hcju/stacksqa huggingface dataset:
Assume given extensions $$ 0 \to \mathcal{K} \to \mathcal{F}'_1 \to \mathcal{F} \to 0 \quad\text{and}\quad 0 \to \mathcal{K} \to \mathcal{F}'_2 \to \mathcal{F} \to 0 $$ with $c_{\mathcal{F}'_1} = c_{\mathcal{F}'_2} = c$. Then the difference (in the extension group, see Homology, Section \ref{homology-section-extensions}) is an extension $$ 0 \to \mathcal{K} \to \mathcal{E} \to \mathcal{F} \to 0 $$ where $\mathcal{E}$ is annihilated by $\mathcal{I}$ (local computation omitted). Hence the sequence is an extension of $\mathcal{O}$-modules, see Modules on Sites, Lemma \ref{sites-modules-lemma-i-star-equivalence}. Conversely, given such an extension $\mathcal{E}$ we can add the extension $\mathcal{E}$ to the $\mathcal{O}'$-extension $\mathcal{F}'$ without affecting the map $c_{\mathcal{F}'}$. Some details omitted.
This is a special case of Lemma \ref{lemma-inf-map-ringed-topoi} but we also give a direct proof. We have short exact sequences of modules $$ 0 \to \mathcal{I} \to \mathcal{O}' \to \mathcal{O} \to 0 \quad\text{and}\quad 0 \to \mathcal{I}\mathcal{G}' \to \mathcal{G}' \to \mathcal{G} \to 0 $$ and similarly for $\mathcal{F}'$. Since $\mathcal{I}$ has square zero the $\mathcal{O}'$-module structure on $\mathcal{I}$ and $\mathcal{I}\mathcal{G}'$ comes from a unique $\mathcal{O}$-module structure. It follows that $$ \Hom_{\mathcal{O}'}(\mathcal{F}', \mathcal{I}\mathcal{G}') = \Hom_\mathcal{O}(\mathcal{F}, \mathcal{I}\mathcal{G}') \quad\text{and}\quad \Hom_{\mathcal{O}'}(\mathcal{F}', \mathcal{G}) = \Hom_\mathcal{O}(\mathcal{F}, \mathcal{G}) $$ The lemma now follows from the exact sequence $$ 0 \to \Hom_{\mathcal{O}'}(\mathcal{F}', \mathcal{I}\mathcal{G}') \to \Hom_{\mathcal{O}'}(\mathcal{F}', \mathcal{G}') \to \Hom_{\mathcal{O}'}(\mathcal{F}', \mathcal{G}) $$ see Homology, Lemma \ref{homology-lemma-check-exactness}.
The map $f^*\mathcal{J} \to \mathcal{I}$ is surjective as $(f, f')$ is a strict morphism of thickenings. Hence the final statement is a consequence of (2). \medskip\noindent Proof of the equivalence of (1) and (2). By definition flatness over $\mathcal{O}_\mathcal{B}$ means flatness over $f^{-1}\mathcal{O}_\mathcal{B}$. Similarly for flatness over $f^{-1}\mathcal{O}_{\mathcal{B}'}$. Note that the strictness of $(f, f')$ and the assumption that $\mathcal{F} = i^*\mathcal{F}'$ imply that $$ \mathcal{F} = \mathcal{F}'/(f^{-1}\mathcal{J})\mathcal{F}' $$ as sheaves on $\mathcal{C}$. Moreover, observe that $f^*\mathcal{J} \otimes_\mathcal{O} \mathcal{F} = f^{-1}\mathcal{J} \otimes_{f^{-1}\mathcal{O}_\mathcal{B}} \mathcal{F}$. Hence the equivalence of (1) and (2) follows from Modules on Sites, Lemma \ref{sites-modules-lemma-flat-over-thickening}.
