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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.

Xiao-Gang Wen
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    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. – Fernando Muro Oct 02 '13 at 14:58
  • 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. – Xiao-Gang Wen Oct 02 '13 at 15:50
  • So $\mathbb N$ is an abelian subgroup of $\mathbb Z$? – Fernando Muro Oct 02 '13 at 21:09
  • I see your point. But here we assume img$(d_n)$ to be an Abelian group. I modified my question to add this assumption. – Xiao-Gang Wen Oct 02 '13 at 23:52
  • I don't see the assumption in your question. – Fernando Muro Oct 02 '13 at 23:54
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    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. – Fernando Muro Oct 02 '13 at 23:57
  • both kernels and images of $d_n$ are groups implies that all the monoids $M_n$ are groups. This will be a great result for me. I can indeed show that both kernels and images are groups and I want to get the usual chain complex. But is the above statement really true? How to prove it? It is not obvious to me. (I am a physicist.) – Xiao-Gang Wen Oct 03 '13 at 00:10
  • A proof is given below. I wish that I had the intuition to see the theorem: if the kernel and the image have a property, then the original set also has the property – Xiao-Gang Wen Oct 04 '13 at 13:38
  • You should look at the papers of Alex Patchkoria listed on MathSciNet for another approach to using abelian modules: he uses abelian semimodules. I might say more later. – Ronnie Brown Oct 09 '13 at 13:57
  • @Ronnie I am also interested in a generalization of cochain complex where $M_n$ are commutative monoids (without the condition that kernels and images of $d_n$ are Abelian groups). This question is asked in http://mathoverflow.net/questions/430/homological-algebra-for-commutative-monoids. But I did find simple reviews there. – Xiao-Gang Wen Oct 10 '13 at 00:01

1 Answers1

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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.

Proof:

  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 !

Xiao-Gang Wen
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