The Stacks project

Remark 91.7.7. Let $f : (X, \mathcal{O}_ X) \to (S, \mathcal{O}_ S)$, $g : (Y, \mathcal{O}_ Y) \to (X, \mathcal{O}_ X)$, $\mathcal{F}$, $\mathcal{G}$, $c : \mathcal{F} \to \mathcal{G}$, $0 \to \mathcal{F} \to \mathcal{O}_{X'} \to \mathcal{O}_ X \to 0$, $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathop{N\! L}\nolimits _{X/S}, \mathcal{F})$, $0 \to \mathcal{G} \to \mathcal{O}_{Y'} \to \mathcal{O}_ Y \to 0$, and $\zeta \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ Y}(\mathop{N\! L}\nolimits _{Y/S}, \mathcal{G})$ be as in Lemma 91.7.6. Using pushout along $c : g^{-1}\mathcal{F} \to \mathcal{G}$ we can construct an extension

Using pullback along $g^\sharp : g^{-1}\mathcal{O}_ X \to \mathcal{O}_ Y$ we can construct an extension

A diagram chase tells us that there exists an $S$-morphism $Y' \to X'$ compatible with $g$ and $c$ if and only if $\mathcal{O}'_1$ is isomorphic to $\mathcal{O}'_2$ as $g^{-1}f^{-1}\mathcal{O}_ S$-algebra extensions of $g^{-1}\mathcal{O}_ X$ by $\mathcal{G}$. By Lemma 91.7.4 these extensions are classified by the LHS of

Here the equality comes from tensor-hom adjunction and the equalities

For the first of these see Modules, Lemma 17.31.3; the second follows from the definition of derived pullback. Thus, in order to see that Lemma 91.7.6 is true, it suffices to show that $\mathcal{O}'_1$ corresponds to the image of $\xi$ and that $\mathcal{O}'_2$ correspond to the image of $\zeta$. The correspondence between $\xi$ and $\mathcal{O}'_1$ is immediate from the construction of the class $\xi$ in Remark 91.7.5. For the correspondence between $\zeta$ and $\mathcal{O}'_2$, we first choose a commutative diagram

such that $g^{-1}f^{-1}\mathcal{O}_ S[\mathcal{E}] \to \mathcal{O}_ Y$ is surjective with kernel $\mathcal{K}$. Next choose a commutative diagram

such that $g^{-1}f^{-1}\mathcal{O}_ S[\mathcal{E}'] \to g^{-1}\mathcal{O}_ X$ is surjective with kernel $\mathcal{J}$. (For example just take $\mathcal{E}' = \mathcal{E} \amalg \mathcal{O}'_2$ as a sheaf of sets.) The map $\varphi$ induces a map of complexes $\mathop{N\! L}\nolimits (\alpha ) \to \mathop{N\! L}\nolimits (\beta )$ (notation as in Modules, Section 17.31) and in particular $\bar\varphi : \mathcal{J}/\mathcal{J}^2 \to \mathcal{K}/\mathcal{K}^2$. Then $\mathop{N\! L}\nolimits (\alpha ) \cong \mathop{N\! L}\nolimits _{Y/S}$ and $\mathop{N\! L}\nolimits (\beta ) \cong \mathop{N\! L}\nolimits _{g^{-1}\mathcal{O}_ X/g^{-1}f^{-1}\mathcal{O}_ S}$ and the map of complexes $\mathop{N\! L}\nolimits (\alpha ) \to \mathop{N\! L}\nolimits (\beta )$ represents the map $Lg^*\mathop{N\! L}\nolimits _{X/S} \to \mathop{N\! L}\nolimits _{Y/S}$ used in the statement of Lemma 91.7.6 (see first part of its proof). Now $\zeta$ corresponds to the class of the map $\mathcal{K}/\mathcal{K}^2 \to \mathcal{G}$ induced by $\beta '$, see Remark 91.7.5. Similarly, the extension $\mathcal{O}'_2$ corresponds to the map $\mathcal{J}/\mathcal{J}^2 \to \mathcal{G}$ induced by $\alpha '$. The commutative diagram above shows that this map is the composition of the map $\mathcal{K}/\mathcal{K}^2 \to \mathcal{G}$ induced by $\beta '$ with the map $\bar\varphi : \mathcal{J}/\mathcal{J}^2 \to \mathcal{K}/\mathcal{K}^2$. This proves the compatibility we were looking for.