The implication (1) $\Rightarrow$ (2) follows from Modules on Sites, Lemma \ref{sites-modules-lemma-local-pullback}. For the converse, assume $\mathcal{F}$ of finite presentation. We may and do assume that $\mathcal{C} = \mathcal{C}'$. By Lemma \ref{lemma-deform-module-ringed-topoi} we have a short exact sequence $$ 0 \to \mathcal{I} \otimes_{\mathcal{O}_X} \mathcal{F} \to \mathcal{F}' \to \mathcal{F} \to 0 $$ Let $U$ be an object of $\mathcal{C}$ such that $\mathcal{F}|_U$ has a presentation $$ \mathcal{O}_U^{\oplus m} \to \mathcal{O}_U^{\oplus n} \to \mathcal{F}|_U \to 0 $$ After replacing $U$ by the members of a covering we may assume the map $\mathcal{O}_U^{\oplus n} \to \mathcal{F}|_U$ lifts to a map $(\mathcal{O}'_U)^{\oplus n} \to \mathcal{F}'|_U$. The induced map $\mathcal{I}^{\oplus n} \to \mathcal{I} \otimes \mathcal{F}$ is surjective by right exactness of $\otimes$. Thus after replacing $U$ by the members of a covering we can find a lift $(\mathcal{O}'|_U)^{\oplus m} \to (\mathcal{O}'|_U)^{\oplus n}$ of the given map $\mathcal{O}_U^{\oplus m} \to \mathcal{O}_U^{\oplus n}$ such that $$ (\mathcal{O}'_U)^{\oplus m} \to (\mathcal{O}'_U)^{\oplus n} \to \mathcal{F}'|_U \to 0 $$ is a complex. Using right exactness of $\otimes$ once more it is seen that this complex is exact.
It is clear from the proof of Lemma \ref{lemma-inf-map-special-ringed-topoi} that the vanishing of the boundary of $\varphi$ via the map $$ \Hom_\mathcal{O}(\mathcal{F}, \mathcal{G}) = \Hom_{\mathcal{O}'}(\mathcal{F}', \mathcal{G}) \longrightarrow \Ext^1_{\mathcal{O}'}(\mathcal{F}', \mathcal{I}\mathcal{G}') $$ is a necessary and sufficient condition for the existence of a lift. We conclude as $$ \Ext^1_{\mathcal{O}'}(\mathcal{F}', \mathcal{I}\mathcal{G}') = \Ext^1_\mathcal{O}(Li^*\mathcal{F}', \mathcal{I}\mathcal{G}') $$ the adjointness of $i_* = Ri_*$ and $Li^*$ on the derived category (Cohomology on Sites, Lemma \ref{sites-cohomology-lemma-adjoint}).
[First proof] This follows from Lemma \ref{lemma-inf-obs-map-special-ringed-topoi} as we claim that under the assumptions of the lemma we have $$ \Ext^1_\mathcal{O}(Li^*\mathcal{F}', \mathcal{I}\mathcal{G}') = \Ext^1_\mathcal{O}(\mathcal{F}, \mathcal{G} \otimes_\mathcal{O} f^*\mathcal{J}) $$ Namely, we have $\mathcal{I}\mathcal{G}' = \mathcal{G} \otimes_\mathcal{O} f^*\mathcal{J}$ by Lemma \ref{lemma-deform-module-ringed-topoi}. On the other hand, observe that $$ H^{-1}(Li^*\mathcal{F}') = \text{Tor}_1^{\mathcal{O}'}(\mathcal{F}', \mathcal{O}) $$ (local computation omitted). Using the short exact sequence $$ 0 \to \mathcal{I} \to \mathcal{O}' \to \mathcal{O} \to 0 $$ we see that this $\text{Tor}_1$ is computed by the kernel of the map $\mathcal{I} \otimes_\mathcal{O} \mathcal{F} \to \mathcal{I}\mathcal{F}'$ which is zero by the final assertion of Lemma \ref{lemma-deform-module-ringed-topoi}. Thus $\tau_{\geq -1}Li^*\mathcal{F}' = \mathcal{F}$. On the other hand, we have $$ \Ext^1_\mathcal{O}(Li^*\mathcal{F}', \mathcal{I}\mathcal{G}') = \Ext^1_\mathcal{O}(\tau_{\geq -1}Li^*\mathcal{F}', \mathcal{I}\mathcal{G}') $$ by the dual of Derived Categories, Lemma \ref{derived-lemma-negative-vanishing}.
