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Consider a commutative diagram of complexes over a ring $R$ $$\require{AMScd}\begin{CD} 0 @>>> X @>>> Y @>>> Z \\ @. @VVfV @VVgV @VVhV \\ 0 @>>> X' @>>> Y' @>>> Z' \end{CD}$$ with exact rows. I know that the complexes $Y$, $Y'$, $Z$, and $Z'$ are exact, hence the maps $g$ and $h$ are automatically quasi-isomorphisms. (i.e., by definition, they induce isomorphisms on the homologies).

My question is: Is the leftmost vertical map $f$ also a quasi isomorphism?

A natural attempt is to use the long exact sequences of homology modules induced by short exact sequence of complexes. However, I'm not able to conclude it.

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  • $\begingroup$ Yes. I corrected it. $\endgroup$
    – H. Ali
    Commented Feb 27, 2022 at 19:50

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No.

Let $X'=Y'=Z'=0$, let $\alpha:Y\to Z$ be any quasi-isomorphism between acyclic complexes whose kernel is not acyclic, and let $X=\ker(\alpha)$.

For example, if $R=\mathbb{Z}$, then $Y$ could be $\cdots\to0\to\mathbb{Z}\stackrel{\sim}{\to}\mathbb{Z}\to0\to\cdots$, with $Z=Y[1]$ and $\alpha$ the obvious map with $\ker(\alpha)$ equal to $\cdots\to0\to0\to\mathbb{Z}\to0\to\cdots$.

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  • $\begingroup$ Thank you Jeremy for your answer. But I'm working with non-zero complexes in both rows. Might it make any difference? $\endgroup$
    – H. Ali
    Commented Feb 27, 2022 at 20:07
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    $\begingroup$ @H.Ali Well, you could just take the direct sum of the diagram I described with one that only involves nonzero complexes. $\endgroup$
    – Jeremy Rickard
    Commented Feb 27, 2022 at 20:10
  • $\begingroup$ You are right; thank you. Are you aware of any extra assumptions that make the answer positive? $\endgroup$
    – H. Ali
    Commented Feb 27, 2022 at 20:16

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