The Stacks project

Proof. The proof of this lemma is identical to the proof of Lemma 91.7.1. We urge the reader to read that proof instead of this one. We will identify the underlying topoi for every thickening in sight (we have already used this convention in the statement). The equalities in the last statement of the lemma are immediate from the definitions. Thus we will work with the groups $\mathop{\mathrm{Ext}}\nolimits ^ k_{\mathcal{O}_2}( Lg^*\mathop{N\! L}\nolimits _{\mathcal{O}_1/\mathcal{O}_{\mathcal{B}_1}}, \mathcal{G}_2)$, $k = 0, 1$ in the rest of the proof. We first argue that we can reduce to the case where the underlying topos of all ringed topoi in the lemma is the same.

To do this, observe that $g^{-1}\mathop{N\! L}\nolimits _{\mathcal{O}_1/\mathcal{O}_{\mathcal{B}_1}}$ is equal to the naive cotangent complex of the homomorphism of sheaves of rings $g^{-1}f_1^{-1}\mathcal{O}_{\mathcal{B}_1} \to g^{-1}\mathcal{O}_1$, see Modules on Sites, Lemma 18.33.5. Moreover, the degree $0$ term of $\mathop{N\! L}\nolimits _{\mathcal{O}_1/\mathcal{O}_{\mathcal{B}_1}}$ is a flat $\mathcal{O}_1$-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 topoi $(\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'_2), \mathcal{O}'_2) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'_1), \mathcal{O}'_1)$ fitting into (91.13.1.1) compatibly with $\nu $ is in one-to-one bijection with the set of homomorphisms of $g^{-1}f_1^{-1}\mathcal{O}_{\mathcal{B}'_1}$-algebras $g^{-1}\mathcal{O}'_1 \to \mathcal{O}'_2$ which are compatible with $f^\sharp $ and $\nu $. In this way we see that we may assume we have a diagram (91.13.1.2) of sheaves on a site $\mathcal{C}$ (with $f_1 = f_2 = \text{id}$ on underlying topoi) and we are looking to find a homomorphism of sheaves of rings $\mathcal{O}'_1 \to \mathcal{O}'_2$ 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 (91.13.1.2) of sheaves on a site $\mathcal{C}$ (with $f_1 = f_2 = \text{id}$ on underlying topoi) and we are looking for homomorphisms of sheaves of rings $\mathcal{O}'_1 \to \mathcal{O}'_2$ fitting into it. As ext groups we will use $\mathop{\mathrm{Ext}}\nolimits ^ k_{\mathcal{O}_1}( \mathop{N\! L}\nolimits _{\mathcal{O}_1/\mathcal{O}_{\mathcal{B}_1}}, \mathcal{G}_2)$, $k = 0, 1$.

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

This comes with a surjective map $\alpha : \mathcal{E} \to \mathcal{O}_1$ and hence we can use $\mathop{N\! L}\nolimits (\alpha )$ instead of $\mathop{N\! L}\nolimits _{\mathcal{O}_1/\mathcal{O}_{\mathcal{B}_1}}$, see Modules on Sites, Lemma 18.35.2. Set

There is a surjection $\mathcal{I}' \to \mathcal{I}$ whose kernel is $\mathcal{J}_1\mathcal{O}_{\mathcal{B}'_1}[\mathcal{E}]$. We obtain two homomorphisms of $\mathcal{O}_{\mathcal{B}'_2}$-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 $\mathcal{J}_1\mathcal{O}_{\mathcal{B}'_1}[\mathcal{E}]$ as maps into $\mathcal{G}_2$ because the left hand square of (91.13.1.2) is commutative. Thus the difference $b|_{\mathcal{I}'} - \nu \circ a|_{\mathcal{I}'}$ induces a well defined $\mathcal{O}_1$-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 $[\xi ] \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_1}(\mathop{N\! L}\nolimits (\alpha ), \mathcal{G}_2)$ be the image (see below).

