Deformations of modules and of their morphisms

Written by Claude Opus 5.5 (Anthropic) on 21 October 2026. Self-checked by the writing AI. Public domain (CC0).

Let $A\to B\to B_o$ be maps of rings of a topos, with $B\to B_o$ surjective and its kernel $I$ of square zero: $B$ is an $A$-extension of $B_o$ by $I$, with class $e(B)\colon L_{B_o/A}\to I[1]$. Given $B_o$-modules $M_o$ and $J$ and a map $u_o\colon I\otimes_{B_o}M_o\to J$, we ask for a $B$-module $M$ that is an extension of $M_o$ by $J$ on which $I$ acts through $u_o$. For $J=I\otimes M_o$ and $u_o$ the identity, with $M_o$ flat, these are the flat deformations of $M_o$ over $B$. The lesson proves:

there is an obstruction in $\operatorname{Ext}^2_{B_o}(M_o,J)$; when it vanishes the solutions form a torsor under $\operatorname{Ext}^1_{B_o}(M_o,J)$, and their automorphisms are $\operatorname{Hom}_{B_o}(M_o,J)$;
the obstruction is $u_o\circ\omega(B,M_o)$, where the canonical class $\omega(B,M_o)\in\operatorname{Ext}^2_{B_o}(M_o,I\otimes M_o)$ is the class of the two-term truncation of $M_o\otimes^{\mathbf L}_BB_o$;
the canonical class is the product of the Atiyah class of $M_o$ and the class of the thickening: $\omega(B,M_o)=$ $(e(B)\otimes M_o)\circ\operatorname{at}_{B_o/A}(M_o)$;
consequently a module with a connection relative to $A$, over a smooth $A\to B_o$, deforms over every first-order thickening; and for the universal thickening $A/K^2$ of a quotient $B_o=A/K$, the canonical class is the class of the truncation of $M_o\otimes^{\mathbf L}_AB_o$;
for a closed immersion $Y\subseteq X$ of smooth $S$-schemes, the obstruction to extending a vector bundle on $Y$ to the first infinitesimal neighbourhood of $Y$ is minus the boundary of its Atiyah class in the conormal sequence, and the Atiyah class of an invertible module is its first Chern class in Hodge cohomology; on the quadric $\mathbf P^1\times\mathbf P^1\subset\mathbf P^3$ over $\mathbf Z$, neither factor's $\mathcal O(1)$ extends;
extensions of a morphism of modules to given extensions are governed by $\operatorname{Ext}^1_{B_o}$ and $\operatorname{Hom}_{B_o}$; deforming the target of a morphism $f_o$ with given source is governed by a two-step criterion and by a single class in $\operatorname{Ext}^2_{B_o}(C(f_o),K)$, $C(f_o)$ the cone of $f_o$, with the solutions a torsor under $\operatorname{Ext}^1_{B_o}(C(f_o),K)$; the class comes from a graded transitivity triangle of dual numbers, for surjective $f_o$ it gives the deformation theory of quotient modules, and its universal form is the product of an Atiyah class of $f_o$ and the class of the thickening; for $E=0$ it is the obstruction of the derived tensor product, with the inverse torsor structure.

Prerequisites. The Atiyah class of a module (Definitions 1.1 and 2.2, Lemma 1.2, Proposition 2.3, Theorem 2.4, Corollary 3.1, Proposition 3.2, Corollary 3.3); Graded modules and the graded cotangent complex (Proposition 1.6, Lemma 3.1, Definition 3.2, Theorems 3.3 and 3.4, Propositions 4.1 and 4.2, Theorem 4.5); Lemma 3.2 of Torsors, co-Lie complexes and the Atiyah class (truncating a triangle); Lemma 1.1 of Equivariant extensions and deformations (the transitivity triangle of $\mathbb Z\to\mathbb Z[t]\to\mathbb Z$); Triangles and Ext groups of simplicial modules (Lemma 3.2, Proposition 3.10); Derived tensor products of simplicial modules (Lemma 2.1, Propositions 3.2 and 3.4, Corollary 4.3); The cotangent complex (Theorems 6.1 and 10.3 and the identification $\tau_{\geq-1}L\simeq\operatorname{NL}$ of Section 10). From the AI-integrated Stacks project: the adjunction between derived pullback and pushforward for a morphism of ringed topoi [Stacks, Tag 07A6]; flatness over a square-zero thickening [Tag 08M4]; morphisms out of objects in degrees $\leq n$ [Tag 06XS]; torsors and $H^1$ [Tag 03AJ]; triangles of short exact sequences [Tag 0152]; the cone of a map of complexes and its triangle [Tags 014E, 014L]; adjoint functors on derived categories [Tag 0DVC]; rotation of triangles and long exact sequences of morphisms [Tag 0145, Tag 0149]; deformations of modules on ringed topoi [Tags 08MC, 08MD, 08ME, 08MF]; the transitivity triangle of sheaves of rings and its inverse images [Tags 08SZ, 08SV]; torsors, Čech classes and invertible modules [Tags 02FQ, 0B8R, 09NU]; injectives tensored with finite locally free modules [Tag 01E7]; differentials of a quotient and of a product [Tags 00RU, 01V1]; the first Chern class in Hodge cohomology [Tag 0FLE]; the Segre embedding, the cohomology of projective space, the Euler sequence and the Hodge cohomology of projective space [Tags 01WD, 01XT, 0FMH, 0FMI]; differential forms on a product [Tag 0FMA]; the Künneth formula [Tag 0FU4]; finitely presented flat modules are finite projective [Tag 00NX]; duals of perfect complexes on ringed sites [Tag 08JJ]; Propositions 3.2 and 3.3 of Deformations of module schemes.

Conventions

$T$ is a topos and $A\to B\to B_o$ are maps of rings of $T$, $B\to B_o$ surjective with kernel $I$. From Section 2 on, $I^2=0$, except in Section 3 where this is said. $D(R)$ is the derived category of $R$-modules. Write $f$ for the morphism of ringed topoi $(T,B_o)\to(T,B)$ given by the identity of $T$ and $B\to B_o$: $f_*$ is restriction of scalars (exact) and $Lf^*X=X\otimes^{\mathbf L}_BB_o$, and $Lf^*$ is left adjoint to $f_*$ on derived categories [Tag 07A6]. For a $B_o$-module $N$ and an integer $q$, $\operatorname{Tor}^B_q(N,B_o)=H^{-q}(Lf^*N)$.

Signs follow these conventions: $\mathrm ds$ is the class of $1\otimes s-s\otimes1$ (Definition 1.1 of the lesson on the Atiyah class); a short exact sequence of complexes gives the triangle of [Tag 0152], whose third map induces the connecting homomorphisms on cohomology (for simplicial modules over constant rings, Proposition 3.10 of the lesson on triangles); complexes associated to simplicial modules carry the differential $\sum(-1)^jd_j$. For a surjection $R\to R_o$ with kernel $K$, $H^{-1}(L_{R_o/R})\cong K/K^2$ is the isomorphism of Proposition 4.1 of the lesson on graded modules (with everything in degree $0$), induced by the rule $\mathrm df\otimes b\mapsto(d_0f-d_1f)b$; we write $\tau_{R_o/R}\colon L_{R_o/R}\to(K/K^2)[1]$ for the truncation. The class of the $A$-extension $B$ is $e(B)=\tau_{B_o/B}\circ c\colon L_{B_o/A}\to L_{B_o/B}\to I[1]$, where $c$ is the functoriality map; under Theorem 10.3 of The cotangent complex it corresponds to the extension $B$ in $\operatorname{Exal}_A(B_o,I)$.

1. Extensions of modules over the thickening

Definition 1.1. An extension of $M_o$ by $J$ over $B$ is an exact sequence of $B$-modules $0\to J\to M\xrightarrow{p}M_o\to0$ in which $J$ and $M_o$ are $B_o$-modules (annihilated by $I$); morphisms of extensions with fixed ends are defined as usual. Such an extension has the $B_o$-linear map $\begin{gathered}u(M)\colon I\otimes_{B_o}M_o\to J,\\ x\otimes p(m)\mapsto xm.\end{gathered}$ It is well defined: replacing $m$ by $m+j$ with $j\in J$ does not change $xm$, since $xj=0$; and $xm\in J$ because $p(xm)=xp(m)=0$.

Lemma 1.2. Let $M$ be an extension of $M_o$ by $J$ over $B$.

The image of $u(M)$ is $IM$. Hence $u(M)$ is surjective if and only if $M\otimes_BB_o\to M_o$ is an isomorphism.
If $M\otimes_BB_o\cong M_o$, then $M$ is flat over $B$ if and only if $M_o$ is flat over $B_o$ and $u(M)$ is an isomorphism.
$u(M)=0$ exactly when $M$ is a $B_o$-module. The extensions of $B_o$-modules form a subgroup of $\operatorname{Ext}^1_B(M_o,J)$ isomorphic to $\operatorname{Ext}^1_{B_o}(M_o,J)$ by restriction of scalars, $u$ is additive for the Baer sum, and its kernel is this subgroup.
A morphism of extensions $M\to M'$ inducing $\varphi\colon M_o\to M'_o$ and $\psi\colon J\to J'$ satisfies $u(M')\circ(1\otimes\varphi)=\psi\circ u(M)$.

