Lemma 91.13.8. Let $t : (\mathop{\mathit{Sh}}\nolimits (\mathcal{B}), \mathcal{O}_\mathcal {B}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{B}'), \mathcal{O}_{\mathcal{B}'})$, $\mathcal{J} = \mathop{\mathrm{Ker}}(t^\sharp )$, $f : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{B}), \mathcal{O}_\mathcal {B})$, $\mathcal{G}$, and $c : \mathcal{J} \to \mathcal{G}$ be as in (91.13.0.1). Denote $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_\mathcal {B}}( \mathop{N\! L}\nolimits _{\mathcal{O}_\mathcal {B}/\mathcal{O}_{\mathcal{B}'}}, \mathcal{J})$ the element corresponding to the extension $\mathcal{O}_{\mathcal{B}'}$ of $\mathcal{O}_\mathcal {B}$ by $\mathcal{J}$ via Lemma 91.13.4. The set of isomorphism classes of solutions is canonically bijective to the fibre of
\[ \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}(\mathop{N\! L}\nolimits _{\mathcal{O}/\mathcal{O}_{\mathcal{B}'}}, \mathcal{G})\to \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}( Lf^*\mathop{N\! L}\nolimits _{\mathcal{O}_\mathcal {B}/\mathcal{O}_{\mathcal{B}'}}, \mathcal{G}) \]
over the image of $\xi $.
Proof.
By Lemma 91.13.4 applied to $t \circ f : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{B}'), \mathcal{O}_{\mathcal{B}'})$ and the $\mathcal{O}$-module $\mathcal{G}$ we see that elements $\zeta $ of $\mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}(\mathop{N\! L}\nolimits _{\mathcal{O}/\mathcal{O}_{\mathcal{B}'}}, \mathcal{G})$ parametrize extensions $0 \to \mathcal{G} \to \mathcal{O}' \to \mathcal{O} \to 0$ of $f^{-1}\mathcal{O}_{\mathcal{B}'}$-algebras. By Lemma 91.13.6 applied to
\[ (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \xrightarrow {f} (\mathop{\mathit{Sh}}\nolimits (\mathcal{B}), \mathcal{O}_\mathcal {B}) \xrightarrow {t} (\mathop{\mathit{Sh}}\nolimits (\mathcal{B}'), \mathcal{O}_{\mathcal{B}'}) \]
and $c : \mathcal{J} \to \mathcal{G}$ we see that there is an morphism
\[ f' : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}') \longrightarrow (\mathop{\mathit{Sh}}\nolimits (\mathcal{B}'), \mathcal{O}_{\mathcal{B}'}) \]
over $(\mathop{\mathit{Sh}}\nolimits (\mathcal{B}'), \mathcal{O}_{\mathcal{B}'})$ compatible with $c$ and $f$ if and only if $\zeta $ maps to $\xi $. Of course this is the same thing as saying $\mathcal{O}'$ is a solution of (91.13.0.1).
$\square$
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