Lemma 91.13.4. Let $f : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{B}), \mathcal{O}_\mathcal {B})$ be a morphism of ringed topoi. Let $\mathcal{G}$ be an $\mathcal{O}$-module. The set of isomorphism classes of extensions of $f^{-1}\mathcal{O}_\mathcal {B}$-algebras

$0 \to \mathcal{G} \to \mathcal{O}' \to \mathcal{O} \to 0$

where $\mathcal{G}$ is an ideal of square zero1 is canonically bijective to $\mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}(\mathop{N\! L}\nolimits _{\mathcal{O}/\mathcal{O}_\mathcal {B}}, \mathcal{G})$.

Proof. To prove this we apply the previous results to the case where (91.13.0.1) is given by the diagram

$\xymatrix{ 0 \ar[r] & \mathcal{G} \ar[r] & {?} \ar[r] & \mathcal{O} \ar[r] & 0 \\ 0 \ar[r] & 0 \ar[u] \ar[r] & f^{-1}\mathcal{O}_\mathcal {B} \ar[u] \ar[r]^{\text{id}} & f^{-1}\mathcal{O}_\mathcal {B} \ar[u] \ar[r] & 0 }$

Thus our lemma follows from Lemma 91.13.3 and the fact that there exists a solution, namely $\mathcal{G} \oplus \mathcal{O}$. (See remark below for a direct construction of the bijection.) $\square$

[1] In other words, the set of isomorphism classes of first order thickenings $i : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}')$ over $(\mathop{\mathit{Sh}}\nolimits (\mathcal{B}), \mathcal{O}_\mathcal {B})$ endowed with an isomorphism $\mathcal{G} \to \mathop{\mathrm{Ker}}(i^\sharp )$ of $\mathcal{O}$-modules.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).