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Let $\mathcal{F}$ be a coherent sheaf on a projective manifold $X$. It is well known that one can construct a resolution of $\mathcal{F}$ by holomorphic vector bundles (locally free sheaves).

Are two such resolutions homotopic? Any reference would be much appreciated.

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  • $\begingroup$ Crossposted at MSE at the same time it was posted here. Please see meta for site rules about cross-posting. $\endgroup$
    – KReiser
    Nov 15, 2021 at 3:28

2 Answers 2

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If this were true, then any short exact sequence of vector bundles would split. Indeed, if $0 \to \mathscr E_1 \to \mathscr E_2 \to \mathscr E_3 \to 0$ is a short exact sequence of vector bundles, then both \begin{align*} K^\bullet = \cdots \to 0 \to \mathscr E_1 \to \mathscr E_2 \to 0 \to \cdots \end{align*} and $L^\bullet = \mathscr E_3[0]$ are resolutions of $\mathscr E_3$. If $g \colon L^\bullet \to K^\bullet$ is a homotopy equivalence (or even a quasi-isomorphism!), then the map $g^0 \colon \mathscr E_3 \to \mathscr E_2$ induces an isomorphism $$\phi \colon \mathscr E_3 \stackrel\sim\to H^0(K^\bullet) = \mathscr E_3.$$ Then $g^0 \circ \phi^{-1} \colon \mathscr E_3 \to \mathscr E_2$ is a splitting of $0 \to \mathscr E_1 \to \mathscr E_2 \to \mathscr E_3 \to 0$. $\square$

(On the other hand, there does always exist a quasi-isomorphism $K^\bullet \to L^\bullet$ in this case, just not in the other direction.)

An example of a short exact sequence of vector bundles that doesn't split is the Koszul sequence $$0 \to \mathcal O_{\mathbf P^1}(-2) \stackrel{\left(\begin{smallmatrix}-y \\ x\end{smallmatrix}\right)}\longrightarrow \mathcal O_{\mathbf P^1}(-1) \oplus \mathcal O_{\mathbf P^1}(-1) \stackrel{\left(x\ \ y\right)}\longrightarrow \mathcal O_{\mathbf P^1} \to 0$$ on $X = \mathbf P^1$. Indeed, $\operatorname{Hom}(\mathcal O_{\mathbf P^1}, \mathcal O_{\mathbf P^1}(-1) \oplus \mathcal O_{\mathbf P^1}(-1)) = 0$.

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No. The simplest example is given by the following two resolutions of the structure sheaf of a point $P \in \mathbb{P}^1$: $$ 0 \to \mathcal{O}_{\mathbb{P}^1}(-1) \to \mathcal{O}_{\mathbb{P}^1} \to \mathcal{O}_{P} \to 0 $$ and $$ 0 \to \mathcal{O}_{\mathbb{P}^1}(-2) \to \mathcal{O}_{\mathbb{P}^1}(-1) \to \mathcal{O}_{P} \to 0 $$ (the second is obtained from the first by a twist).

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