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Let $X$ be a smooth projective variety (say $\mathbb P^1$), and $Q$ be the functor from the category of complexes of coherent sheaves to its derived category that formally inverts quasi-isomorphisms. Let $f\colon F\to F$ be a chain map of bounded complex of locally free sheaves on $X$. My question is:

If $Q(f)=0$, is it true that $f$ is homotopic to zero?

If $X$ is affine, then this is true since $F$ is a complex of projective objects. I guess this cannot globalize, but it would be good to have a counter-example. I also noticed this post so we just need an $F$ such that $Q(F)=0$ and not zero in the homotopy category, but I don't know if it exists for this bounded locally free setting.

Thanks in advance!

PS: This is motivated by my previous question.

Martin Brandenburg
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Cyrist
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1 Answers1

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Take any short exact sequence of vector bundles that does not split. For example the following on $\Bbb P^1$:

$$0 \to \mathcal O(-2) \to \mathcal O(-1)^2 \to \mathcal O \to 0$$

This defines a bounded complex $F$ of locally free sheaves on $\Bbb P^1$. Because it is acyclic, it is zero in the derived category. But a short exact sequence is contractible as a complex iff it splits. The above sequence does not split because $\mathcal O(-1)^2$ has no nonzero global section, whereas $\mathcal O(-2) \oplus \mathcal O$ does. Thus the identity on $F$ is a counterexample to the question.

Lukas Heger
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