Lemma 90.7.8. Let $t : (S, \mathcal{O}_ S) \to (S', \mathcal{O}_{S'})$, $\mathcal{J} = \mathop{\mathrm{Ker}}(t^\sharp )$, $f : (X, \mathcal{O}_ X) \to (S, \mathcal{O}_ S)$, $\mathcal{G}$, and $c : \mathcal{J} \to \mathcal{G}$ be as in (90.7.0.1). Denote $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ S}(\mathop{N\! L}\nolimits _{S/S'}, \mathcal{J})$ the element corresponding to the extension $\mathcal{O}_{S'}$ of $\mathcal{O}_ S$ by $\mathcal{J}$ via Lemma 90.7.4. The set of isomorphism classes of solutions is canonically bijective to the fibre of

\[ \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathop{N\! L}\nolimits _{X/S'}, \mathcal{G}) \to \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(Lf^*\mathop{N\! L}\nolimits _{S/S'}, \mathcal{G}) \]

over the image of $\xi $.

**Proof.**
By Lemma 90.7.4 applied to $X \to S'$ and the $\mathcal{O}_ X$-module $\mathcal{G}$ we see that elements $\zeta $ of $\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathop{N\! L}\nolimits _{X/S'}, \mathcal{G})$ parametrize extensions $0 \to \mathcal{G} \to \mathcal{O}_{X'} \to \mathcal{O}_ X \to 0$ of $f^{-1}\mathcal{O}_{S'}$-algebras. By Lemma 90.7.6 applied to $X \to S \to S'$ and $c : \mathcal{J} \to \mathcal{G}$ we see that there is an $S'$-morphism $X' \to S'$ compatible with $c$ and $f : X \to S$ if and only if $\zeta $ maps to $\xi $. Of course this is the same thing as saying $\mathcal{O}_{X'}$ is a solution of (90.7.0.1).
$\square$

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