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Given two diagram of chain complex $C_*,D_*$(with field coefficient) and a map $f$ between them, assume there are filtration $F_1^*,F_2^*$ on this two diagram and $f$ respect filtration. Let $Gr(C_*),Gr(D_*)$ be the associated graded module of each filtration, it's easy to see they are also diagram of chain compex and $f$ induce a map between them, denoted by $Gr(f)$. If $Gr(f)$ is homotopy equivalence of two diagram, can we say $f$ is also a homotopy equivalence ?

(It's easy to see that $f$ is quasi-isomorphism on every position of diagram by spectral sequence, and is also a homotopy equivalence on every position because we use field coefficient, this do not imply homotopy equivalence of diagram. But I think $Gr(f)$ be homotopy equivalence of diagram is a more strong condition, so may be $f$ is also a homotopy equivalence?)

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    $\begingroup$ If by 'diagram of chain complexes' you mean a chain complex in a functor category $[\mathscr J,\mathbf{Vec}_k]$ for some small category $\mathscr J$, then I suspect the answer is negative. For example for $\mathscr J$ the monoid $\mathbf N$ viewed as a one-object category, the category $[\mathscr J,\mathbf{Vec}_k]$ is equivalent to $\mathbf{Mod}_{k[x]}$. If you take the $(x)$-adic filtration, then the associated graded pieces are $k$-vector spaces, for which homotopy equivalence coincides with quasi-isomorphism. The same is not true for $k[x]$-modules, so this should give counterexamples. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Nov 10, 2019 at 4:46
  • $\begingroup$ I expect there should be similar counterexamples for easier index categories, possibly already if $\mathscr J = (\bullet \to \bullet)$, but you'd need to think a little harder. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Nov 10, 2019 at 4:47
  • $\begingroup$ @R. van Dobben de Bruyn, thank you for your comment, I think there are also a small gap in your counterexample. I believe it is easy to give two chain of $k[x]$-module that is quasi-isomorphic but not homotopy equivalence, but if we want it to be a counterexample of my question, the associated graded pieces of (x)-adic filtration should be quasi-isomorphic. $\endgroup$
    – J.D.Chern
    Commented Nov 10, 2019 at 6:48
  • $\begingroup$ @R. van Dobben de Bruyn, Then then question is : can we build two chain complex of $k[x]$-module and a map $f$ between them, that $f$ induce quasi-isomorphism on associated graded pieces of $(x)$-adic filtration and $f$ itself is not a homotopy equivalence. May be this kind of example is easy? $\endgroup$
    – J.D.Chern
    Commented Nov 10, 2019 at 6:49
  • $\begingroup$ you also run into completeness issues for this question. If you consider the filtered $k$-modules $k[x]$ and $k[[x]]$ (formal power series) (both filtered by powers of the ideal $(x)$), you get a map of filtered modules that induces an isomorphism on each subquotient and thus it also induces isomorphisms on the associated graded modules, but the two modules are not isomorphic. Viewing these modules as chain complexes concentrated in one degree gives to non-homotopy equivalent chain complexes, and a filtration preserving map between them which induces an iso on the associated graded. $\endgroup$
    – HenrikRüping
    Commented Nov 11, 2019 at 17:32

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