If we assume there exists one such module, then the canonical map $$ f^*\mathcal{J} \otimes_\mathcal{O} \mathcal{F} \to \mathcal{I} \otimes_\mathcal{O} \mathcal{F} $$ is an isomorphism by Lemma \ref{lemma-deform-module-ringed-topoi}. Apply Lemma \ref{lemma-inf-ext-ringed-topoi} with $\mathcal{K} =  \mathcal{I} \otimes_\mathcal{O} \mathcal{F}$ and $c = 1$. By Lemma \ref{lemma-deform-module-ringed-topoi} the corresponding extensions $\mathcal{F}'$ are all flat over $\mathcal{O}_{\mathcal{B}'}$.
Part (1) follows from Lemma \ref{lemma-inf-obs-ext-rel-ringed-topoi} as we have seen above that $\mathcal{I} = f^*\mathcal{J}$. Part (2) follows from Lemma \ref{lemma-inf-ext-rel-ringed-topoi}. Part (3) follows from Lemma \ref{lemma-inf-map-rel-ringed-topoi}.
Note that both $\pi^*\mathcal{F}$ and $h^*\mathcal{F}'_2$ restrict to $\mathcal{F}$ on $(\Sh(\mathcal{C}), \mathcal{O})$ and that the kernels of $\pi^*\mathcal{F} \to \mathcal{F}$ and $h^*\mathcal{F}'_2 \to \mathcal{F}$ are given by $f^*\mathcal{J}_1 \otimes_\mathcal{O} \mathcal{F}$. Hence flatness by Lemma \ref{lemma-deform-module-ringed-topoi}. Taking the boundary makes sense as the sequence of modules $$ 0 \to f^*\mathcal{J}_3 \otimes_\mathcal{O} \mathcal{F} \to f^*\mathcal{J}_2 \otimes_\mathcal{O} \mathcal{F} \to f^*\mathcal{J}_1 \otimes_\mathcal{O} \mathcal{F} \to 0 $$ is short exact due to the assumptions in Situation \ref{situation-ses-flat-thickenings-ringed-topoi} and the fact that $\mathcal{F}$ is flat over $\mathcal{O}_\mathcal{B}$. The statement on the obstruction class is a direct translation of the result of Remark \ref{remark-complex-thickenings-and-ses-modules-ringed-topoi} to this particular situation.
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 $\NL(\alpha) = (\mathcal{I}/\mathcal{I}^2 \to  \Omega_{\mathcal{A}[\mathcal{E}]/\mathcal{A}} \otimes_{\mathcal{A}[\mathcal{E}]} \mathcal{B})$ is canonically isomorphic to $\NL_{\mathcal{B}/\mathcal{A}}$, see Modules on Sites, Lemma \ref{sites-modules-lemma-NL-up-to-qis}. 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 $\mathcal{E} \to \Omega$. \medskip\noindent 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 $\Ext^1_\mathcal{B}(\Omega, \mathcal{G})$. 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 $\mathcal{I}' = \Ker(\mathcal{A}[\mathcal{E}'] \to \mathcal{B})$ and $\Omega' = \Omega_{\mathcal{A}[\mathcal{E}']/\mathcal{A}} \otimes_{\mathcal{A}[\mathcal{E}']} \mathcal{B}$. The pullback of $\xi$ under the quasi-isomorphism $\NL(\alpha') \to \NL(\alpha)$ maps to zero in $\Ext^1_\mathcal{B}(\Omega', \mathcal{G})$ 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.
