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Let $\mathcal{A}$ be an abelian category, $D(\mathcal{A})$ be its derived category and $X,Y$ be complexes with morphisms in $\mathcal{A}$. I am trying to understand what does it mean to say that $X$ and $Y$ are isomorphic in $D(\mathcal{A})$. Obviously, if there is a quasi-isomorphism $X \longrightarrow Y$, then these complexes are isomorphic in $D(\mathcal{A})$, but does the converse hold? I have two questions:

  1. If $X$ and $Y$ are isomorphic in $D(\mathcal{A})$, then is it true that there is a quasi-isomorphism $X \longrightarrow Y$?
  2. If $X$ and $Y$ are isomorphic in $D(\mathcal{A})$, then there is a (left) fraction $(s,f) : X \longrightarrow Y$, where $s : Z \longrightarrow X$ is a quasi-isomorphism and $f: Z \longrightarrow Y$ is a morphism of complexes (modulo homotopy), such that $(s,f)$ is an isomorphism in $D(\mathcal{A})$. In this case, is it true that $f$ is also a quasi-isomorphism?

I will be also grateful for any explanation or idea which may clarify the meaning of two complexes being isomorphic in the derived category.

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    $\begingroup$ 1. No. Consider $X:0\to 0\to \mathbb Z/2\to 0,Y:0\to \mathbb Z\stackrel{\cdot 2}{\to}\mathbb Z\to 0$ (padded with zeros in both directions) in $D(Ab)$. There is no nonzero $X\to Y$, hence no quasiisomorphism $X\to Y$, yet they are isomorphic in $D(Ab)$ since there is a quasiisomorphism $Y\to X$. We can strengthen it to show there needn't be a quasiisomorphism either way, by making a complex $X'$ which contians a copy of $X$ and $Y$, and $Y'$ which contains copies of $Y$ and $X$ in corresponding spaces. $\endgroup$
    – Wojowu
    Jun 25, 2020 at 20:29
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    $\begingroup$ 2. If we assume the fraction $(s,f)$ is an isomorphism in $D(\mathcal A)$ (which you didn't explicitly state), then yes, as then also $f$ is an isomorphism in $D(\mathcal A)$, hence (as cohomology is a well-defined functor on $D(\mathcal A)$) it induces isomorphism on all cohomology. $\endgroup$
    – Wojowu
    Jun 25, 2020 at 20:32
  • $\begingroup$ @Wojowu Thank you, I edited the question. $\endgroup$
    – user144185
    Jun 25, 2020 at 20:36
  • $\begingroup$ 1. In fact, $X$ and $Y$ are isomorphic in $D( \mathcal(A))$ if there exist a finite zig-zag $X \stackrel{\sim }{\rightarrow} X' \stackrel{\sim }{\leftarrow} ... \stackrel{\sim }{\rightarrow} Y' \stackrel{\sim }{\leftarrow} Y$. If the category has enough projectives or enough injectives, you can using the theory of model categories. Then you can control the length of the zigzag (it's always of length three). $\endgroup$
    – MoreauT
    Jun 26, 2020 at 15:47
  • $\begingroup$ @MoreatuT I guess the composition of roofs (or zig-zags as you call them) $X\leftarrow X'\to\cdots\leftarrow Y'\to Y$ is another roof $X\leftarrow Z\to Y$. So $X\cong Y$ in $D(\mathcal{A})$ if and only if there is a roof $X\leftarrow Z\to Y$ in which both morphisms are quasi-isomorphisms. $\endgroup$ Feb 5, 2022 at 19:54

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