The Stacks project

Proof. A pair $(K, \rho _ U)$ is called a solution in the text above. The uniqueness follows from Lemma 20.45.4. If $X$ has a finite covering by elements of $\mathcal{B}$ (for example if $X$ is quasi-compact), then the theorem is a consequence of Lemma 20.45.6. In the general case we argue in exactly the same manner, using transfinite induction and Lemma 20.45.7.

First we use transfinite recursion to choose opens $W_\alpha \subset X$ for any ordinal $\alpha $. Namely, we set $W_0 = \emptyset $. If $\alpha = \beta + 1$ is a successor, then either $W_\beta = X$ and we set $W_\alpha = X$ or $W_\beta \not= X$ and we set $W_\alpha = W_\beta \cup U_\alpha $ where $U_\alpha \in \mathcal{B}$ is not contained in $W_\beta $. If $\alpha $ is a limit ordinal we set $W_\alpha = \bigcup _{\beta < \alpha } W_\beta $. Then for large enough $\alpha $ we have $W_\alpha = X$. Observe that for every $\alpha $ the open $W_\alpha $ is a union of elements of $\mathcal{B}$. Hence if $\mathcal{B}_\alpha = \{ U \in \mathcal{B}, U \subset W_\alpha \} $, then

is a system as in Lemma 20.45.4 on the ringed space $W_\alpha $.

We will show by transfinite induction that for every $\alpha $ the system $S_\alpha $ has a solution. This will prove the theorem as this system is the system given in the theorem for large $\alpha $.

The case where $\alpha = \beta + 1$ is a successor ordinal. (This case was already treated in the proof of the lemma above but for clarity we repeat the argument.) Recall that $W_\alpha = W_\beta \cup U_\alpha $ for some $U_\alpha \in \mathcal{B}$ in this case. By induction hypothesis we have a solution $(K_{W_\beta }, \{ \rho ^{W_\beta }_ U\} _{U \in \mathcal{B}_\beta })$ for the system $S_\beta $. Then we can consider the collection $\mathcal{B}_\alpha ^* = \mathcal{B}_\alpha \cup \{ W_\beta \} $ of opens of $W_\alpha $ and we see that we obtain a system $(\{ K_ U\} _{U \in \mathcal{B}_\alpha ^*}, \{ \rho _ V^ U\} _{V \subset U\text{ with }U, V \in \mathcal{B}_\alpha ^*})$. Note that this new system also satisfies condition (3) by Lemma 20.45.4 applied to the solution $K_{W_\beta }$. For this system we have $W_\alpha = W_\beta \cup U_\alpha $. This reduces us to the case handled in Lemma 20.45.6.

The case where $\alpha $ is a limit ordinal. Recall that $W_\alpha = \bigcup _{\beta < \alpha } W_\beta $ in this case. For $\beta < \alpha $ let $(K_{W_\beta }, \{ \rho ^{W_\beta }_ U\} _{U \in \mathcal{B}_\beta })$ be the solution for $S_\beta $. For $\gamma < \beta < \alpha $ the restriction $K_{W_\beta }|_{W_\gamma }$ endowed with the maps $\rho ^{W_\beta }_ U$, $U \in \mathcal{B}_\gamma $ is a solution for $S_\gamma $. By uniqueness we get unique isomorphisms $\rho _{W_\gamma }^{W_\beta } : K_{W_\beta }|_{W_\gamma } \to K_{W_\gamma }$ compatible with the maps $\rho ^{W_\beta }_ U$ and $\rho ^{W_\gamma }_ U$ for $U \in \mathcal{B}_\gamma $. These maps compose in the correct manner, i.e., $\rho _{W_\delta }^{W_\gamma } \circ \rho _{W_\gamma }^{W_\beta }|_{W_\delta } = \rho ^{W_\delta }_{W_\beta }$ for $\delta < \gamma < \beta < \alpha $. Thus we may apply Lemma 20.45.7 (note that the vanishing of negative exts is true for $K_{W_\beta }$ by Lemma 20.45.4 applied to the solution $K_{W_\beta }$) to obtain $K_{W_\alpha }$ and isomorphisms

compatible with the maps $\rho _{W_\gamma }^{W_\beta }$ for $\gamma < \beta < \alpha $.

To show that $K_{W_\alpha }$ is a solution we still need to construct the isomorphisms $\rho _ U^{W_\alpha } : K_{W_\alpha }|_ U \to K_ U$ for $U \in \mathcal{B}_\alpha $ satisfying certain compatibilities. We choose $\rho _ U^{W_\alpha }$ to be the unique map such that for any $\beta < \alpha $ and any $V \in \mathcal{B}_\beta $ with $V \subset U$ the diagram

commutes. This makes sense because

is a system as in Lemma 20.45.4 on the ringed space $U$ and because $(K_ U, \rho ^ U_ V)$ and $(K_{W_\alpha }|_ U, \rho _ V^{W_\beta }\circ \rho _{W_\beta }^{W_\alpha }|_ V)$ are both solutions for this system. This gives existence and uniqueness. We omit the proof that these maps satisfy the desired compatibilities (it is just bookkeeping). $\square$