We observe right away that given two solutions $\mathcal{O}'_1$ and $\mathcal{O}'_2$ to (\ref{equation-to-solve-ringed-topoi}) we obtain by Lemma \ref{lemma-huge-diagram-ringed-topoi} an obstruction element $o(\mathcal{O}'_1, \mathcal{O}'_2) \in \Ext^1_\mathcal{O}( \NL_{\mathcal{O}/\mathcal{O}_\mathcal{B}}, \mathcal{G})$ to the existence of a map $\mathcal{O}'_1 \to \mathcal{O}'_2$. Clearly, this element is the obstruction to the existence of an isomorphism, hence separates the isomorphism classes. To finish the proof it therefore suffices to show that given a solution $\mathcal{O}'$ and an element $\xi \in \Ext^1_\mathcal{O}( \NL_{\mathcal{O}/\mathcal{O}_\mathcal{B}}, \mathcal{G})$ we can find a second solution $\mathcal{O}'_\xi$ such that $o(\mathcal{O}', \mathcal{O}'_\xi) = \xi$. \medskip\noindent Pick $\alpha : \mathcal{E} \to \mathcal{O}$ as in Lemma \ref{lemma-NL-represent-ext-class-ringed-topoi} for the class $\xi$. Consider the surjection $f^{-1}\mathcal{O}_\mathcal{B}[\mathcal{E}] \to \mathcal{O}$ with kernel $\mathcal{I}$ and corresponding naive cotangent complex $\NL(\alpha) = (\mathcal{I}/\mathcal{I}^2 \to \Omega_{f^{-1}\mathcal{O}_\mathcal{B}[\mathcal{E}]/ f^{-1}\mathcal{O}_\mathcal{B}} \otimes_{f^{-1}\mathcal{O}_\mathcal{B}[\mathcal{E}]} \mathcal{O})$. By the lemma $\xi$ is the class of a morphism $\delta : \mathcal{I}/\mathcal{I}^2 \to \mathcal{G}$. After replacing $\mathcal{E}$ by $\mathcal{E} \times_\mathcal{O} \mathcal{O}'$ we may also assume that $\alpha$ factors through a map $\alpha' : \mathcal{E} \to \mathcal{O}'$. \medskip\noindent These choices determine an $f^{-1}\mathcal{O}_{\mathcal{B}'}$-algebra map $\varphi : \mathcal{O}_{\mathcal{B}'}[\mathcal{E}] \to \mathcal{O}'$. Let $\mathcal{I}' = \Ker(\varphi)$. Observe that $\varphi$ induces a map $\varphi|_{\mathcal{I}'} : \mathcal{I}' \to \mathcal{G}$ and that $\mathcal{O}'$ is the pushout, as in the following diagram $$ \xymatrix{ 0 \ar[r] & \mathcal{G} \ar[r] & \mathcal{O}' \ar[r] & \mathcal{O} \ar[r] & 0 \\ 0 \ar[r] & \mathcal{I}' \ar[u]^{\varphi|_{\mathcal{I}'}} \ar[r] & f^{-1}\mathcal{O}_{\mathcal{B}'}[\mathcal{E}] \ar[u] \ar[r] & \mathcal{O} \ar[u]_{=} \ar[r] & 0 } $$ Let $\psi : \mathcal{I}' \to \mathcal{G}$ be the sum of the map $\varphi|_{\mathcal{I}'}$ and the composition $$ \mathcal{I}' \to \mathcal{I}'/(\mathcal{I}')^2 \to \mathcal{I}/\mathcal{I}^2 \xrightarrow{\delta} \mathcal{G}. $$ Then the pushout along $\psi$ is an other ring extension $\mathcal{O}'_\xi$ fitting into a diagram as above. A calculation (omitted) shows that $o(\mathcal{O}', \mathcal{O}'_\xi) = \xi$ as desired.
To prove this we apply the previous results where we work over the base ringed topos $(\Sh(*), \mathbf{Z})$ with trivial thickening. Part (1) follows from Lemma \ref{lemma-choices-ringed-topoi} and the fact that there exists a solution, namely $\mathcal{J} \oplus \mathcal{O}_\mathcal{B}$. Part (2) follows from Lemma \ref{lemma-huge-diagram-ringed-topoi} and a compatibility between the constructions in the proofs of Lemmas \ref{lemma-choices-ringed-topoi} and \ref{lemma-huge-diagram-ringed-topoi} whose statement and proof we omit.
