Remark 91.7.5. Let $f : (X, \mathcal{O}_ X) \to (S, \mathcal{O}_ S)$ and $\mathcal{G}$ be as in Lemma 91.7.4. Consider an extension $0 \to \mathcal{G} \to \mathcal{O}_{X'} \to \mathcal{O}_ X \to 0$ as in the lemma. We can choose a sheaf of sets $\mathcal{E}$ and a commutative diagram

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

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

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

See Modules, Section 17.31 and in particular Lemma 17.31.2. Of course $\alpha '$ determines a map $f^{-1}\mathcal{O}_ S[\mathcal{E}] \to \mathcal{O}_{X'}$ which in turn determines a map

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

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

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