Remark 91.13.5. Let $f : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathcal{B}, \mathcal{O}_\mathcal {B})$ and $\mathcal{G}$ be as in Lemma 91.13.4. Consider an extension $0 \to \mathcal{G} \to \mathcal{O}' \to \mathcal{O} \to 0$ as in the lemma. We can choose a sheaf of sets $\mathcal{E}$ and a commutative diagram

$\xymatrix{ \mathcal{E} \ar[d]_{\alpha '} \ar[rd]^\alpha \\ \mathcal{O}' \ar[r] & \mathcal{O} }$

such that $f^{-1}\mathcal{O}_\mathcal {B}[\mathcal{E}] \to \mathcal{O}$ is surjective with kernel $\mathcal{J}$. (For example you can take any sheaf of sets surjecting onto $\mathcal{O}'$.) Then

$\mathop{N\! L}\nolimits _{\mathcal{O}/\mathcal{O}_\mathcal {B}} \cong \mathop{N\! L}\nolimits (\alpha ) = \left( \mathcal{J}/\mathcal{J}^2 \longrightarrow \Omega _{f^{-1}\mathcal{O}_\mathcal {B}[\mathcal{E}]/ f^{-1}\mathcal{O}_\mathcal {B}} \otimes _{f^{-1}\mathcal{O}_\mathcal {B}[\mathcal{E}]} \mathcal{O}\right)$

See Modules on Sites, Section 18.35 and in particular Lemma 18.35.2. Of course $\alpha '$ determines a map $f^{-1}\mathcal{O}_\mathcal {B}[\mathcal{E}] \to \mathcal{O}'$ which in turn determines a map

$\mathcal{J}/\mathcal{J}^2 \longrightarrow \mathcal{G}$

which in turn determines the element of $\mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}(\mathop{N\! L}\nolimits (\alpha ), \mathcal{G}) = \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}(\mathop{N\! L}\nolimits _{\mathcal{O}/\mathcal{O}_\mathcal {B}}, \mathcal{G})$ corresponding to $\mathcal{O}'$ by the bijection of the lemma.

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