Let $X$ be a topological space. Define two cochain complexes $\mathcal{C}$ and $\mathcal{D}$ by

$\mathcal{C}=\{C^k(X; \mathbb{Q}), \partial^k\}, \qquad\mathcal{D}=\{C^k(X; \mathbb{R}), \partial^k\},$

where $C^k$ is the free abelian group of (singular) cochains on $X$. The inclusion of coefficients $j:\mathbb{Q}\to\mathbb{R}$ induces a chain map $i:\mathcal{C}\to\mathcal{D}$.

My question is: is the homology of the cochain complex

$\mathcal{B}:=\{C^k(X; \mathbb{R})\times C^{k-1}(X; \mathbb{Q}), \partial_j^k\},$

where $\partial_j^k(u, v)=(\partial u+j(v), -\partial v)$, equal to $H^*(X; \mathbb{R}/\mathbb{Q})$?

Of course $\mathcal{B}$ seems to be the mapping cone complex of $i:\mathcal{C}\to\mathcal{D}$, but I am not sure, since the mapping cone complex seems to be about maps between spaces.

• The mapping cone of chain complexes is something that can be defined purely algebraically. It is analogous to the mapping cone of spaces but there is no reason to define one in terms of the other. – Zhen Lin Dec 5 '14 at 21:13
• So is my claim coorect or not？ – GRR Dec 6 '14 at 2:50
• Yes. The mapping cone induces a long exact sequence and you can use the five lemma. There are some details in the introduction of Weibel's book on homological algebra. – Zhen Lin Dec 6 '14 at 9:08
• thx.....oh...i just found that how much algebraic topology I forgot! – GRR Dec 6 '14 at 10:34