The Stacks project

Proof. The naive cotangent complex $\mathop{N\! L}\nolimits _{X_1/S_1}$ is defined in Modules, Definition 17.31.6. The equalities in the last statement of the lemma follow from the fact that $g^*$ is adjoint to $g_*$, the fact that $H^0(\mathop{N\! L}\nolimits _{X_1/S_1}) = \Omega _{X_1/S_1}$ (by construction of the naive cotangent complex) and the fact that $Lg^*$ is the left derived functor of $g^*$. Thus we will work with the groups $\mathop{\mathrm{Ext}}\nolimits ^ k_{\mathcal{O}_{X_2}}(Lg^*\mathop{N\! L}\nolimits _{X_1/S_1}, \mathcal{G}_2)$, $k = 0, 1$ in the rest of the proof. We first argue that we can reduce to the case where the underlying topological spaces of all ringed spaces in the lemma is the same.

To do this, observe that $g^{-1}\mathop{N\! L}\nolimits _{X_1/S_1}$ is equal to the naive cotangent complex of the homomorphism of sheaves of rings $g^{-1}f_1^{-1}\mathcal{O}_{S_1} \to g^{-1}\mathcal{O}_{X_1}$, see Modules, Lemma 17.31.3. Moreover, the degree $0$ term of $\mathop{N\! L}\nolimits _{X_1/S_1}$ is a flat $\mathcal{O}_{X_1}$-module, hence the canonical map

induces an isomorphism on cohomology sheaves in degrees $0$ and $-1$. Thus we may replace the Ext groups of the lemma with

The set of morphism of ringed spaces $X'_2 \to X'_1$ fitting into (91.7.1.1) compatibly with $\nu $ is in one-to-one bijection with the set of homomorphisms of $g^{-1}f_1^{-1}\mathcal{O}_{S'_1}$-algebras $g^{-1}\mathcal{O}_{X'_1} \to \mathcal{O}_{X'_2}$ which are compatible with $f^\sharp $ and $\nu $. In this way we see that we may assume we have a diagram (91.7.1.2) of sheaves on $X$ and we are looking to find a homomorphism of sheaves of rings $\mathcal{O}_{X'_1} \to \mathcal{O}_{X'_2}$ fitting into it.

In the rest of the proof of the lemma we assume all underlying topological spaces are the same, i.e., we have a diagram (91.7.1.2) of sheaves on a space $X$ and we are looking for homomorphisms of sheaves of rings $\mathcal{O}_{X'_1} \to \mathcal{O}_{X'_2}$ fitting into it. As ext groups we will use $\mathop{\mathrm{Ext}}\nolimits ^ k_{\mathcal{O}_{X_1}}( \mathop{N\! L}\nolimits _{\mathcal{O}_{X_1}/\mathcal{O}_{S_1}}, \mathcal{G}_2)$, $k = 0, 1$.

Step 1. Construction of the obstruction class. Consider the sheaf of sets

This comes with a surjective map $\alpha : \mathcal{E} \to \mathcal{O}_{X_1}$ and hence we can use $\mathop{N\! L}\nolimits (\alpha )$ instead of $\mathop{N\! L}\nolimits _{\mathcal{O}_{X_1}/\mathcal{O}_{S_1}}$, see Modules, Lemma 17.31.2. Set

There is a surjection $\mathcal{I}' \to \mathcal{I}$ whose kernel is $\mathcal{J}_1\mathcal{O}_{S'_1}[\mathcal{E}]$. We obtain two homomorphisms of $\mathcal{O}_{S'_2}$-algebras

which induce maps $a|_{\mathcal{I}'} : \mathcal{I}' \to \mathcal{G}_1$ and $b|_{\mathcal{I}'} : \mathcal{I}' \to \mathcal{G}_2$. Both $a$ and $b$ annihilate $(\mathcal{I}')^2$. Moreover $a$ and $b$ agree on $\mathcal{J}_1\mathcal{O}_{S'_1}[\mathcal{E}]$ as maps into $\mathcal{G}_2$ because the left hand square of (91.7.1.2) is commutative. Thus the difference $b|_{\mathcal{I}'} - \nu \circ a|_{\mathcal{I}'}$ induces a well defined $\mathcal{O}_{X_1}$-linear map

which sends the class of a local section $f$ of $\mathcal{I}$ to $a(f') - \nu (b(f'))$ where $f'$ is a lift of $f$ to a local section of $\mathcal{I}'$. We let $[\xi ] \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_{X_1}}(\mathop{N\! L}\nolimits (\alpha ), \mathcal{G}_2)$ be the image (see below).

Step 2. Vanishing of $[\xi ]$ is necessary. Let us write $\Omega = \Omega _{\mathcal{O}_{S_1}[\mathcal{E}]/\mathcal{O}_{S_1}} \otimes _{\mathcal{O}_{S_1}[\mathcal{E}]} \mathcal{O}_{X_1}$. Observe that $\mathop{N\! L}\nolimits (\alpha ) = (\mathcal{I}/\mathcal{I}^2 \to \Omega )$ fits into a distinguished triangle

Thus we see that $[\xi ]$ is zero if and only if $\xi $ is a composition $\mathcal{I}/\mathcal{I}^2 \to \Omega \to \mathcal{G}_2$ for some map $\Omega \to \mathcal{G}_2$. Suppose there exists a homomorphisms of sheaves of rings $\varphi : \mathcal{O}_{X'_1} \to \mathcal{O}_{X'_2}$ fitting into (91.7.1.2). In this case consider the map $\mathcal{O}_{S'_1}[\mathcal{E}] \to \mathcal{G}_2$, $f' \mapsto b(f') - \varphi (a(f'))$. A calculation shows this annihilates $\mathcal{J}_1\mathcal{O}_{S'_1}[\mathcal{E}]$ and induces a derivation $\mathcal{O}_{S_1}[\mathcal{E}] \to \mathcal{G}_2$. The resulting linear map $\Omega \to \mathcal{G}_2$ witnesses the fact that $[\xi ] = 0$ in this case.

Step 3. Vanishing of $[\xi ]$ is sufficient. Let $\theta : \Omega \to \mathcal{G}_2$ be a $\mathcal{O}_{X_1}$-linear map such that $\xi $ is equal to $\theta \circ (\mathcal{I}/\mathcal{I}^2 \to \Omega )$. Then a calculation shows that

annihilates $\mathcal{I}'$ and hence defines a map $\mathcal{O}_{X'_1} \to \mathcal{O}_{X'_2}$ fitting into (91.7.1.2).

Proof of (2) in the special case above. Omitted. Hint: This is exactly the same as the proof of (2) of Lemma 91.2.1. $\square$