Proof. (1) $IM$ is generated by the $xm$. And $M\otimes_BB_o=M/IM\to M_o=M/J$ is an isomorphism exactly when $IM=J$. (2) With $J=IM$, the map $I\otimes_BM=I\otimes_{B_o}M_o\to M$ has image $IM$ and corestriction $u(M)$; apply [Tag 08M4]. (3) $u(M)=0$ means $IM=0$. A splitting over $B$ of an extension of $B_o$-modules is $B_o$-linear, so the map $\operatorname{Ext}^1_{B_o}\to\operatorname{Ext}^1_B$ is injective, and Baer sums are formed by fibre products and pushouts, which $u$ respects. (4) $\psi(xm)=x\,m'$ for the image $m'$ of $m$. $\square$

The next result is in the AI-integrated Stacks project [Tags 08MC, 08MD]; we include it because Section 4 uses its proof.

Lemma 1.3 (morphisms of extensions). Let $M$, $M'$ be extensions of $M_o$, $M'_o$ by $J$, $J'$ over $B$, and $\varphi\colon M_o\to M'_o$, $\psi\colon J\to J'$ maps of $B_o$-modules with $u(M')\circ(1\otimes\varphi)=\psi\circ u(M)$. A morphism of extensions inducing $\varphi$ and $\psi$ exists if and only if $\psi_*[M]-\varphi^*[M']\in\operatorname{Ext}^1_B(M_o,J')$ vanishes; this element lies in the subgroup $\operatorname{Ext}^1_{B_o}(M_o,J')$. The morphisms, when they exist, form a torsor under $\operatorname{Hom}_{B_o}(M_o,J')$.

Proof. A morphism inducing $\varphi,\psi$ is the same as an isomorphism of extensions of $M_o$ by $J'$ from the pushout $\psi_*M$ to the pullback $\varphi^*M'$, which exists exactly when their classes agree. By Lemma 1.2 (4), $u(\psi_*M)=\psi u(M)$ and $u(\varphi^*M')=u(M')(1\otimes\varphi)$, which are equal, so the difference has $u=0$ and lies in $\operatorname{Ext}^1_{B_o}$ by Lemma 1.2 (3). Two morphisms differ by a map $M\to M_o\to J'\to M'$, with the middle map in $\operatorname{Hom}_B(M_o,J')=\operatorname{Hom}_{B_o}(M_o,J')$. $\square$

2. The canonical class and the obstruction

Lemma 2.1. Let $M_o$ be a $B_o$-module, $I^2=0$.

The connecting homomorphism $\delta_I\colon\operatorname{Tor}^B_1(M_o,B_o)\to M_o\otimes_BI$ of the sequence $0\to I\to B\to B_o\to0$, that is, $H^{-1}$ of $M_o\otimes^{\mathbf L}_B-$ applied to its triangle, is an isomorphism. Write $\gamma=$ $\gamma_{M_o}\colon$ $I\otimes_{B_o}M_o\to$ $\operatorname{Tor}^B_1(M_o,B_o)$ for its inverse composed with the exchange of factors.
For every extension $M$ of $M_o$ by $J$ over $B$, let $\delta^M\colon\operatorname{Tor}^B_1(M_o,B_o)\to J\otimes_BB_o=J$ be the connecting homomorphism of the triangle $Lf^*$ of the triangle of $0\to J\to M\to M_o\to0$. Then $\delta^M\circ\gamma=-u(M)$.

Proof. (1) The sequence $\operatorname{Tor}^B_1(M_o,B)=$ $0\to$ $\operatorname{Tor}^B_1(M_o,B_o)\to$ $M_o\otimes_BI\to$ $M_o\otimes_BB=$ $M_o$ is exact, and the last map $m\otimes x\mapsto xm$ is zero.

(2) First let $\Phi\colon E\to M_o$ be a surjection from a flat $B$-module (for instance the free module on the underlying sheaf of $M$, mapping to $M$ and then to $M_o$), $Z=\ker\Phi$, and $M_\Phi=E/IZ$, an extension of $M_o$ by $Z/IZ$ over $B$, with $u(M_\Phi)(x\otimes m)=x\tilde e\bmod IZ$ for a local lift $\tilde e\in E$ of $m$. Choose a flat resolution $G\to M_o$ with $G_0=E$ and $G_1\to Z$ surjective. The triangle of $0\to Z\to E\to M_o\to0$ is $Z\xrightarrow{j}E\xrightarrow{\Phi}M_o\xrightarrow{\delta}Z[1]$. In terms of $G$: the sequence of complexes $0\to G_0\to G\to G_{\geq1}\to0$ has the triangle $G_0\to G\to G_{\geq1}\xrightarrow{\delta'}G_0[1]$, with $\delta'$ inducing $G_1\to G_0$ on cohomology; rotating it backward [Tag 0145] gives the distinguished triangle $(G_{\geq1}[-1],G_0,G,-\delta'[-1],\mathrm{incl},\mathrm{pr})$. With the augmentations $a\colon G_{\geq1}[-1]\to Z$ and $q\colon G\to M_o$, $(-a,\mathrm{id},q)$ is an isomorphism from this triangle to the triangle of the sequence, because $\delta'[-1]=j\circ a$ on cohomology; the third square gives $\delta=-a[1]\circ\mathrm{pr}\circ q^{-1}$. Computing $Lf^*$ with the flat models $G$ of $M_o$ and $G_{\geq1}[-1]$ of $Z$, in which $a$ becomes the identity, $\delta^Z:=$ $H^{-1}(Lf^*\delta)\colon$ $\operatorname{Tor}_1^B(M_o,B_o)=$ $H_1(G\otimes_BB_o)\to$ $H^0(G_{\geq1}[-1]\otimes_BB_o)=$ $Z\otimes_BB_o$ sends the class of a cycle $g\otimes1$, $g\in G_1$, to $-[\mathrm dg\otimes1]$.

Now let $x\in I$ and $m\in M_o$, with a local lift $\tilde e\in E$. Then $x\tilde e\in Z$, so locally $x\tilde e=\mathrm dg$ with $g\in G_1$; $g\otimes1$ is a cycle of $G\otimes_BB_o$, since $x\tilde e\otimes1=\tilde e\otimes\bar x=0$. The connecting map $\delta_I$ is computed by the sequence of complexes $0\to G\otimes_BI\to G\to G\otimes_BB_o\to0$: lifting $g\otimes1$ to $g$ and applying the differential gives $x\tilde e\in G_0\otimes_BI$, whose class is $m\otimes x$. So $\gamma(x\otimes m)=[g\otimes1]$. The map of sequences $(0\to Z\to E\to M_o\to0)\to$ $(0\to Z/IZ\to M_\Phi\to M_o\to0)$, given by the quotient maps, gives $\delta^{M_\Phi}=(Z\otimes B_o\to Z/IZ)\circ\delta^Z$, so $\delta^{M_\Phi}(\gamma(x\otimes m))=$ $-[x\tilde e]=$ $-u(M_\Phi)(x\otimes m)$.

For a general extension $M$, take $E$ free on the underlying sheaf of $M$, with $\Phi$ the composite $E\to M\to M_o$. The evaluation $E\to M$ maps $Z$ into $J$ and $IZ$ into $IJ=0$, so it induces a morphism of extensions $M_\Phi\to M$ which is the identity on $M_o$, with some $\psi\colon Z/IZ\to J$. Connecting maps and $u$ are natural (Lemma 1.2 (4)), so $\delta^M\gamma=\psi\delta^{M_\Phi}\gamma=-\psi u(M_\Phi)=-u(M)$. $\square$

Definition 2.2. Let $I^2=0$ and $M_o$ be a $B_o$-module. Put $Y=t_{[-1}(Lf^*M_o)$, the truncation of $M_o\otimes^{\mathbf L}_BB_o$ to degrees $[-1,0]$; its cohomology is $H^0=M_o\otimes_BB_o=M_o$ and $H^{-1}=\operatorname{Tor}_1^B(M_o,B_o)$. The truncation triangle $H^{-1}(Y)[1]\to Y\to M_o\xrightarrow{\kappa}H^{-1}(Y)[2]$ defines $\kappa$, and the canonical class is $\begin{aligned}&\omega(B,M_o) =\gamma^{-1}[2]\circ\kappa\\&\quad \in\operatorname{Ext}^2_{B_o}(M_o,I\otimes_{B_o}M_o).\end{aligned}$ For a map $u_o\colon I\otimes_{B_o}M_o\to J$ put $\omega(B,u_o)=$ $u_o\circ\omega(B,M_o)\in$ $\operatorname{Ext}^2_{B_o}(M_o,J)$.

Theorem 2.3 (obstruction, classification, automorphisms). Let $I^2=0$, $M_o$, $J$ be $B_o$-modules and $u_o\colon I\otimes_{B_o}M_o\to J$.

There is an extension $M$ of $M_o$ by $J$ over $B$ with $u(M)=u_o$ if and only if $\omega(B,u_o)=0$.
If so, the isomorphism classes of such extensions form a torsor under $\operatorname{Ext}^1_{B_o}(M_o,J)$, acting by Baer sum.
The automorphisms of such an extension inducing the identity on $M_o$ and $J$ form the group $\operatorname{Hom}_{B_o}(M_o,J)$.
In particular, if $M_o$ is flat over $B_o$, there is a flat $B$-module $M$ with $M\otimes_BB_o\cong M_o$ if and only if $\omega(B,M_o)=0$, and then the isomorphism classes of such $(M,M\otimes_BB_o\cong M_o)$ form a torsor under $\operatorname{Ext}^1_{B_o}(M_o,I\otimes M_o)$.

