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Consider a homomorphism $f$ between two complexes of vector bundles over a fixed smooth manifold $M$. $$ \cdots \to V_{i - 1} \xrightarrow{\delta_{i-1}} V_i \xrightarrow{\delta_i} V_{i + 1} \to \cdots $$ $$ \cdots \to W_{i - 1} \xrightarrow{\Delta_{i-1}} W_i \xrightarrow{\Delta_i} W_{i + 1} \to \cdots $$ When people (e.g. those working in derived algebraic geometry) say that $f$ is a quasi-isomorphism, does that simply mean that we have isomorphisms $\frac{\ker \delta_i}{\operatorname{im} \delta_{i-1}}|_x \to \frac{\ker \Delta_i}{\operatorname{im} \Delta_{i-1}}|_x$ for all $x \in M$?

I'm worried since $\frac{\ker \delta_i}{\operatorname{im} \delta_{i-1}}$ may not form a smooth vector bundle.

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    $\begingroup$ en.wikipedia.org/wiki/Quasi-isomorphism $\endgroup$
    – Sasha
    Commented May 3, 2023 at 12:44
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    $\begingroup$ Dear @Sasha, his question seems reasonable to me. The category of vector bundles is not abelian, so I would imagine that some care must be taken to make sense of quasi-isomorphisms. $\endgroup$
    – Gabriel
    Commented May 3, 2023 at 12:48
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    $\begingroup$ @Gabriel: But the category of coherent sheaves is, and a morphism of complexes of vector bundles is a quasiisomorphis if and only if it is a quasiisomorphism when considered as a morphism of complexes of coherent sheaves. $\endgroup$
    – Sasha
    Commented May 3, 2023 at 13:07
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    $\begingroup$ Perfect, so this seems to answer his question. $\endgroup$
    – Gabriel
    Commented May 3, 2023 at 13:31
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    $\begingroup$ I think it would help to clarify the precise context and the meaning of the notation. Generally speaking, in either differential or algebraic geometry, I would say that a quasi-isomorphism of vector bundles means a quasi-isomorphism of the corresponding sheaves of sections. This means that $f$ induces isomorphisms on homology sheaves. This latter condition can be checked pointwise on stalks of the homology sheaves, but I guess the pointwise condition in mentioned in the question is supposed to refer to the fibers at $x$ rather than stalks (which is not sufficient to check isomorphisms.) $\endgroup$ Commented May 3, 2023 at 20:41

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