**Proof.**
Uniqueness was seen in Lemma 20.41.4. We may prove the lemma by induction on $n$. The case $n = 1$ is immediate.

The case $n = 2$. Consider the isomorphism $\rho _{U_1, U_2} : K_{U_1}|_{U_1 \cap U_2} \to K_{U_2}|_{U_1 \cap U_2}$ constructed in Remark 20.41.5. By Lemma 20.41.1 we obtain an object $K$ in $D(\mathcal{O}_ X)$ and isomorphisms $\rho _{U_1} : K|_{U_1} \to K_{U_1}$ and $\rho _{U_2} : K|_{U_2} \to K_{U_2}$ compatible with $\rho _{U_1, U_2}$. Take $U \in \mathcal{B}$. We will construct an isomorphism $\rho _ U : K|_ U \to K_ U$ and we will leave it to the reader to verify that $(K, \rho _ U)$ is a solution. Consider the set $\mathcal{B}'$ of elements of $\mathcal{B}$ contained in either $U \cap U_1$ or contained in $U \cap U_2$. Then $(K_ U, \rho ^ U_{U'})$ is a solution for the system $(\{ K_{U'}\} _{U' \in \mathcal{B}'}, \{ \rho _{V'}^{U'}\} _{V' \subset U'\text{ with }U', V' \in \mathcal{B}'})$ on the ringed space $U$. We claim that $(K|_ U, \tau _{U'})$ is another solution where $\tau _{U'}$ for $U' \in \mathcal{B}'$ is chosen as follows: if $U' \subset U_1$ then we take the composition

\[ K|_{U'} \xrightarrow {\rho _{U_1}|_{U'}} K_{U_1}|_{U'} \xrightarrow {\rho ^{U_1}_{U'}} K_{U'} \]

and if $U' \subset U_2$ then we take the composition

\[ K|_{U'} \xrightarrow {\rho _{U_2}|_{U'}} K_{U_2}|_{U'} \xrightarrow {\rho ^{U_2}_{U'}} K_{U'}. \]

To verify this is a solution use the property of the map $\rho _{U_1, U_2}$ described in Remark 20.41.5 and the compatibility of $\rho _{U_1}$ and $\rho _{U_2}$ with $\rho _{U_1, U_2}$. Having said this we apply Lemma 20.41.4 to see that we obtain a unique isomorphism $K|_{U'} \to K_{U'}$ compatible with the maps $\tau _{U'}$ and $\rho ^ U_{U'}$ for $U' \in \mathcal{B}'$.

The case $n > 2$. Consider the open subspace $X' = U_1 \cup \ldots \cup U_{n - 1}$ and let $\mathcal{B}'$ be the set of elements of $\mathcal{B}$ contained in $X'$. Then we find a system $(\{ K_ U\} _{U \in \mathcal{B}'}, \{ \rho _ V^ U\} _{U, V \in \mathcal{B}'})$ on the ringed space $X'$ to which we may apply our induction hypothesis. We find a solution $(K_{X'}, \rho ^{X'}_ U)$. Then we can consider the collection $\mathcal{B}^* = \mathcal{B} \cup \{ X'\} $ of opens of $X$ and we see that we obtain a system $(\{ K_ U\} _{U \in \mathcal{B}^*}, \{ \rho _ V^ U\} _{V \subset U\text{ with }U, V \in \mathcal{B}^*})$. Note that this new system also satisfies condition (3) by Lemma 20.41.4 applied to the solution $K_{X'}$. For this system we have $X = X' \cup U_ n$. This reduces us to the case $n = 2$ we worked out above.
$\square$

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