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I am currently reading Robert Friedman's notes on sheaf cohomology and hypercohomology http://www.math.columbia.edu/~rf/cohomology.pdf

In it he defines quasi-isomorphism of complexes of sheaves by saying the $i^{th}$ cohomology sheaves $\mathcal{H}^i$ are isomorphic.

I can't seem to find a precise definition of these. Does he mean that the cohomology of the $i^{th}$ sheaf in one complex is the same as the cohomology of the $i^{th}$ sheaf in the other complex, for all $i$.

I would be grateful if someone could provide a reference?

The reason i am asking is motivated by the case when the complexes are unbounded (which if i'm not mistaken is not covered in these notes).

Splatenstein in http://www.numdam.org/article/CM_1988__65_2_121_0.pdf introduces $K$-injective resolutions of complexes of objects in an abelian category (not necessarily bounded) and defines cohomology for them as the cohomology of the complex of global sections of a $K$ injective resolution.

At the start of the paper he says quasi-isomorphism of two complexes $A^*$, $B^*$ is a map between the complexes inducing isomorphism between $H^*(A^*)$ and $H^*(B^*)$. How does this look like in the case when $A^*$ and $B^*$ are complexes of sheaves?

amd1234
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  • Your second paragraph is wrong. He defines a quasi-isomorphism in the correct and standard way, as a map which induces isomorphisms on cohomology (sheaves). This is the standard definition for quasi-isomorphism of complexes of objects in any abelian category, specialized to the abelian category of sheaves. – Qiaochu Yuan Aug 29 '22 at 19:15
  • okay thanks this makes sense, so we want the isomorphisms to be between the co-kernels of the map $Im(\partial)\rightarrow Ker(\partial)}$ (for the two complxes) – amd1234 Aug 30 '22 at 06:40

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