sd$ then if $x\in Ker(sd)$ then I can show that $sdx+dsx=x$ but if $x\in Ker(sd-id)$ I cant prove that $dsx=0$. I would appreciate any help.
[Math] Definition of exact split complex
homological-algebra
Best Answer
Let $B_n$ denote the boundaries in $C_n$. The previous exercise implies that the complex $C_*$ is isomorphic to the complex with degree $n$ term $B_n\oplus B_{n-1}$. The map $B_n\oplus B_{n-1}\to B_{n-1}\oplus B_{n-2}$ is the map that sends $B_n$ to zero and $B_{n-1}$ to itself.
In this form, the maps $s_n:B_n\oplus B_{n-1}\to B_{n+1}\oplus B_n$ are just $(b,c)\mapsto (0,b)$. You can easily check that $1 = ds + sd$.