It appears that we can generalize cochain complex to quasi-cochain complex, that still allow us to define cohomology.

Definition: A quasi-cochain complex is a sequence of commutative monoids $M_n$ connected by monoid-homomorphisms $d_n$: \begin{align} \cdots \overset{d_{n-1}}{\rightarrow} M_n \overset{d_n}{\rightarrow} M_{n+1} \overset{d_{n+1}}{\rightarrow} M_{n+2} \overset{d_{n+2}}{\rightarrow} \cdots , \end{align} such that $d_{n+1}d_n$ maps $M_n$ to $0_{n+2}$ (the identity in $M_{n+2}$), the subset of $M_n$, $A_n=\{a_n|d_n(a_n)=0,a_n\in M_n\}$ is an Abelian group, --Edit-- and the Img$(d_n)$ is also an Abelian group.

In the quasi-cochain complex, we can define the cohomology classes since both $\text{Ker}(d_n)$ and $\text{Img}(d_{n-1})$ are Abelian groups: $H^n=\text{Ker}(d_n)/\text{Img}(d_{n-1})$.

I wonder

(1) if the above definition is OK

(2) Has any one studied such a quasi-cochain complex.

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    $\begingroup$ I don't see why the image should be an abelian group. Also, a map between different monoids cannot be the identity, I bet you mean zero. $\endgroup$ Oct 2, 2013 at 14:58
  • $\begingroup$ Since $d_{n+1}d_n$ maps $M_n$ to the identity of $M_{n+2}$, img$(d_n)$ is a subset of ker$(d_{n+1})$. ker$(d_{n+1})$ is an Abelian group, and thus img$(d_n)$ is an Abelian subgroup. $\endgroup$ Oct 2, 2013 at 15:50
  • $\begingroup$ So $\mathbb N$ is an abelian subgroup of $\mathbb Z$? $\endgroup$ Oct 2, 2013 at 21:09
  • $\begingroup$ I see your point. But here we assume img$(d_n)$ to be an Abelian group. I modified my question to add this assumption. $\endgroup$ Oct 2, 2013 at 23:52
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    $\begingroup$ BTW, if you assume that both kernels and images are groups then so are all the monoids, so you end up with a chain complex. $\endgroup$ Oct 2, 2013 at 23:57

1 Answer 1


Kong Liang told me a proof for the statement if both kernels and images of $d_n$ are Abelian groups then all the monoids $M_n$ are Abelian groups.

Let $f: A \to B$ be a surjective morphism between two commutative monoids. Surjectivity means $B =$ Img$(f)$. Assume that Ker$(f)$ is an Abelian group and $B =$ Img$(f)$ is an Abelian group.

Then $A$ is automatically an Abelian group.


  1. it is enough to show that any element $a$ in $A$ has a right inverse $a'$, i.e. $a a' = 1$. (by the commutativity, $a' a = a a' = 1$.) Notice that if such $a'$ exists, it must be unique. Otherwise, let $a'$ and $a''$ be such that $a a' =1 = a a''$. Then we have $a' = a' a a'' = a''$ (use commutativity)

  2. For any $a$ in $A$, let $b$ be the inverse of $f( a )$ and let $c$ be an element in $A$ such that $f( c ) = b$. Then we have $f(a c) = f( a ) f( c ) = f( a ) b =1$. Therefore, $a c$ is in the Ker$(f)$. Since Ker$(f)$ is an Abelian group, there is an element $d \in A$ such that $acd = 1$. Hence $cd$ is the right inverse of $a$.

Therefore the quasi-cochain complex is the usual cochain complex, and there is a lot of work on cochain complex, as well as a lot of theorems. :-) Thanks, @ Fernando Muro !

  • $\begingroup$ You're welcome. I'd say this is rather elementary from a mathematical point of view, though. $\endgroup$ Oct 3, 2013 at 11:10
  • $\begingroup$ Indeed, it is elementary. But the result is very important to me. $\endgroup$ Oct 3, 2013 at 11:16

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