Lemma 20.39.4. In Situation 20.39.3 assume

1. $X = \bigcup _{U \in \mathcal{B}} U$ and for $U, V \in \mathcal{B}$ we have $U \cap V = \bigcup _{W \in \mathcal{B}, W \subset U \cap V} W$,

2. for any $U \in \mathcal{B}$ we have $\mathop{\mathrm{Ext}}\nolimits ^ i(K_ U, K_ U) = 0$ for $i < 0$.

If a solution $(K, \rho _ U)$ exists, then it is unique up to unique isomorphism and moreover $\mathop{\mathrm{Ext}}\nolimits ^ i(K, K) = 0$ for $i < 0$.

Proof. Let $(K, \rho _ U)$ and $(K', \rho '_ U)$ be a pair of solutions. Let $f : X \to Y$ be the continuous map constructed in Topology, Lemma 5.5.6. Set $\mathcal{O}_ Y = f_*\mathcal{O}_ X$. Then $K, K'$ and $\mathcal{B}$ are as in Lemma 20.39.2 part (2). Hence we obtain the vanishing of negative exts for $K$ and we see that the rule

$V \longmapsto \mathop{\mathrm{Hom}}\nolimits (K|_{f^{-1}V}, K'|_{f^{-1}V})$

is a sheaf on $Y$. As both $(K, \rho _ U)$ and $(K', \rho '_ U)$ are solutions the maps

$(\rho '_ U)^{-1} \circ \rho _ U : K|_ U \longrightarrow K'|_ U$

over $U = f^{-1}(f(U))$ agree on overlaps. Hence we get a unique global section of the sheaf above which defines the desired isomorphism $K \to K'$ compatible with all structure available. $\square$

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