Lemma 91.13.2. Let $\mathcal{C}$ be a site. Let $\mathcal{A} \to \mathcal{B}$ be a homomorphism of sheaves of rings on $\mathcal{C}$. Let $\mathcal{G}$ be a $\mathcal{B}$-module. Let $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {B}(\mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}}, \mathcal{G})$. There exists a map of sheaves of sets $\alpha : \mathcal{E} \to \mathcal{B}$ such that $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {B}(\mathop{N\! L}\nolimits (\alpha ), \mathcal{G})$ is the class of a map $\mathcal{I}/\mathcal{I}^2 \to \mathcal{G}$ (see proof for notation).
Proof. Recall that given $\alpha : \mathcal{E} \to \mathcal{B}$ such that $\mathcal{A}[\mathcal{E}] \to \mathcal{B}$ is surjective with kernel $\mathcal{I}$ the complex $\mathop{N\! L}\nolimits (\alpha ) = (\mathcal{I}/\mathcal{I}^2 \to \Omega _{\mathcal{A}[\mathcal{E}]/\mathcal{A}} \otimes _{\mathcal{A}[\mathcal{E}]} \mathcal{B})$ is canonically isomorphic to $\mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}}$, see Modules on Sites, Lemma 18.35.2. Observe moreover, that $\Omega = \Omega _{\mathcal{A}[\mathcal{E}]/\mathcal{A}} \otimes _{\mathcal{A}[\mathcal{E}]} \mathcal{B}$ is the sheaf associated to the presheaf $U \mapsto \bigoplus _{e \in \mathcal{E}(U)} \mathcal{B}(U)$. In other words, $\Omega $ is the free $\mathcal{B}$-module on the sheaf of sets $\mathcal{E}$ and in particular there is a canonical map $\mathcal{E} \to \Omega $.
Having said this, pick some $\mathcal{E}$ (for example $\mathcal{E} = \mathcal{B}$ as in the definition of the naive cotangent complex). The obstruction to writing $\xi $ as the class of a map $\mathcal{I}/\mathcal{I}^2 \to \mathcal{G}$ is an element in $\mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {B}(\Omega , \mathcal{G})$. Say this is represented by the extension $0 \to \mathcal{G} \to \mathcal{H} \to \Omega \to 0$ of $\mathcal{B}$-modules. Consider the sheaf of sets $\mathcal{E}' = \mathcal{E} \times _\Omega \mathcal{H}$ which comes with an induced map $\alpha ' : \mathcal{E}' \to \mathcal{B}$. Let $\mathcal{I}' = \mathop{\mathrm{Ker}}(\mathcal{A}[\mathcal{E}'] \to \mathcal{B})$ and $\Omega ' = \Omega _{\mathcal{A}[\mathcal{E}']/\mathcal{A}} \otimes _{\mathcal{A}[\mathcal{E}']} \mathcal{B}$. The pullback of $\xi $ under the quasi-isomorphism $\mathop{N\! L}\nolimits (\alpha ') \to \mathop{N\! L}\nolimits (\alpha )$ maps to zero in $\mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {B}(\Omega ', \mathcal{G})$ because the pullback of the extension $\mathcal{H}$ by the map $\Omega ' \to \Omega $ is split as $\Omega '$ is the free $\mathcal{B}$-module on the sheaf of sets $\mathcal{E}'$ and since by construction there is a commutative diagram
\[ \xymatrix{ \mathcal{E}' \ar[r] \ar[d] & \mathcal{E} \ar[d] \\ \mathcal{H} \ar[r] & \Omega } \]
This finishes the proof. $\square$
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