Proof. The adjunction [Tag 07A6] gives $\operatorname{Ext}^1_B(M_o,J)=$ $\operatorname{Hom}_{D(B)}(M_o,f_*J[1])=$ $\operatorname{Hom}_{D(B_o)}(Lf^*M_o,J[1])=$ $\operatorname{Hom}_{D(B_o)}(Y,J[1])$; the last equality holds because the cone of $Lf^*M_o\to Y$ has cohomology in degrees $\leq-1$, so maps from it and from its shift by $1$ into $J[1]$ vanish [Tag 06XS]. Apply $\operatorname{Hom}(-,J[1])$ to the truncation triangle of $Y$: $\begin{aligned}&\operatorname{Hom}(M_o,J[1]) \to\operatorname{Hom}(Y,J[1])\\&\quad \xrightarrow{r}\operatorname{Hom}(\operatorname{Tor}_1[1],J[1])\\&\quad \xrightarrow{\partial}\operatorname{Hom}(M_o,J[2])\\&\quad \to\operatorname{Hom}(Y,J[2]),\end{aligned}$ which is exact [Tags 0145, 0149]: rotating the truncation triangle backward gives the distinguished triangle $M_o[-1]\xrightarrow{-\kappa[-1]}\operatorname{Tor}_1[1]\to Y\to M_o$, so $\partial(h)=-h[1]\circ\kappa$. The class of an extension $M$ corresponds to $\varepsilon_J[1]\circ Lf^*(\delta_M)$, where $\delta_M\colon M_o\to J[1]$ is the third map of its triangle and $\varepsilon_J\colon Lf^*f_*J\to J$ the counit; its image under $r$ is the map induced on $H^{-1}$, namely $\delta^M$ followed by $H^0(\varepsilon_J)=\mathrm{id}_J$. By Lemma 2.1, $r([M])=-u(M)\circ\gamma^{-1}$. So $u_o$ is of the form $u(M)$ if and only if $-u_o\gamma^{-1}$ lies in the image of $r$, which is $\ker\partial$, and $\partial(-u_o\gamma^{-1}[1])=$ $u_o[2]\circ\gamma^{-1}[2]\circ\kappa=$ $\omega(B,u_o)$. This is (1). The first map is restriction of scalars from $\operatorname{Ext}^1_{B_o}$, because the counit $Lf^*f_*M_o\to M_o$ is the truncation $Y\to M_o$; with Lemma 1.2 (3) this gives (2): the fibres of the additive map $u$ are the cosets of its kernel $\operatorname{Ext}^1_{B_o}(M_o,J)$. (3) An automorphism is $m\mapsto m+h(p(m))$ with $h\in\operatorname{Hom}_B(M_o,J)=\operatorname{Hom}_{B_o}(M_o,J)$. (4) By Lemma 1.2 (1), (2), the flat $M$ with $M\otimes B_o\cong M_o$ are the extensions of $M_o$ by $I\otimes M_o$ with $u(M)=\mathrm{id}$. $\square$

Remark 2.4. The AI-integrated Stacks project constructs a class $o(\mathcal F,\mathcal K,c)\in\operatorname{Ext}^2$ with the property of Theorem 2.3 (1) by embedding the coefficients into an injective module [Tag 08MF], and proves (2) and (3) directly [Tags 08ME, 08MC]. The class of Definition 2.2 is canonical and depends only on $B\to B_o$ and $M_o$; this is what allows its comparison with the Atiyah class.

Example 2.5 (locally free modules and vector groups). Let $S\to S'$ be a first-order thickening of schemes with ideal $I$, take for $T$ the Zariski topos of $S$, $B=\mathcal O_{S'}$ and $B_o=\mathcal O_S$, and let $E$ be a finite locally free $\mathcal O_S$-module. A flat $\mathcal O_{S'}$-module $E'$ with $E'\otimes\mathcal O_S\cong E$ is finite locally free: it is finitely presented, as in the proof of Proposition 3.3 of Deformations of module schemes, and flat [Tag 00NX]. By Theorem 2.3 (4) and (3), such deformations exist if and only if $\omega(B,E)=0$ in $\operatorname{Ext}^2_{\mathcal O_S}(E,I\otimes E)$, they form a torsor under $\operatorname{Ext}^1_{\mathcal O_S}(E,I\otimes E)$, and the automorphisms of each form $\operatorname{Hom}(E,I\otimes E)$. As $E$ is finite locally free, these groups are $H^j(S,\mathcal End(E)\otimes I)$ [Tag 08JJ]. By Proposition 3.3 of the lesson on deformations of module schemes, the same deformations, with the same isomorphisms, are the deformations of the vector group $\mathbf V(E)$ as a module scheme over $\mathcal O$, for which Proposition 3.2 there finds the same groups through the Lie complex. The homomorphisms $\mathbf V(E_1)\to\mathbf V(E_2)$ are the module maps $E_1\to E_2$. So Proposition 4.1 below describes the extensions of homomorphisms of vector groups to given deformations, and Proposition 4.2, applied to the transposes $E_2^\vee\to E_1^\vee$, the deformations of a homomorphism of vector groups with a given deformation of its target.

3. The canonical class and the Atiyah class

In this section $B\to B_o$ is any surjection with kernel $I$, not necessarily of square zero, and $P=P_B(B_o)$ is the standard resolution, a simplicial $B$-algebra with flat terms resolving $B_o$; so $Lf^*M_o=M_o\otimes^{\mathbf L}_BB_o$ is computed by $M_o\otimes_BP$, and also by $P\otimes_BM_o$, the exchange of factors being an isomorphism of simplicial modules. Lemma 2.1 (1) holds with $I$ replaced by $I/I^2$: $\delta_I\colon$ $\operatorname{Tor}^B_1(M_o,B_o)\cong$ $M_o\otimes_BI=$ $M_o\otimes_{B_o}I/I^2$, and $\gamma$ is defined in the same way; Lemma 2.1 (2) and its proof hold without change.

For a constant $x\in I\subseteq B\subseteq P_0=B[B_o]$, let $z_x=[x]-[0]\in P_1=B[U P_0]$, the difference of the variables labelled by the elements $x$ and $0$ of $P_0$. Then $d_0z_x=x$ and $d_1z_x=[\bar x]-[\bar0]=0$ (Construction 2.1 of the lesson on graded modules), so the image of $z_x$ in $P\to B_o$ is zero.

Lemma 3.1 (principal parts compute the truncation). Let $M_o$ be a $B_o$-module, and $\alpha\colon P\otimes_BM_o\to P^1_{P/B}(M_o)$ the quotient map of Definition 1.1 of the lesson on the Atiyah class ($P^1_{P/B}(M_o)=(P\otimes_BM_o)/J^2(P\otimes_BM_o)$), a map of simplicial $P$-modules compatible with the maps to $M_o$.

$\alpha$ induces isomorphisms on $\pi_0$ and $\pi_1$.
The composite isomorphism $\begin{aligned}&I/I^2\otimes M_o\\&\quad \cong H^{-1}(L_{B_o/B})\otimes M_o\\&\quad =H^{-1}(L_{B_o/B}\otimes^{\mathbf L}M_o)\\&\quad \xrightarrow{\iota}H^{-1}(\operatorname{PP}_{B_o/B}(M_o))\\&\quad \xleftarrow{\alpha}\operatorname{Tor}^B_1(M_o,B_o),\end{aligned}$ where $\iota$ is the first map of the triangle (2.2) of the lesson on the Atiyah class, is $\gamma$.

Proof. (1) Put $\overline P=P\otimes_BB_o$, a simplicial $B_o$-algebra with $\overline P_n=B_o[X_n]$, and $\overline I=\ker(\overline P\to B_o)$, the multiplication $p\otimes b\mapsto\bar pb$. Then $P\otimes_BM_o=\overline P\otimes_{B_o}M_o$, and $P^1_{P/B}(M_o)=(\overline P/\overline I^2)\otimes_{B_o}M_o$, because $J$ maps onto $\overline I$ (it is generated by the $1\otimes p-p\otimes1$, which map to $1\otimes\bar p-p\otimes1$, and these generate $\overline I$). The unit $B_o\to\overline P$ splits the multiplication, so $\overline P=B_o\oplus\overline I$, and $\alpha=\mathrm{id}_{M_o}\oplus\big((\overline I\to\overline I/\overline I^2)\otimes M_o\big)$. In each degree $\overline I_n$ is the ideal of $B_o[X_n]$ generated by the $x-\bar x$, a free $B_o$-module on the monomials of positive degree in these elements, and $\overline I/\overline I^2=\Omega_{P/B}\otimes_PB_o=L_{B_o/B}$ is free on the $x-\bar x$. Both are connected: $\pi_0(\overline I)=$ $\ker(\pi_0\overline P\to B_o)=$ $\ker(B_o\otimes_BB_o\to B_o)=$ $0$, and $\pi_0(L_{B_o/B})=\Omega_{B_o/B}=0$. For a connected simplicial $B_o$-module $X$ with flat terms, $\pi_1(X\otimes_{B_o}M_o)=\pi_1(X)\otimes_{B_o}M_o$: $X\otimes M_o$ computes $X\otimes^{\mathbf L}M_o$, $X\to H_1(X)[1]$ is the truncation, and its cone lies in degrees $\leq-2$ (cohomologically), so the map on $H^{-1}$ of $-\otimes^{\mathbf L}M_o$ is an isomorphism onto $H^{-1}(H_1(X)[1]\otimes^{\mathbf L}M_o)=H_1(X)\otimes M_o$. Hence it suffices to treat $M_o=B_o$, where we must show that $\pi_1(\overline I)\to\pi_1(\overline I/\overline I^2)$ is an isomorphism; $\pi_0$ is clear in all cases.