Step 2. Vanishing of $[\xi ]$ is necessary. Let us write $\Omega = \Omega _{\mathcal{O}_{\mathcal{B}_1}[\mathcal{E}]/\mathcal{O}_{\mathcal{B}_1}} \otimes _{\mathcal{O}_{\mathcal{B}_1}[\mathcal{E}]} \mathcal{O}_1$. 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 $\varphi : \mathcal{O}'_1 \to \mathcal{O}'_2$ fitting into (91.13.1.2). In this case consider the map $\mathcal{O}'_1[\mathcal{E}] \to \mathcal{G}_2$, $f' \mapsto b(f') - \varphi (a(f'))$. A calculation shows this annihilates $\mathcal{J}_1\mathcal{O}_{\mathcal{B}'_1}[\mathcal{E}]$ and induces a derivation $\mathcal{O}_{\mathcal{B}_1}[\mathcal{E}] \to \mathcal{G}_2$. 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 $\mathcal{O}_1$-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 $\mathcal{O}'_1 \to \mathcal{O}'_2$ fitting into (91.13.1.2).

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. $\square$

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

This finishes the proof. $\square$

Proof. We observe right away that given two solutions $\mathcal{O}'_1$ and $\mathcal{O}'_2$ to (91.13.0.1) we obtain by Lemma 91.13.1 an obstruction element $o(\mathcal{O}'_1, \mathcal{O}'_2) \in \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}( \mathop{N\! L}\nolimits _{\mathcal{O}/\mathcal{O}_\mathcal {B}}, \mathcal{G})$ 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 $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}( \mathop{N\! L}\nolimits _{\mathcal{O}/\mathcal{O}_\mathcal {B}}, \mathcal{G})$ we can find a second solution $\mathcal{O}'_\xi $ such that $o(\mathcal{O}', \mathcal{O}'_\xi ) = \xi $.

Pick $\alpha : \mathcal{E} \to \mathcal{O}$ as in Lemma 91.13.2 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 $\mathop{N\! L}\nolimits (\alpha ) = (\mathcal{I}/\mathcal{I}^2 \to \Omega _{f^{-1}\mathcal{O}_\mathcal {B}[\mathcal{E}]/ f^{-1}\mathcal{O}_\mathcal {B}} \otimes _{f^{-1}\mathcal{O}_\mathcal {B}[\mathcal{E}]} \mathcal{O})$. 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 $\alpha ' : \mathcal{E} \to \mathcal{O}'$.

These choices determine an $f^{-1}\mathcal{O}_{\mathcal{B}'}$-algebra map $\varphi : \mathcal{O}_{\mathcal{B}'}[\mathcal{E}] \to \mathcal{O}'$. Let $\mathcal{I}' = \mathop{\mathrm{Ker}}(\varphi )$. 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

Let $\psi : \mathcal{I}' \to \mathcal{G}$ be the sum of the map $\varphi |_{\mathcal{I}'}$ and the composition

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. $\square$

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

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.) $\square$

Proof. The stament makes sense as we have the maps

using the map $Lg^*\mathcal{F} \to g^*\mathcal{F} \xrightarrow {c} \mathcal{G}$ and

using the map $Lg^*\mathop{N\! L}\nolimits _{\mathcal{O}_\mathcal {C}/\mathcal{O}_\mathcal {B}} \to \mathop{N\! L}\nolimits _{\mathcal{O}_\mathcal {D}/\mathcal{O}_\mathcal {B}}$. The statement of the lemma can be deduced from Lemma 91.13.1 applied to the diagram

and a compatibility between the constructions in the proofs of Lemmas 91.13.4 and 91.13.1 whose statement and proof we omit. (See remark below for a direct argument.) $\square$

Proof. By Lemma 91.13.4 applied to $t \circ f : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{B}'), \mathcal{O}_{\mathcal{B}'})$ and the $\mathcal{O}$-module $\mathcal{G}$ we see that elements $\zeta $ of $\mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}(\mathop{N\! L}\nolimits _{\mathcal{O}/\mathcal{O}_{\mathcal{B}'}}, \mathcal{G})$ parametrize extensions $0 \to \mathcal{G} \to \mathcal{O}' \to \mathcal{O} \to 0$ of $f^{-1}\mathcal{O}_{\mathcal{B}'}$-algebras. By Lemma 91.13.6 applied to

and $c : \mathcal{J} \to \mathcal{G}$ we see that there is an morphism

over $(\mathop{\mathit{Sh}}\nolimits (\mathcal{B}'), \mathcal{O}_{\mathcal{B}'})$ compatible with $c$ and $f$ if and only if $\zeta $ maps to $\xi $. Of course this is the same thing as saying $\mathcal{O}'$ is a solution of (91.13.0.1). $\square$