We stress that this is not a triviality. \medskip\noindent Assume (1). By More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-thickening-equivalence} the morphism $i$ induces an equivalence of small \'etale sites and in particular of topoi. Of course $i^\sharp$ is surjective with locally nilpotent kernel by definition of thickenings. \medskip\noindent Assume (2). (This direction is less important and more of a curiosity.) For any \'etale morphism $Y' \to Z'$ we see that $Y = Z \times_{Z'} Y'$ has the same \'etale topos as $Y'$. In particular, $Y'$ is quasi-compact if and only if $Y$ is quasi-compact because being quasi-compact is a topos theoretic notion (Sites, Lemma \ref{sites-lemma-quasi-compact}). Having said this we see that $Y'$ is quasi-compact and quasi-separated if and only if $Y$ is quasi-compact and quasi-separated (because you can characterize $Y'$ being quasi-separated by saying that for all $Y'_1, Y'_2$ quasi-compact algebraic spaces \'etale over $Y'$ we have that $Y'_1 \times_{Y'} Y'_2$ is quasi-compact). Take $Y'$ affine. Then the algebraic space $Y$ is quasi-compact and quasi-separated. For any quasi-coherent $\mathcal{O}_Y$-module $\mathcal{F}$ we have $H^q(Y, \mathcal{F}) = H^q(Y', (Y \to Y')_*\mathcal{F})$ because the \'etale topoi are the same. Then $H^q(Y', (Y \to Y')_*\mathcal{F}) = 0$ because the pushforward is quasi-coherent (Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-pushforward}) and $Y$ is affine. It follows that $Y'$ is affine by Cohomology of Spaces, Proposition \ref{spaces-cohomology-proposition-vanishing-affine} (there surely is a proof of this direction of the lemma avoiding this reference). Hence $i$ is an affine morphism. In the affine case it follows easily from the conditions in Section \ref{section-thickenings-ringed-topoi} that $i$ is a thickening of algebraic spaces.
We will apply the material on deformations of ringed topoi to the small \'etale topoi of the algebraic spaces in the lemma. We may think of $X$ as a closed subspace of $X'$ so that $(f, f') : (X \subset X') \to (Y \subset Y')$ is a morphism of first order thickenings. By Lemma \ref{lemma-match-thickenings} this translates into a morphism of thickenings of ringed topoi. Then we see from More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-deform} (or from the more general Lemma \ref{lemma-deform-module-ringed-topoi}) that the ideal sheaf of $X$ in $X'$ is equal to $f^*\mathcal{C}_{Y'/Y}$ and this is in fact equivalent to flatness of $X'$ over $Y'$. Hence we have a commutative diagram $$ \xymatrix{ 0 \ar[r] & f^*\mathcal{C}_{Y/Y'} \ar[r] & \mathcal{O}_{X'} \ar[r] & \mathcal{O}_X \ar[r] & 0 \\ 0 \ar[r] & f_{small}^{-1}\mathcal{C}_{Y/Y'} \ar[u] \ar[r] & f_{small}^{-1}\mathcal{O}_{Y'} \ar[u] \ar[r] & f_{small}^{-1}\mathcal{O}_Y \ar[u] \ar[r] & 0 } $$ Please compare with (\ref{equation-to-solve-ringed-topoi}). Observe that automorphisms $\varphi$ as in (2) give automorphisms $\varphi^\sharp : \mathcal{O}_{X'} \to \mathcal{O}_{X'}$ fitting in the diagram above. Conversely, an automorphism $\alpha : \mathcal{O}_{X'} \to \mathcal{O}_{X'}$ fitting into the diagram of sheaves above is equal to $\varphi^\sharp$ for some automorphism $\varphi$ as in (2) by More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-first-order-thickening-maps}. Finally, by More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-first-order-thickening} if we find another sheaf of rings $\mathcal{A}$ on $X_\etale$ fitting into the diagram $$ \xymatrix{ 0 \ar[r] & f^*\mathcal{C}_{Y/Y'} \ar[r] & \mathcal{A} \ar[r] & \mathcal{O}_X \ar[r] & 0 \\ 0 \ar[r] & f_{small}^{-1}\mathcal{C}_{Y/Y'} \ar[u] \ar[r] & f_{small}^{-1}\mathcal{O}_{Y'} \ar[u] \ar[r] & f_{small}^{-1}\mathcal{O}_Y \ar[u] \ar[r] & 0 } $$ then there exists a first order thickening $X \subset X''$ with $\mathcal{O}_{X''} = \mathcal{A}$ and applying  More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-first-order-thickening-maps} once more, we obtain a morphism $(f, f'') : (X \subset X'') \to (Y \subset Y')$ with all the desired properties. Thus part (1) follows from Lemma \ref{lemma-choices-ringed-topoi} and part (2) from part (2) of Lemma \ref{lemma-huge-diagram-ringed-topoi}. (Note that $\NL_{X/Y}$ as defined for a morphism of algebraic spaces in More on Morphisms of Spaces, Section \ref{spaces-more-morphisms-section-netherlander} agrees with $\NL_{X/Y}$ as used in Section \ref{section-deformations-ringed-topoi}.)
