Lemma 91.12.5. In Situation 91.12.4 the modules $\pi ^*\mathcal{F}$ and $h^*\mathcal{F}'_2$ are $\mathcal{O}'_1$-modules flat over $\mathcal{O}_{\mathcal{B}'_1}$ restricting to $\mathcal{F}$ on $(\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O})$. Their difference (Lemma 91.12.1) is an element $\theta$ of $\mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}(\mathcal{F}, f^*\mathcal{J}_1 \otimes _\mathcal {O} \mathcal{F})$ whose boundary in $\mathop{\mathrm{Ext}}\nolimits ^2_\mathcal {O}(\mathcal{F}, f^*\mathcal{J}_3 \otimes _\mathcal {O} \mathcal{F})$ equals the obstruction (Lemma 91.12.1) to lifting $\mathcal{F}$ to an $\mathcal{O}'_3$-module flat over $\mathcal{O}_{\mathcal{B}'_3}$.

Proof. Note that both $\pi ^*\mathcal{F}$ and $h^*\mathcal{F}'_2$ restrict to $\mathcal{F}$ on $(\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O})$ and that the kernels of $\pi ^*\mathcal{F} \to \mathcal{F}$ and $h^*\mathcal{F}'_2 \to \mathcal{F}$ are given by $f^*\mathcal{J}_1 \otimes _\mathcal {O} \mathcal{F}$. Hence flatness by Lemma 91.11.2. Taking the boundary makes sense as the sequence of modules

$0 \to f^*\mathcal{J}_3 \otimes _\mathcal {O} \mathcal{F} \to f^*\mathcal{J}_2 \otimes _\mathcal {O} \mathcal{F} \to f^*\mathcal{J}_1 \otimes _\mathcal {O} \mathcal{F} \to 0$

is short exact due to the assumptions in Situation 91.12.4 and the fact that $\mathcal{F}$ is flat over $\mathcal{O}_\mathcal {B}$. The statement on the obstruction class is a direct translation of the result of Remark 91.10.11 to this particular situation. $\square$

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