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The Stacks project

Proof. Let 0 \to \mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3 \to 0 be a short exact sequence of \mathcal{O}-modules with \mathcal{F}_1 and \mathcal{F}_3 as in (18.30.7.2). Choose presentations

\bigoplus A_{V_ j} \to \bigoplus A_{U_ i} \to \mathcal{F}_1 \to 0 \quad \text{and}\quad \bigoplus A_{T_ j} \to \bigoplus A_{W_ i} \to \mathcal{F}_3 \to 0

In this proof the direct sums are always finite, and we write A_ U = j_{U!}\mathcal{O}_ U for U \in \mathcal{B}. Since \mathcal{F}_2 \to \mathcal{F}_3 is surjective, we can choose coverings \{ W_{ik} \to W_ i\} with W_{ik} \in \mathcal{B} such that A_{W_{ik}} \to \mathcal{F}_3 lifts to a map A_{W_{ik}} \to \mathcal{F}_2. By Lemma 18.30.9 we may replace our collection \{ W_ i\} by a finite subcollection of the collection \{ W_{ik}\} and assume the map \bigoplus A_{W_ i} \to \mathcal{F}_3 lifts to a map into \mathcal{F}_2. Consider the kernel

\mathcal{K}_2 = \mathop{\mathrm{Ker}}(\bigoplus A_{U_ i} \oplus \bigoplus A_{W_ i} \longrightarrow \mathcal{F}_2)

By the snake lemma this kernel surjects onto \mathcal{K}_3 = \mathop{\mathrm{Ker}}(\bigoplus A_{W_ i} \to \mathcal{F}_3). Thus, arguing as above, after replacing each T_ j by a finite family of elements of \mathcal{B} (permissible by Lemma 18.30.2) we may assume there is a map \bigoplus A_{T_ j} \to \mathcal{K}_2 lifting the given map \bigoplus A_{T_ j} \to \mathcal{K}_3. Then \bigoplus A_{V_ j} \oplus \bigoplus A_{T_ j} \to \mathcal{K}_2 is surjective which finishes the proof. \square


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