Recall that $\NL_{X/B}$ is an object of $D_\QCoh(\mathcal{O}_X)$, see More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-netherlander-quasi-coherent}. Hence our assumption implies the maps $$ \Ext^i_X(\NL_{X/B}, \mathcal{F}) \longrightarrow \Ext^i_X(\NL_{X/B}, \mathcal{G}) $$ are isomorphisms for all $i$. This implies our functor is fully faithful by Lemma \ref{lemma-huge-diagram-ringed-topoi}. On the other hand, the functor is essentially surjective by Lemma \ref{lemma-choices-ringed-topoi} because we have the solutions $\mathcal{O}_X \oplus \mathcal{F}$ and $\mathcal{O}_X \oplus \mathcal{G}$ in both categories.
A solution of (\ref{equation-to-solve-ringed-topoi}) for $\mathcal{F}$ in particular gives an extension of $f^{-1}\mathcal{O}_{B'}$-algebras $$ 0 \to \mathcal{F} \to \mathcal{O}' \to \mathcal{O}_X \to 0 $$ where $\mathcal{F}$ is an ideal of square zero. Similarly for $\mathcal{G}$. Moreover, given such an extension, we obtain a map $c_{\mathcal{O}'} : f^{-1}\mathcal{J} \to \mathcal{F}$. Thus we are looking at the full subcategory of such extensions of $f^{-1}\mathcal{O}_{B'}$-algebras with $c = c_{\mathcal{O}'}$. Clearly, if $\mathcal{O}'' = F(\mathcal{O}')$ where $F$ is the equivalence of Lemma \ref{lemma-thickening-space-quasi-coherent} (applied to $X \to B'$ this time), then $c_{\mathcal{O}''}$ is the composition of $c_{\mathcal{O}'}$ and the map $\mathcal{F} \to \mathcal{G}$. This proves the lemma.
Choose a distinguished triangle $$ K \to L \to C \to K[1] $$ In case (2) we see that $H^i(C) = 0$ for $i \geq n$. Hence $H^i(C \otimes_\mathcal{O}^\mathbf{L} \mathcal{F}) = 0$ for $i \geq n$ by (the dual of) Derived Categories, Lemma \ref{derived-lemma-negative-vanishing}. This in turn shows that $H^i(\alpha \otimes_\mathcal{O}^\mathbf{L} \text{id}_\mathcal{F})$ is an isomorphism for $i > n$ and surjective for $i = n$. In case (1) we moreover see that $H^{n - 1}(L) \to H^{n - 1}(C)$ is surjective. Considering the diagram $$ \xymatrix{ H^{n - 1}(L) \otimes_\mathcal{O} \mathcal{F} \ar[r] \ar[d] & H^{n - 1}(C) \otimes_\mathcal{O} \mathcal{F} \ar@{=}[d] \\ H^{n - 1}(L \otimes_\mathcal{O}^\mathbf{L} \mathcal{F}) \ar[r] & H^{n - 1}(C \otimes_\mathcal{O}^\mathbf{L} \mathcal{F}) } $$ we conclude the lower horizontal arrow is surjective. Combined with what was said before this implies that $H^n(\alpha \otimes_\mathcal{O}^\mathbf{L} \text{id}_\mathcal{F})$ is an isomorphism.
Set $\mathcal{F}_0^\bullet = \mathcal{F}^\bullet \otimes_\mathcal{O} \mathcal{O}_0$. By Derived Categories, Lemma \ref{derived-lemma-make-surjective} there exists a factorization $$ \mathcal{K}_0^\bullet \to \mathcal{L}_0^\bullet \to \mathcal{F}_0^\bullet $$ of the given map such that the first arrow has an inverse up to homotopy and the second arrow is termwise split surjective. Hence we may assume that $\mathcal{K}_0^\bullet \to \mathcal{F}_0^\bullet$ is termwise surjective. In that case we take $$ \mathcal{G}^n = \mathcal{F}^n \times_{\mathcal{F}^n_0} \mathcal{K}_0^n $$ and everything is clear.
