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I have $A$ and $B$ two graded vector spaces, and $D: A \oplus B \to A \oplus B$ with $D(a + b) = d_0(a) + d_1(a) + d_0(b)$ for $a \in A$ and $b \in B$, where $d_0 : A \to A$, $d_0 : B \to B$ and $d_1 : A \to B$ all of degree +1 (cohomological convention) such that $D^2 = 0$.

This implies that $d_0^2 = 0$ and $d_1$ is a chain map for $d_0$ up to sing. So it makes sense to talk about $d_1^\bullet = H^\bullet(d_1) : H^\bullet(A,d_0) \to H^\bullet(B,d_0)$.

My question is: is it true that $H^\bullet(A\oplus B, D) \simeq ker(d_1^\bullet) \oplus coker(d_1^\bullet)$ as graded vector spaces?

I found a particular instance of this statement while reading this article http://arxiv.org/abs/1411.1685, equation 4.3, and was wondering if it was true in general or if it was true for the particular cochain complexes they are dealing with.

I thaught it would be easy, at least to find a reference that does it, but I had hard time trying...

Najib Idrissi
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jeanmfischer
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1 Answers1

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The complex $(A\oplus B,D)$ is the mapping cone of the chain map $$d_1: (A[-1],d_0)\to(B,d_0),$$ and the fact you want follows from the resulting long exact sequence in cohomology: $$\dots H^{i+1}(A)\to H^i(B)\to H^i(A\oplus B)\to H^i(A)\to H^{i-1}(B)\to\dots$$

Jeremy Rickard
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  • Thank you! Is it true if I drop the vector space requirement and replace it by abelian groups? – jeanmfischer Apr 09 '15 at 15:49
  • @jeanmfischer No. You'll still have a short exact sequence $$0\to\operatorname{coker}(d_1^\bullet)\to H^\bullet(A\oplus B,D)\to\operatorname{ker}(d_1^\bullet)\to0,$$ but it won't necessarily be split. – Jeremy Rickard Apr 09 '15 at 16:17
  • @jeanmfischer For example, let $A\oplus B$ be the complex $$\dots\to0\to\mathbb{Z}\to\mathbb{Z}\oplus\mathbb{Z}\to0\to\dots$$ with $$A=\dots\to0\to\mathbb{Z}\to\mathbb{Z}\oplus 0\to0\to\dots,$$ $$B=\dots\to0\to0\to0\oplus\mathbb{Z}\to0\to\dots$$ and differential $\begin{pmatrix}2&1\end{pmatrix}$. – Jeremy Rickard Apr 09 '15 at 17:16