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.


1 Answer 1


I think it's more natural to view the definition of a split short exact sequence as a special case of the definition of a split chain complex.

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?.

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    $\begingroup$ 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. $\endgroup$
    – elidiot
    Sep 24, 2020 at 15:04
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    $\begingroup$ It is not necessarily true. To get ds + sd = id you need the complex to be split exact. Just split doesn't guarantee that. $\endgroup$ Nov 3, 2020 at 12:56

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