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I have a question, how do I prove the following proposition? A and B are filtered chain complex , $f: A \rightarrow B $ is a filtered chain map if indicated mapping on the associated graded object of $f: A \rightarrow B $ be quasi-isomorphism then $f: A \rightarrow B $ is quasi-isomorphism.

fateme
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    Are you sure this is true? What if you take two chain complexes which are just a copy of $\mathbb{Z}$ in degree 0 and trivial elsewhere, filter the first one by $0 \leq 4\mathbb{Z} \leq \mathbb{Z}$ and the second one by $0\leq 2\mathbb{Z} \leq \mathbb{Z}$, and consider the identity map; the identity respects the filtration, is a quasi-iso (which is equivalent to iso in this case), but the associated gradeds are not isomorphic. – Aleksandar Milivojević Jan 27 '19 at 19:16
  • @AleksandarMilivojevic thanks, do you have a proof for other direction? – fateme Jan 27 '19 at 23:13
  • You could consider multiplication by 3 from $\mathbb{Z}$ to $\mathbb{Z}$, and filter both of them by $\cdots \leq 16\mathbb{Z} \leq 8\mathbb{Z} \leq 4\mathbb{Z} \leq 2\mathbb{Z} \leq \mathbb{Z}$; then the induced map on associated gradeds is an isomorphism. – Aleksandar Milivojević Jan 28 '19 at 00:47

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This question and other direction are true. You can use from Five-lemma and following proposition to proof them. The chain map $f: C \rightarrow C'$ is quasi-isomorphism if and only if homology of mapping cone of f be trivial.

hussein
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