# Definition of split chain complex

I'm trying to reconcile two definitions in homological algebra. One of them is standard and the other is found in Weibel's Homological Algebra (so probably also standard, but I can't say from my own experience.)

First, an exact sequence $0\to A\to B\to C\to 0$ in an abelian category is split if any of the following equivalent conditions hold:

1. the map $A\to B$ has a section,
2. the map $B\to C$ has a retract,
3. $B=A\oplus C'$ for some subobject $C'$ of $B$, and
4. $B=A'\oplus C$ for some quotient object $A'$ of $B$.

Second, Weibel defines a chain complex $C_\bullet$ to be split if there are maps $s_n:C_n\to C_{n+1}$ such that $d_{n+1}s_nd_{n+1}=d_{n+1}$. Such a chain complex has very nice properties: the maps $s_n$ induce splittings of $C_n$ and $Z_n(C_\bullet)$.

Although Weibel's definition has nice properties, it comes from nowhere. Is there some way to view his definition as a particular case of the first? It seems like if you just constructed the right short exact sequence of chain complexes, perhaps something involving a shifted complex like $C[-1]_\bullet$, then these two definitions could be reconciled. I can't see how to do it, though.

Say you have a chain complex $$\cdots\xrightarrow{\ \ \ }C_{n+2}\xrightarrow{\ d_{n+2} \ }C_{n+1}\xrightarrow{\ d_{n+1} \ }C_n\xrightarrow{\ d_n \ }C_{n-1}\xrightarrow{\ d_{n-1} \ }C_{n-2}\xrightarrow{\ \ \ }\cdots$$ which is split. Then there exist maps $\{s_n:C_n\to C_{n+1}\}$ satisfying $d_n\circ s_{n-1}\circ d_n=d_n$ for all $n$. Now, suppose that $C_{n+2}=C_{n-2}=0$, and we are exact at $n-1$, $n$, and $n+1$. Then, we have a short exact sequence: $$0\xrightarrow{\ \ \ }C_{n+1}\xrightarrow{\ d_{n+1}\ }C_n\xrightarrow{\ d_n \ }C_{n-1}\xrightarrow{\ \ \ }0.$$ Since $d_{n+1}\circ s_n\circ d_{n+1} = d_{n+1}$ and $d_{n+1}$ is injective, this implies that $s_n\circ d_{n+1}=id_{C_{n+1}}$, which gives the first condition. Similarly, we have $d_n\circ s_{n-1}\circ d_n=d_n$, and as $d_n$ is surjective, it has a right-inverse, and composing on the right with this inverse implies $d_n\circ s_{n-1}=id_{C_{n-1}}$, giving the second condition. Lastly, it can be checked that the map $$\phi:C_n\to C_{n+1}\oplus C_{n-1},\quad \phi(x) = (s_n(x),d_n(x))$$ is an isomorphism. I'll let you workout the other two equivalent statements.
Now as you said, Weibel's definition of a split chain complex doesn't come with much motivation so I'll briefly motivate that as well. As explained in this question the condition $d_n=d_n\circ s_{n-1}\circ d_n$ for all $n$ is equivalent to $id = d_{n+1}\circ s_{n}+s_{n-1}\circ d_n$ for all $n$. So, being split exact is analogous to the topological property of being contractible, since the condition $id = d_{n+1}\circ s_{n}+s_{n-1}\circ d_n$ is really just saying that the identity map is null-homotopic. This can be made even more concrete using the ideas in the question: Can we think of a chain homotopy as a homotopy?.
• I think the equivalence you mention at the end is not true; it is true though if you can pick different functions $s$ for each condition. If the complex is split thanks to "sections" $s$ then $h:=s-\partial ss$ can serve as a nullhomotopy. Sep 24, 2020 at 15:04