Let $M_o=B_o$. The classes $[z_x\otimes1]\in\pi_1(P\otimes_BB_o)$ satisfy $\delta_I[z_x\otimes1]=1\otimes x$: computing $\delta_I$ with the resolution $P$ of the first factor, the sequence of complexes $0\to P\otimes I\to P\to P\otimes B_o\to0$ sends $z_x\otimes1$, lifted to $z_x$, to $d_0z_x-d_1z_x=x\in P_0\otimes I$, of class $1\otimes x$. So these classes span $\pi_1(P\otimes B_o)=\pi_1(\overline I)$, and $\delta_I$ is an isomorphism (Lemma 2.1 (1)). In $P^1_{P/B}(B_o)=P\otimes_BB_o/(\cdots)$, the element $\iota(\mathrm dz_x\otimes1)=(1\otimes z_x-z_x\otimes1)\otimes1$ is $1\otimes\overline{z_x}-z_x\otimes1=-z_x\otimes1$. So $\alpha_*[z_x\otimes1]=-\iota_*[\mathrm dz_x\otimes1]$, and $[\mathrm dz_x]\in H_1(L_{B_o/B})$ corresponds to $d_0z_x-d_1z_x=x\in I/I^2$ (Conventions). Hence the map $\pi_1(\overline I)\to\pi_1(\overline I/\overline I^2)\cong I/I^2$ sends $[z_x\otimes1]$ to $-x$: it is $-\delta_I$ (followed by $B_o\otimes I=I/I^2$), an isomorphism.

(2) Let $\gamma'$ be the composite isomorphism of the statement (well defined by (1)). By the computation just made, for general $M_o$, $x\in I$ and $m\in M_o$, $\gamma'(\bar x\otimes m)=-[z_x\otimes m]$ in $\pi_1(P\otimes_BM_o)$. We compute $\delta^{M_\Phi}$ for the extensions $M_\Phi=E/IZ$ of the proof of Lemma 2.1 with the flat resolution $P$ of the second factor: $Lf^*$ is computed by $-\otimes_BP$, which carries the sequence $0\to Z/IZ\to M_\Phi\to M_o\to0$ to a short exact sequence of simplicial modules; its connecting map sends the class of $m\otimes z_x$, lifted to $\tilde e\otimes z_x$, to the class of $\tilde e\otimes(d_0z_x-d_1z_x)=x\tilde e\otimes1$, that is, $\delta^{M_\Phi}[m\otimes z_x]=u(M_\Phi)(x\otimes m)$. The exchange of factors identifies $[z_x\otimes m]$ with $[m\otimes z_x]$. So $\delta^{M_\Phi}\gamma'=-u(M_\Phi)=\delta^{M_\Phi}\gamma$ by Lemma 2.1 (2) (whose proof did not use $I^2=0$). Finally $\delta^{M_\Phi}$ is injective: it is $\delta^Z$ followed by the identity of $Z/IZ$, and $\delta^Z$ is injective because $\operatorname{Tor}^B_1(E,B_o)=0$. Hence $\gamma'=\gamma$. $\square$

Theorem 3.2 (the canonical class is the product of the Atiyah class and the class of the thickening). Let $A\to B\to B_o$ be maps of rings of $T$, $B\to B_o$ surjective with kernel $I$ of square zero, and $M_o$ a $B_o$-module. Then $\begin{aligned}&\omega(B,M_o)\\&\quad =\big(e(B)\otimes^{\mathbf L}1_{M_o}\big)[1]\circ\operatorname{at}_{B_o/A}(M_o),\end{aligned}$ followed by $I\otimes^{\mathbf L}_{B_o}M_o[2]\to I\otimes_{B_o}M_o[2]$. In particular, if $M_o$ is flat, $\omega(B,M_o)$ is the composite of $\operatorname{at}_{B_o/A}(M_o)$ and $e(B)\otimes M_o$.

Proof. By Proposition 2.3 (2) of the lesson on the Atiyah class, applied to $A\to B\to B_o$, $\operatorname{at}_{B_o/B}(M_o)=(c\otimes1)[1]\circ\operatorname{at}_{B_o/A}(M_o)$, and $e(B)=\tau_{B_o/B}\circ c$. So it suffices to show that $\omega(B,M_o)$ is $\tau_{B_o/B}\otimes1$ composed with $\operatorname{at}_{B_o/B}(M_o)$, followed by the projection onto $H^{-1}$. Write $X=L_{B_o/B}\otimes^{\mathbf L}M_o$; $H^0(X)=0$ and $H^{-1}(X)=I\otimes M_o$, and $X[1]\to t_{[-1}X[1]=H^{-1}(X)[2]$ is $\tau_{B_o/B}\otimes1$ followed by that projection. The triangle $X\to\operatorname{PP}_{B_o/B}(M_o)\to M_o\xrightarrow{\operatorname{at}}X[1]$ has its third term in $D^{\geq-1}$, so by Lemma 3.2 of the lesson on torsors (truncating a triangle) $\begin{aligned}&H^{-1}(X)[1] \to t_{[-1}\operatorname{PP}(M_o)\\&\quad \to M_o \to H^{-1}(X)[2]\end{aligned}$ is distinguished, its third map being $\operatorname{at}$ followed by the truncation. Its first two maps are the truncation maps of $Y'=t_{[-1}\operatorname{PP}(M_o)$, up to the isomorphism $H^{-1}(X)\cong H^{-1}(Y')$ induced by $\iota$; two distinguished triangles with the same first two maps and third term in degree $0$ have the same third map (the comparison map $\gamma_3$ with $\gamma_3\circ v=v$ for the surjection $v$ on $H^0$ is the identity [Tag 06XS]). So the truncation class of $Y'$, read in $I\otimes M_o$ through $\iota$ and $\tau_{B_o/B}$, is $(\tau_{B_o/B}\otimes1)\circ\operatorname{at}_{B_o/B}(M_o)$ followed by the projection. By Lemma 3.1 (1), $\alpha$ induces an isomorphism $Y=t_{[-1}(P\otimes_BM_o)\to Y'$ which is the identity on $H^0=M_o$; by Lemma 3.1 (2), on $H^{-1}$ it carries $\gamma$ to the identification through $\iota$ and $\tau$. Truncation classes are natural for such isomorphisms, so $\omega(B,M_o)=\gamma^{-1}[2]\kappa$ equals that class. If $M_o$ is flat, the projection is the identity. $\square$

Corollary 3.3 (the universal thickening). Let $A\to B_o$ be a surjection with kernel $K$, and $M_o$ a $B_o$-module. Then $\operatorname{Tor}^A_1(M_o,B_o)\cong K/K^2\otimes M_o$ by $\gamma$, and the canonical class $\omega(A/K^2,M_o)$ is the class of the truncation of $M_o\otimes^{\mathbf L}_AB_o$: $\omega(A/K^2,M_o)=$ $\gamma^{-1}[2]\circ\kappa(t_{[-1}(M_o\otimes^{\mathbf L}_AB_o))$.

Proof. Apply Lemma 3.1 to the surjection $A\to B_o$ (with $K$ in place of $I$) and argue as in the proof of Theorem 3.2: the truncation class of $M_o\otimes^{\mathbf L}_AB_o$, read in $K/K^2\otimes M_o$ through $\gamma$, is $(\tau_{B_o/A}\otimes1)\circ\operatorname{at}_{B_o/A}(M_o)$ followed by the projection. The $A$-extension $A/K^2$ of $B_o$ by $K/K^2$ has class $e(A/K^2)=\tau_{B_o/(A/K^2)}\circ c$, and $c\colon L_{B_o/A}\to L_{B_o/(A/K^2)}$ induces on $H^{-1}$ the map $K/K^2\to(K/K^2)/(K/K^2)^2=K/K^2$ induced by $A\to A/K^2$, the identity, by the naturality in Proposition 4.1 of the lesson on graded modules. A map $L_{B_o/A}\to(K/K^2)[1]$ is determined by its effect on $H^{-1}$ [Tag 06XS], so $e(A/K^2)=\tau_{B_o/A}$. Theorem 3.2 for $B=A/K^2$ gives the same composite. $\square$

Corollary 3.4 (the truncated graded cotangent complex of the dual numbers). Let $A\to B$ be a surjection of rings and $M$ a $B$-module, $C=B\oplus M$ with $M$ in degree one. Then $k^1(t_{[-1}L^{\mathrm{gr}}_{C/A})\cong t_{[-1}(M\otimes^{\mathbf L}_AB)$ in $D(B)$, compatibly with the maps to $H^0=M$.

Proof. By Theorem 2.4 of the lesson on the Atiyah class, $k^1(L^{\mathrm{gr}}_{C/A})\cong\operatorname{PP}_{B/A}(M)$ compatibly with the maps to $M$; $k^1$ commutes with truncation (it is exact); and $t_{[-1}\operatorname{PP}_{B/A}(M)\cong t_{[-1}(M\otimes^{\mathbf L}_AB)$ by Lemma 3.1 (1) for the surjection $A\to B$. $\square$

Corollary 3.5 (connections). Let $A\to B_o$ be such that $L_{B_o/A}\to\Omega_{B_o/A}$ is a quasi-isomorphism and $\Omega_{B_o/A}$ is flat (for instance a smooth morphism of schemes, on the Zariski site). If a flat $B_o$-module $M_o$ has a connection relative to $A$, then for every $A$-extension $B$ of $B_o$ by an ideal of square zero there is a flat $B$-module $M$ with $M\otimes_BB_o\cong M_o$.

Proof. By Corollary 3.3 of the lesson on the Atiyah class, $\operatorname{at}_{B_o/A}(M_o)=0$; so $\omega(B,M_o)=0$ by Theorem 3.2, and Theorem 2.3 (4) applies. $\square$

Remark 3.6. For $M_o$ locally free of finite rank over a local ring, the local deformations of $M_o$ to locally free $B$-modules form a gerbe banded by $\mathcal End(M_o)\otimes I$, whose class in $H^2(T,\mathcal End(M_o)\otimes I)$ is a second obstruction to a global deformation. Comparing this class with $\omega(B,M_o)$ is a question posed by Illusie; it is kept in the scope of this course as an open point and is not settled here.

