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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