Finding $\alpha : K \to L$ lifing $\alpha_0$ is the same as finding $\alpha : K \to L$ such that the composition $K \xrightarrow{\alpha} L \to L_0$ is equal to the composition $K \to K_0 \xrightarrow{\alpha_0} L_0$. The short exact sequence $0 \to \mathcal{I} \to \mathcal{O} \to \mathcal{O}_0 \to 0$ gives rise to a canonical distinguished triangle $$ L \otimes_\mathcal{O}^\mathbf{L} \mathcal{I} \to L \to L_0 \to (L \otimes_\mathcal{O}^\mathbf{L} \mathcal{I})[1] $$ in $D(\mathcal{O})$. By Derived Categories, Lemma \ref{derived-lemma-representable-homological} the composition $$ K \to K_0 \xrightarrow{\alpha_0} L_0 \to (L \otimes_\mathcal{O}^\mathbf{L} \mathcal{I})[1] $$ is zero if and only if we can find $\alpha : K \to L$ lifting $\alpha_0$. The composition is an element in $$ \Hom_{D(\mathcal{O})}(K, (L \otimes_\mathcal{O}^\mathbf{L} \mathcal{I})[1]) = \Hom_{D(\mathcal{O}_0)}(K_0, (L \otimes_\mathcal{O}^\mathbf{L} \mathcal{I})[1]) = \Ext^1_{\mathcal{O}_0}(K_0, L_0 \otimes_{\mathcal{O}_0}^\mathbf{L} \mathcal{I}) $$ by adjunction.
We will prove this by using the explicit description of the quotient stacks given in Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-quotient-stack-objects}. However, we strongly urge the reader to find their own proof. First, we may view $(U, R, s, t, c)$ as a groupoid in algebraic spaces over $X$, hence we obtain a map $f : [U/R] \to \mathcal{S}_X$, see Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-quotient-stack-arrows}. Similarly, we have $f' : [U'/R'] \to X'$. \medskip\noindent An object of the $2$-fibre product $\mathcal{S}_{X'} \times_{\mathcal{S}_X} [U/R]$ over a scheme $T$ over $S$ is the same as a morphism $x' : T \to X'$ and an object $y$ of $[U/R]$ over $T$ such that such that the composition $g \circ x'$ is equal to $f(y)$. This makes sense because objects of $\mathcal{S}_X$ over $T$ are morphisms $T \to X$. By Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-quotient-stack-objects} we may assume $y$ is given by a $[U/R]$-descent datum $(u_i, r_{ij})$ relative to an fppf covering $\{T_i \to T\}$. The agreement of $g \circ x' = f(y)$ means that the diagrams $$ \vcenter{ \xymatrix{ T_i \ar[rr]_{u_i} \ar[d] & & U \ar[d] \\ T \ar[r]^{x'} & X' \ar[r]^g & X } } \quad\text{and}\quad \vcenter{ \xymatrix{ T_i \times_T T_j \ar[rr]_{r_{ij}} \ar[d] & & R \ar[d] \\ T \ar[r]^{x'} & X' \ar[r]^g & X } } $$ are commutative. \medskip\noindent On the other hand, an object $y'$ of $[U'/R']$ over a scheme $T$ over $S$ by Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-quotient-stack-objects} is given by a $[U'/R']$-descent datum $(u'_i, r'_{ij})$ relative to an fppf covering $\{T_i \to T\}$. Setting $f'(y') = x' : T \to X'$ we see that the diagrams $$ \vcenter{ \xymatrix{ T_i \ar[r]_{u'_i} \ar[d] & U' \ar[d] \\ T \ar[r]^{x'} & X' } } \quad\text{and}\quad \vcenter{ \xymatrix{ T_i \times_T T_j \ar[r]_{r'_{ij}} \ar[d] & U' \ar[d] \\ T \ar[r]^{x'} & X' } } $$ are commutative. \medskip\noindent With this notation in place, we define a functor $$ [U'/R'] \longrightarrow \mathcal{S}_{X'} \times_{\mathcal{S}_X} [U/R] $$ by sending $y' = (u'_i, r'_{ij})$ as above to the object $(x', (u_i, r_{ij}))$ where $x' = f'(y')$, where $u_i$ is the composition $T_i \to U' \to U$, and where $r_{ij}$ is the composition $T_i \times_T T_j \to R' \to R$. Conversely, given an object $(x', (u_i, r_{ij})$ of the right hand side we can send this to the object $((x', u_i), (x', r_{ij}))$ of the left hand side. We omit the discussion of what to do with morphisms (works in exactly the same manner).