Lemma 3.7 (boundaries through local lifts). Let $0\to\mathcal F\to\mathcal G\xrightarrow{q}\mathcal H\to0$ be an exact sequence of abelian sheaves on a topological space $Z$, let $h\in\mathcal H(Z)$, and let $t_i\in\mathcal G(U_i)$ satisfy $q(t_i)=h|_{U_i}$, for an open covering $Z=\bigcup_iU_i$. Then the boundary of $h$ in $H^1(Z,\mathcal F)$ is the image of the Čech class of the cocycle $(t_{i_1}-t_{i_0})$.

Proof. The local sections of $\mathcal G$ lifting $h$ form an $\mathcal F$-torsor $\mathcal T$, on which $\mathcal F$ acts by addition. The homomorphism $\mathbf Z[\mathcal T]\to\mathcal G$, $[t]\mapsto t$, sends $[t]-[t']$ to $t-t'$. On the kernel of the augmentation $\mathbf Z[\mathcal T]\to\mathbf Z$ it is therefore the map $a$ of the proof of [Tag 02FQ], and it lies over the map $\mathbf Z\to\mathcal H$, $1\mapsto h$. So it induces a morphism, equal to the identity on $\mathcal F$, from the pushout extension $0\to\mathcal F\to\mathcal E\to\mathbf Z\to0$ of that proof to the given sequence. By naturality of boundary maps, the boundary of $h$ is the boundary of $1$ for $\mathcal E$, which is the class of the torsor $\mathcal T$ [Tag 02FQ]. The sections $t_i$ trivialize $\mathcal T$ over the $U_i$, and $(t_{i_1}-t_{i_0})+t_{i_0}=t_{i_1}$; by [Tag 0B8R] the class of $\mathcal T$ is the image of the Čech class of $(t_{i_1}-t_{i_0})$. $\square$

Proposition 3.8 (the Atiyah class of an invertible module). Let $Y\to S$ be a smooth morphism of schemes and $L$ an invertible $\mathcal O_Y$-module. Through $L_{Y/S}\simeq\Omega_{Y/S}$ and $\mathcal End(L)=\mathcal O_Y$, the Atiyah class $\operatorname{at}_{Y/S}(L)$ lies in $\operatorname{Ext}^1_{\mathcal O_Y}(L,\Omega_{Y/S}\otimes L)=H^1(Y,\Omega_{Y/S})$. It is the first Chern class $c_1^{\mathrm{Hodge}}(L)$ of [Tag 0FLE], the image of the class of $L$ in $H^1(Y,\mathcal O_Y^*)$ [Tag 09NU] under $\mathrm d\log\colon g\mapsto g^{-1}\mathrm dg$. In particular $\operatorname{at}_{Y/S}(L'\otimes L'')=$ $\operatorname{at}_{Y/S}(L')+\operatorname{at}_{Y/S}(L'')$ for invertible $L',L''$.

Proof. For a finite locally free $\mathcal O_Y$-module $E$, $\mathcal Hom(E,-)$ is exact and carries injective modules to injective modules [Tag 01E7]. So $\operatorname{Ext}^q(E,\mathcal M)=H^q(Y,\mathcal Hom(E,\mathcal M))$, compatibly with boundary maps. By Corollary 3.3 of the lesson on the Atiyah class, $\operatorname{at}_{Y/S}(L)$ is the third map of the triangle of the principal parts extension $0\to\Omega_{Y/S}\otimes L\xrightarrow{\iota}P^1_{Y/S}(L)\xrightarrow{\pi}L\to0 .$ It is the image of $\mathrm{id}_L$ under the boundary map $\operatorname{Hom}(L,L)\to\operatorname{Ext}^1(L,\Omega_{Y/S}\otimes L)$, that is, the boundary of $\mathrm{id}_L\in H^0(Y,\mathcal Hom(L,L))$ in $H^1(Y,\mathcal Hom(L,\Omega_{Y/S}\otimes L))=$ $H^1(Y,\Omega_{Y/S})$.

Choose an open covering $Y=\bigcup U_i$ and trivializing sections $s_i$ of $L$ over $U_i$, and put $f_{i_0i_1}=s_{i_1}s_{i_0}^{-1}$ over $U_{i_0i_1}$. The class of $L$ is the image of the Čech class of $(f_{i_0i_1})$, and $c_1^{\mathrm{Hodge}}(L)$ is that of $(\mathrm d\log f_{i_0i_1})$ [Tag 0FLE]. Let $\nabla_i$ be the connection on $L|_{U_i}$ with $\nabla_i(gs_i)=\mathrm dg\otimes s_i$, and $\sigma_i=\mathrm d^1-\iota\nabla_i$ the corresponding section of $\pi$ (Proposition 3.2 of the lesson on the Atiyah class). The $\sigma_i$ are local lifts of $\mathrm{id}_L$, and $\sigma_{i_1}-\sigma_{i_0}=\iota(\nabla_{i_0}-\nabla_{i_1})$. Over $U_{i_0i_1}$, $\begin{aligned}(\nabla_{i_0}-\nabla_{i_1})(s_{i_1})&=\nabla_{i_0}(f_{i_0i_1}s_{i_0})\\&=\mathrm df_{i_0i_1}\otimes s_{i_0}\\&=\mathrm d\log(f_{i_0i_1})\otimes s_{i_1},\end{aligned}$ so $\nabla_{i_0}-\nabla_{i_1}$ is multiplication by $\mathrm d\log(f_{i_0i_1})$. Lemma 3.7 gives the first assertion. The second follows, because the class of $L'\otimes L''$ is the product of the classes and $\mathrm d\log$ is a homomorphism. $\square$

Lemma 3.9 (the conormal map of the transitivity triangle). Let $A\to B\to B_o$ be maps of rings of $T$, with $B\to B_o$ surjective with kernel $I$, and let $K\colon L_{B_o/B}\to L_{B/A}\otimes^{\mathbf L}_BB_o[1]$ be the third map of their transitivity triangle [Tag 08SZ]. Identify $H^{-1}(L_{B_o/B})$ with $I/I^2$ as in the Conventions, and $H^0(L_{B/A}\otimes^{\mathbf L}_BB_o)$ with $\Omega_{B/A}\otimes_BB_o$. Then $H^{-1}(K)$ is the map $\bar x\mapsto\mathrm dx\otimes1$.

Proof. The identification of the Conventions sends the class of a cycle $\mathrm dy\otimes1$ of degree $-1$ to the element of $I/I^2$ whose image under the map of presentations is the class of $d_0y-d_1y$. So it is the identification of formula (6.5) of Theorem 6.1 of The cotangent complex, and Lemma 1.1 of Equivariant extensions and deformations shows that $K(\bar t)=\mathrm dt$ for the maps $\mathbb Z\to\mathbb Z[t]\to\mathbb Z$, $t\mapsto0$, of the punctual topos. In $T$, the constant triple $\mathbb Z_T\to\mathbb Z_T[t]\to\mathbb Z_T$ is the inverse image of this one along $T\to\mathrm{pt}$, which carries the transitivity triangle and the identifications along [Tag 08SV]; so $K(\bar t)=\mathrm dt$ there. Let $f$ be a section of $I$ over an object $U$ of $T$. The homomorphism $\mathbb Z_U[t]\to B_U$, $t\mapsto f$, and the structure maps form a map of triples from $\mathbb Z_U\to\mathbb Z_U[t]\to\mathbb Z_U$ to $A_U\to B_U\to B_{o,U}$, since $f$ maps to $0$ in $B_o$. It sends $\bar t$ to $\bar f$ and $\mathrm dt\otimes1$ to $\mathrm df\otimes1$. All the maps involved are natural in the triple, so $K(\bar f)=\mathrm df\otimes1$. $\square$

Theorem 3.10 (extending vector bundles to the first infinitesimal neighbourhood). Let $X$ and $Y$ be smooth schemes over a scheme $S$, and $i\colon Y\to X$ a closed immersion over $S$ with ideal $\mathcal I$. Let $Y_1\subseteq X$ be the closed subscheme defined by $\mathcal I^2$, the first infinitesimal neighbourhood of $Y$, and $N=\mathcal I/\mathcal I^2$.

The conormal sequence (3.1) $0\to N\to i^*\Omega_{X/S}\to\Omega_{Y/S}\to0,$ whose first map $\mathrm d$ sends $\bar x$ to $\mathrm dx\otimes1$, is exact. Let $\delta\colon\Omega_{Y/S}\to N[1]$ be the third map of its triangle.
Through $L_{Y/S}\simeq\Omega_{Y/S}$, the class of the $S$-extension $\mathcal O_{Y_1}$ of $\mathcal O_Y$ by $N$ is $e(\mathcal O_{Y_1})=-\delta$.
Let $L$ be a locally free $\mathcal O_Y$-module of finite rank and $\omega(i,L)=\omega(\mathcal O_{Y_1},L)$, an element of $\operatorname{Ext}^2(L,N\otimes L)=$ $H^2(Y,\mathcal End(L)\otimes N)$. Then $L$ extends to a locally free $\mathcal O_{Y_1}$-module if and only if $\omega(i,L)=0$, and $\omega(i,L)=-\partial\operatorname{at}_{Y/S}(L),$ where $\partial\colon H^1(Y,\mathcal End(L)\otimes\Omega_{Y/S})\to$ $H^2(Y,\mathcal End(L)\otimes N)$ is the boundary map of (3.1) tensored with $\mathcal End(L)$.
For invertible $L',L''$, $\omega(i,L'\otimes L'')=\omega(i,L')+\omega(i,L'')$. The homomorphism $\operatorname{Pic}(Y)\to H^2(Y,N)$, $L\mapsto\omega(i,L)$, is $-\partial\circ c_1^{\mathrm{Hodge}}$: minus the boundary map of the extension of $\mathcal O_Y^*$ by $N$ obtained from (3.1) by pullback along $\mathrm d\log$.

Proof. Work on the Zariski topos of $Y$, which is that of $Y_1$. Put $A=f^{-1}\mathcal O_S$, $B=\mathcal O_{Y_1}$ and $B_o=\mathcal O_Y$. Then $B\to B_o$ is surjective with kernel $N$, and $N^2=0$. As $Y\to S$ is smooth, the augmentation $\beta\colon L_{B_o/A}\to\Omega_{Y/S}$ is a quasi-isomorphism and $\Omega_{Y/S}$ is locally free.

(1) Consider the transitivity triangle of $A\to B\to B_o$, $\begin{aligned}&L_{B/A}\otimes^{\mathbf L}_BB_o\xrightarrow{\varphi}L_{B_o/A}\\&\quad\xrightarrow{c}L_{B_o/B}\xrightarrow{K}L_{B/A}\otimes^{\mathbf L}_BB_o[1].\end{aligned}$ Here $H^{-1}(L_{B_o/A})=0$, $H^0(L_{B_o/B})=0$, and $H^{-1}(L_{B_o/B})\cong N$ as in the Conventions. The long exact cohomology sequence gives the exact sequence $\begin{aligned}0\to N&\to\Omega_{B/A}\otimes_BB_o\\&\xrightarrow{H^0(\varphi)}\Omega_{Y/S}\to0,\end{aligned}$ whose first map is $x\mapsto\mathrm dx\otimes1$ by Lemma 3.9, and whose second map is the natural one. By [Tag 00RU] for $\mathcal O_X\to\mathcal O_X/\mathcal I^2$, $\Omega_{Y_1/S}$ is the quotient of $\Omega_{X/S}\otimes\mathcal O_{Y_1}$ by the image of $\mathcal I^2$ under $\mathrm d$. This image dies in $\Omega_{X/S}\otimes\mathcal O_Y$, since $\mathrm d(ab)=a\,\mathrm db+b\,\mathrm da$. So $\Omega_{B/A}\otimes_BB_o=i^*\Omega_{X/S}$, and the sequence is (3.1).

(2) Put $E=i^*\Omega_{X/S}$, with $p\colon E\to\Omega_{Y/S}$. Rotating the triangle of (3.1) gives the distinguished triangle $E\xrightarrow{p}\Omega_{Y/S}\xrightarrow{\delta}N[1]\xrightarrow{-\mathrm d[1]}E[1].$ Let $\alpha\colon L_{B/A}\otimes^{\mathbf L}_BB_o\to E$ be the truncation to $H^0$. Then $p\alpha=\beta\varphi$: both are maps from a complex in degrees $\leq0$ to a module in degree $0$, which are determined by their effect on $H^0$ [Tag 06XS], and both induce $H^0(\varphi)$. By the axiom TR3 there is $\gamma\colon L_{B_o/B}\to N[1]$ such that $(\alpha,\beta,\gamma)$ is a morphism of triangles: $\gamma c=\delta\beta$ and $(-\mathrm d[1])\gamma=\alpha[1]K$. As $L_{B_o/B}$ has no cohomology in degrees $>-1$, a map from it to $N[1]$ is determined by its effect on $H^{-1}$ [Tag 06XS]. On $H^{-1}$ the second relation reads $-\mathrm d\circ H^{-1}(\gamma)=\mathrm d$ under the identification of $H^{-1}(L_{B_o/B})$ with $N$, by Lemma 3.9. As $\mathrm d$ is injective, $H^{-1}(\gamma)=-1$, so $\gamma=-\tau_{B_o/B}$. The first relation then gives $e(B)=\tau_{B_o/B}\circ c=-\delta\beta$.

(3) A flat $\mathcal O_{Y_1}$-module $L'$ with $L'\otimes\mathcal O_Y\cong L$ is locally free: locally, a lift $B^n\to L'$ of a basis of $L$ has a cokernel $Q$ with $Q=NQ=N^2Q=0$, and a kernel $R$ with $R\otimes_BB_o=0$, as $L'$ is flat, so $R=NR=N^2R=0$. Conversely locally free modules are flat. So the first assertion is Theorem 2.3 (4). By Theorem 3.2 and (2), $\omega(i,L)=-\big((\delta\beta)\otimes1_L\big)[1]\circ\operatorname{at}_{B_o/A}(L),$ and $(\beta\otimes1)[1]\circ\operatorname{at}_{B_o/A}(L)$ is $\operatorname{at}_{Y/S}(L)$, the class of the principal parts extension of $L$ (Corollary 3.3 of the lesson on the Atiyah class). As $L$ is flat, $\delta\otimes1_L$ is the third map of the triangle of (3.1) tensored with $L$. Under $\operatorname{Ext}^q(L,\mathcal M)=H^q(Y,\mathcal Hom(L,\mathcal M))$ (proof of Proposition 3.8) and $\mathcal Hom(L,\mathcal M\otimes L)=\mathcal End(L)\otimes\mathcal M$, composition with $(\delta\otimes1_L)[1]$ is therefore $\partial$.

(4) By (3) and Proposition 3.8, $\omega(i,L)=-\partial\,c_1^{\mathrm{Hodge}}(L)$, and $c_1^{\mathrm{Hodge}}$ is a homomorphism. The pullback of (3.1) along $\mathrm d\log\colon\mathcal O_Y^*\to\Omega_{Y/S}$ is an exact sequence $0\to N\to E'\to\mathcal O_Y^*\to0$ of abelian sheaves, with a morphism to (3.1) that is the identity on $N$. By naturality, its boundary map is $\partial\circ H^1(\mathrm d\log)=\partial\circ c_1^{\mathrm{Hodge}}$. $\square$

Example 3.11 (the quadric surface). Let $S=\operatorname{Spec}\mathbf Z$ and $X=\mathbf P^3_{\mathbf Z}$ with coordinates $T_0,\dots,T_3$. Let $Y=\mathbf P^1_{\mathbf Z}\times_{\mathbf Z}\mathbf P^1_{\mathbf Z}$, with projections $p_1,p_2$ and coordinates $(s_0:s_1)$, $(t_0:t_1)$, and let $i\colon Y\to X$ be the Segre embedding $(T_0,T_1,T_2,T_3)=(s_0t_0,s_0t_1,s_1t_0,s_1t_1)$ [Tag 01WD], defined by $L_1\otimes L_2$, where $L_j=p_j^*\mathcal O(1)$. Put $e_j=c_1^{\mathrm{Hodge}}(L_j)=\operatorname{at}_{Y/\mathbf Z}(L_j)$.

$H^1(Y,\Omega_{Y/\mathbf Z})$ is free with basis $e_1,e_2$.
The boundary map $\partial\colon H^1(Y,\Omega_{Y/\mathbf Z})\to H^2(Y,N)$ of (3.1) is surjective with kernel $\mathbf Z(e_1+e_2)$. So $H^2(Y,N)$ is free of rank one with basis $\partial e_1=-\partial e_2$, and the cohomology sequence of (3.1) gives the exact sequence $\begin{aligned}0&\to H^0(Y,\mathcal O_Y)\\&\to H^1(Y,\Omega_{Y/\mathbf Z})\\&\xrightarrow{\partial}H^2(Y,N)\to0,\end{aligned}$ whose first map sends $1$ to $e_1+e_2$.
$\omega(i,L_1)=-\omega(i,L_2)=-\partial e_1$, a basis of $H^2(Y,N)$. So neither $L_1$ nor $L_2$ extends to an invertible module on the first infinitesimal neighbourhood $Y_1$ of $Y$ in $\mathbf P^3_{\mathbf Z}$, while $L_1\otimes L_2\cong\mathcal O_Y(1)$ extends, as the restriction of $\mathcal O_X(1)$.
The module $\Omega^2_{Y/\mathbf Z}$ is $p_1^*\Omega_{\mathbf P^1/\mathbf Z}\otimes p_2^*\Omega_{\mathbf P^1/\mathbf Z}$, and $H^2(Y,\Omega^2_{Y/\mathbf Z})$ is free with basis the cup product $e_1e_2$ of $e_1\in H^1(Y,p_1^*\Omega_{\mathbf P^1/\mathbf Z})$ and $e_2\in H^1(Y,p_2^*\Omega_{\mathbf P^1/\mathbf Z})$. The modules $\Omega^2_{Y/\mathbf Z}$ and $N$ are isomorphic, by exactly two isomorphisms, opposite to each other; they carry $e_1e_2$ to $\omega(i,L_1)$ and to $-\omega(i,L_1)$. In this sense $H^2(Y,N)$ is free with basis $e_1e_2$, and $\omega(i,L_1)=-\omega(i,L_2)=\pm e_1e_2$.

Proof. The quadric. Put $Q=T_0T_3-T_1T_2$. Over $D_+(T_0)=\operatorname{Spec}\mathbf Z[x,y,z]$, with $x=T_1/T_0$, $y=T_2/T_0$, $z=T_3/T_0$, the preimage is $D_+(s_0)\times D_+(t_0)=\operatorname{Spec}\mathbf Z[u,v]$, with $u=s_1/s_0$ and $v=t_1/t_0$, and $i$ is given by $x\mapsto v$, $y\mapsto u$, $z\mapsto uv$. This is a closed immersion with ideal $(z-xy)=(Q/T_0^2)$. Exchanging $s_0$ with $s_1$, or $t_0$ with $t_1$, changes $Q$ into $-Q$, and these exchanges and their composite carry $T_0$ to each of the $T_k$; so the same holds over every $D_+(T_k)$. Hence $\mathcal I=Q\cdot\mathcal O_X(-2)\cong\mathcal O_X(-2)$, and $N\cong\mathcal O_Y(-2)\cong L_1^{\otimes-2}\otimes L_2^{\otimes-2}$.

Cohomology. For integers $a,b$ write $\mathcal O(a,b)=p_1^*\mathcal O(a)\otimes p_2^*\mathcal O(b)$. The groups $H^q(\mathbf P^1_{\mathbf Z},\mathcal O(a))$ are free: $\mathbf Z^{a+1}$ for $q=0\leq a$, $\mathbf Z^{-a-1}$ for $q=1$ and $a\leq-2$, and $0$ otherwise [Tag 01XT]. A complex of abelian groups whose cohomology groups are free is isomorphic in $D(\mathbf Z)$ to the sum of its shifted cohomology groups: lifting bases of the cohomology groups to cycles gives a quasi-isomorphism from that sum. With the Künneth formula [Tag 0FU4] ($\mathbf P^1_{\mathbf Z}$ is separated and quasi-compact, and the modules are flat over $\mathbf Z$) this gives $\begin{aligned}&H^n(Y,p_1^*\mathcal F\otimes p_2^*\mathcal G)\\&\quad\cong\bigoplus_kH^k(\mathcal F)\otimes H^{n-k}(\mathcal G)\end{aligned}$ for finite locally free $\mathcal F,\mathcal G$ on $\mathbf P^1_{\mathbf Z}$ with free cohomology. Hence $H^*(Y,\mathcal O(-1,-1))=0$; $H^0(Y,\mathcal O_Y)=\mathbf Z$ and $H^q(Y,\mathcal O_Y)=0$ for $q>0$; and $H^1(Y,N)=0$. Moreover $\Omega_{Y/\mathbf Z}=p_1^*\Omega_{\mathbf P^1/\mathbf Z}\oplus p_2^*\Omega_{\mathbf P^1/\mathbf Z}$ [Tag 01V1]. As $H^0(\mathbf P^1_{\mathbf Z},\Omega)=0$ and $H^1(\mathbf P^1_{\mathbf Z},\Omega)$ is free with basis $c=c_1^{\mathrm{Hodge}}(\mathcal O(1))$ [Tag 0FMI], $H^1(Y,p_j^*\Omega_{\mathbf P^1/\mathbf Z})\cong\mathbf Z$ for $j=1,2$.

(1) The Čech description of $c_1^{\mathrm{Hodge}}$ [Tag 0FLE] shows that it commutes with pullback, and $\mathrm d\log(p_1^*g)=p_1^*\mathrm d\log(g)$. So $e_1$ is the image of $p_1^*c\in H^1(Y,p_1^*\Omega_{\mathbf P^1/\mathbf Z})$ in the first summand of $H^1(Y,\Omega_{Y/\mathbf Z})$. Let $\sigma\colon\mathbf P^1_{\mathbf Z}\to Y$ be the section $x\mapsto(x,(1:0))$ of $p_1$. If $g$ is a basis of $H^1(Y,p_1^*\Omega_{\mathbf P^1/\mathbf Z})$ and $p_1^*c=mg$, then $c=\sigma^*p_1^*c=m\,\sigma^*g$, so $m=\pm1$ and $p_1^*c$ is a basis. The same holds for $e_2$.

(2) The Euler sequence $0\to\Omega_{X/\mathbf Z}\to\mathcal O_X(-1)^{\oplus4}\to\mathcal O_X\to0$ [Tag 0FMH] consists of finite locally free modules, so it stays exact on $Y$. As $\mathcal O_Y(-1)\cong\mathcal O(-1,-1)$ has no cohomology, the boundary maps $H^k(Y,\mathcal O_Y)\to H^{k+1}(Y,i^*\Omega_{X/\mathbf Z})$ are isomorphisms. So $H^2(Y,i^*\Omega_{X/\mathbf Z})=0$, and $H^1(Y,i^*\Omega_{X/\mathbf Z})$ is free with basis the boundary $\xi$ of $1$. In [Tag 0FMH], $\mathcal O_X(-1)^{\oplus4}=\bigoplus_j\mathcal O_X(-1)\mathrm dT_j$ maps to $\mathcal O_X$ by $\mathrm dT_j\mapsto T_j$, and over $D_+(T_k)$ the section $\mathrm d(T_j/T_k)$ of $\Omega_{X/\mathbf Z}$ maps to $T_k^{-1}\mathrm dT_j-T_jT_k^{-2}\mathrm dT_k$. Over $i^{-1}D_+(T_k)$, $1$ lifts to $T_k^{-1}\mathrm dT_k$, and $T_{k_1}^{-1}\mathrm dT_{k_1}-T_{k_0}^{-1}\mathrm dT_{k_0}$ is the image of $\mathrm d\log(T_{k_1}/T_{k_0})$. By Lemma 3.7, $\xi$ is the class of the cocycle $(\mathrm d\log(T_{k_1}/T_{k_0}))$, restricted to $Y$. Its image in $H^1(Y,\Omega_{Y/\mathbf Z})$ is $c_1^{\mathrm{Hodge}}(\mathcal O_Y(1))$, computed with the trivializing sections $i^*T_k$ of $\mathcal O_Y(1)$. As $\mathcal O_Y(1)\cong L_1\otimes L_2$ and $c_1^{\mathrm{Hodge}}$ is a homomorphism, it is $e_1+e_2$. The cohomology sequence of (3.1) reads $\begin{aligned}&0=H^1(Y,N)\to H^1(Y,i^*\Omega_{X/\mathbf Z})\\&\quad\to H^1(Y,\Omega_{Y/\mathbf Z})\xrightarrow{\partial}H^2(Y,N)\\&\quad\to H^2(Y,i^*\Omega_{X/\mathbf Z})=0.\end{aligned}$ So $\partial$ is surjective with kernel $\mathbf Z(e_1+e_2)$, and $H^2(Y,N)\cong(\mathbf Ze_1\oplus\mathbf Ze_2)/\mathbf Z(e_1+e_2)$ is free with basis $\partial e_1=-\partial e_2$. Composing the boundary isomorphism $H^0(Y,\mathcal O_Y)\to H^1(Y,i^*\Omega_{X/\mathbf Z})$, $1\mapsto\xi$, with the next map gives the exact sequence of (2).

(3) By Theorem 3.10 (3), $\omega(i,L_j)=-\partial e_j$, so $\omega(i,L_1)=-\partial e_1$ and $\omega(i,L_2)=-\partial e_2=\partial e_1$. Both are nonzero, so neither $L_j$ extends.

(4) The first assertion is [Tag 0FMA] for $n=2$, as $\Omega^2_{\mathbf P^1/\mathbf Z}=0$. Over $D_+(T_0)$ and $D_+(T_1)$ of $\mathbf P^1_{\mathbf Z}$, with $x=T_1/T_0$, the rules $T_0^{-2}\mapsto\mathrm dx$ and $T_1^{-2}\mapsto-\mathrm d(x^{-1})$ agree on the overlap, since $T_1^{-2}=x^{-2}T_0^{-2}$ and $-\mathrm d(x^{-1})=x^{-2}\mathrm dx$. So $\Omega_{\mathbf P^1/\mathbf Z}\cong\mathcal O(-2)$, and $\Omega^2_{Y/\mathbf Z}\cong\mathcal O(-2,-2)\cong N$. The Künneth isomorphism [Tag 0FU4] is built from the pullbacks and the cup product, so on cohomology it sends $\xi\otimes\zeta$ to $p_1^*\xi\cup p_2^*\zeta$. In degree $2$, as $H^0$ and $H^2$ of $\Omega_{\mathbf P^1/\mathbf Z}$ vanish, it identifies $H^1(\mathbf P^1,\Omega)\otimes H^1(\mathbf P^1,\Omega)=\mathbf Z\,c\otimes c$ with $H^2(Y,p_1^*\Omega\otimes p_2^*\Omega)$. So $e_1e_2=p_1^*c\cup p_2^*c$ is a basis. As $H^0(Y,\mathcal O_Y)=\mathbf Z$, the automorphisms of an invertible $\mathcal O_Y$-module are $\pm1$, so the isomorphisms $\Omega^2_{Y/\mathbf Z}\cong N$ are some $\phi$ and $-\phi$. They carry the basis $e_1e_2$ to bases of $H^2(Y,N)=\mathbf Z\,\omega(i,L_1)$, that is, to $\omega(i,L_1)$ and to $-\omega(i,L_1)$. $\square$

4. Morphisms

Proposition 4.1 (extending a morphism). Let $E$ and $F$ be extensions of $E_o$ by $J$ and of $F_o$ by $K$ over $B$ ($I^2=0$), and $f_o\colon E_o\to F_o$, $f_1\colon J\to K$ maps of $B_o$-modules with $f_1\circ u(E)=u(F)\circ(1\otimes f_o)$. A morphism of extensions $f\colon E\to F$ inducing $f_o$ and $f_1$ exists if and only if $f_{1*}[E]-f_o^*[F]=0$; this element lies in $\operatorname{Ext}^1_{B_o}(E_o,K)$. The solutions form a torsor under $\operatorname{Hom}_{B_o}(E_o,K)$.

Proof. This is Lemma 1.3 with $\varphi=f_o$, $\psi=f_1$. $\square$

Proposition 4.2 (deforming a morphism with given source). Let $E$ be an extension of $E_o$ by $J$ over $B$ ($I^2=0$), $F_o$ and $K$ $B_o$-modules, and $f_o\colon E_o\to F_o$, $f_1\colon J\to K$, $u_o\colon I\otimes F_o\to K$ maps with $f_1\circ u(E)=u_o\circ(1\otimes f_o)$. A solution is a pair $(F,f)$ of an extension $F$ of $F_o$ by $K$ over $B$ with $u(F)=u_o$ and a morphism $f\colon E\to F$ inducing $f_o,f_1$; isomorphisms of solutions are isomorphisms of extensions compatible with the maps from $E$.

A solution exists if and only if $\omega(B,u_o)=0$ and, for one (equivalently, every) extension $F'$ with $u(F')=u_o$, the class of $f_{1*}[E]-f_o^*[F']\in\operatorname{Ext}^1_{B_o}(E_o,K)$ vanishes in the cokernel of $f_o^*\colon\operatorname{Ext}^1_{B_o}(F_o,K)\to$ $\operatorname{Ext}^1_{B_o}(E_o,K)$.
The automorphisms of a solution form the group $\operatorname{Hom}_{B_o}(\operatorname{coker}f_o,K)$.
If solutions exist, the classes $[F]$ of the extensions occurring in solutions form a coset of the kernel of $f_o^*\colon\operatorname{Ext}^1_{B_o}(F_o,K)\to$ $\operatorname{Ext}^1_{B_o}(E_o,K)$ in the torsor of Theorem 2.3 (2); and for each such $F$, the isomorphism classes of solutions $(F',f')$ with $F'\cong F$ form a torsor under the cokernel of $f_o^*\colon\operatorname{Hom}_{B_o}(F_o,K)\to$ $\operatorname{Hom}_{B_o}(E_o,K)$.

Proof. (1) A solution requires an $F$ with $u(F)=u_o$, which exists iff $\omega(B,u_o)=0$ (Theorem 2.3), and then a morphism, which exists iff $f_{1*}[E]-f_o^*[F]=0$ (Proposition 4.1). The extensions $F$ with $u(F)=u_o$ are $F'+y$, $y\in\operatorname{Ext}^1_{B_o}(F_o,K)$, and $f_o^*[F'+y]=f_o^*[F']+f_o^*y$. So a solution exists iff $f_{1*}[E]-f_o^*[F']\in f_o^*\operatorname{Ext}^1_{B_o}(F_o,K)$.

(2) An automorphism of $F$ as an extension is $\mathrm{id}+h\circ p$ with $h\in\operatorname{Hom}_{B_o}(F_o,K)$; it is compatible with $f$ iff $h\circ p\circ f=h\circ f_o\circ p_E=0$, that is, $hf_o=0$, $p_E$ being surjective.

(3) By the proof of (1), $F$ occurs in a solution exactly when $u(F)=u_o$ and $f_o^*[F]=f_{1*}[E]$; if $F_1$ is one such, the others are the $F_1+y$ with $f_o^*y=0$. Fix $F$. The morphisms $f\colon E\to F$ inducing $f_o,f_1$ form a torsor under $\operatorname{Hom}_{B_o}(E_o,K)$, acting by $f\mapsto f+k\circ h\circ p_E$ with $k\colon K\to F$ the inclusion (Proposition 4.1). Two solutions $(F,f)$, $(F,f')$ are isomorphic exactly when an automorphism $\mathrm{id}+k\,g\,p$ of the extension $F$, $g\in\operatorname{Hom}_{B_o}(F_o,K)$, carries $f$ to $f'$, that is, when $f'-f=k\,g\,p\,f=k\,(gf_o)\,p_E$; so the isomorphism classes are the orbits of $\operatorname{Hom}_{B_o}(E_o,K)$ modulo the image of $f_o^*$. An isomorphism $F'\cong F$ of extensions transports solutions with $F'$ to solutions with $F$. $\square$

Remark 4.3. The two conditions of Proposition 4.2 (1) combine into one class in $\operatorname{Ext}^2_{B_o}(C(f_o),K)$, $C(f_o)$ the cone of $f_o$, and the solutions form a torsor under $\operatorname{Ext}^1_{B_o}(C(f_o),K)$ (Theorem 5.3 and Exercise 6.5). The long exact sequence of $\operatorname{Ext}_{B_o}(-,K)$ for the triangle of the cone, $E_o\to F_o\to C(f_o)\to E_o[1]$, shows that $\operatorname{Ext}^1_{B_o}(C(f_o),K)$ is an extension of the kernel of $f_o^*\colon\operatorname{Ext}^1_{B_o}(F_o,K)\to$ $\operatorname{Ext}^1_{B_o}(E_o,K)$ by the cokernel of $f_o^*\colon\operatorname{Hom}_{B_o}(F_o,K)\to$ $\operatorname{Hom}_{B_o}(E_o,K)$, the two groups of Proposition 4.2 (3).

5. The obstruction as a single class

This section obtains the obstruction of Proposition 4.2 as one class, by turning modules into graded rings. For a ring $R$ of $T$ and an $R$-module $M$, $R\oplus M$ denotes the graded ring with $R$ in degree $0$, $M$ in degree $1$ and $M\cdot M=0$. An extension $E$ of $E_o$ by $J$ over $B$ is the degree-one part of the graded ring $B\oplus E$, and a morphism of extensions is the degree-one part of a map of such graded rings. Graded rings and modules, the functors $k^n$, the modules $F\langle n\rangle$, the derived categories $D(-)_{\mathrm{gr}}$ and the graded cotangent complex $L^{\mathrm{gr}}$ are those of the lesson on graded modules, and $\operatorname{Ext}^i_P(X,Y)_{\mathrm{gr}}=\operatorname{Hom}_{D(P)_{\mathrm{gr}}}(X,Y[i])$.

Lemma 5.1 (graded extensions over a thickening). Let $R'\to R$ be a surjective map of graded rings whose kernel $N$ has square zero, $R\to P$ a map of graded rings, $M$ a graded $P$-module and $c\colon N\to M$ a graded $R$-linear map. A solution is a graded $R'$-algebra $X$ with a surjective map $X\to P$ of graded $R'$-algebras whose kernel has square zero and is identified with $M$ as a graded $P$-module, such that the structure map $R'\to X$ carries $N$ into $M$ by $c$; an isomorphism of solutions is an isomorphism of graded $R'$-algebras compatible with the maps to $P$ and inducing the identity on $M$. Let $\begin{aligned}&D\to L^{\mathrm{gr}}_{P/R'}\\&\quad\to L^{\mathrm{gr}}_{P/R}\xrightarrow{\ \delta\ }D[1]\end{aligned}$ be the transitivity triangle of $R'\to R\to P$ (Theorem 3.4 of the lesson on graded modules), $D=L^{\mathrm{gr}}_{R/R'}\otimes^{\mathbf L}_RP$. Then $D$ has no cohomology in degrees $>-1$, $H^{-1}(D)=N\otimes_RP$, and there is a unique map $\tilde c\colon D\to M[1]$ inducing on $H^{-1}$ the $P$-linear extension $N\otimes_RP\to M$ of $c$.

A solution exists if and only if the class $\operatorname{ob}(c)=\tilde c[1]\circ\delta$ in $\operatorname{Ext}^2_P(L^{\mathrm{gr}}_{P/R},M)_{\mathrm{gr}}$ vanishes.
If it vanishes, the isomorphism classes of solutions form a torsor under $\operatorname{Ext}^1_P(L^{\mathrm{gr}}_{P/R},M)_{\mathrm{gr}}$, and the automorphisms of a solution form the group $\operatorname{Ext}^0_P(L^{\mathrm{gr}}_{P/R},M)_{\mathrm{gr}}=\operatorname{Der}_R(P,M)_{\mathrm{gr}}$.
For a graded $P$-linear map $\varphi\colon M\to M'$, $\operatorname{ob}(\varphi c)=\varphi[2]\circ\operatorname{ob}(c)$.

Proof. By Proposition 4.1 of the lesson on graded modules, $L=L^{\mathrm{gr}}_{R/R'}$ has $H^0(L)=0$ and $H^{-1}(L)\cong N/N^2=N$. Its terms are free, so $D$ is computed by $L\otimes_RP$ (Proposition 1.6 (3) there). As $H^0(L)=0$, the differential $L^{-1}\to L^0$ is surjective; its kernel $Z$ is flat (after $\Phi$, the kernel of a surjection of flat modules), so $0\to Z\to L^{-1}\to L^0\to0$ stays exact after $-\otimes_RP$, and $H^{-1}(L)$ is the cokernel of $L^{-2}\to Z$. By right exactness of $-\otimes_RP$, $D$ has no cohomology in degrees $>-1$ and $H^{-1}(D)=N\otimes_RP$. Hence $\operatorname{Hom}(D,M)_{\mathrm{gr}}=0$, and the maps $D\to M[1]$ correspond to the maps $H^{-1}(D)\to M$ [Tag 06XS]: $\operatorname{Ext}^1(D,M)_{\mathrm{gr}}=$ $\operatorname{Hom}_P(N\otimes_RP,M)_{\mathrm{gr}}=$ $\operatorname{Hom}_R(N,M)_{\mathrm{gr}}$. This gives $\tilde c$. The sequence $\begin{aligned}&0\to\operatorname{Ext}^1(L^{\mathrm{gr}}_{P/R},M)_{\mathrm{gr}}\\&\quad\to\operatorname{Ext}^1(L^{\mathrm{gr}}_{P/R'},M)_{\mathrm{gr}}\\&\quad\xrightarrow{\ r\ }\operatorname{Hom}_R(N,M)_{\mathrm{gr}}\\&\quad\xrightarrow{\ \partial\ }\operatorname{Ext}^2(L^{\mathrm{gr}}_{P/R},M)_{\mathrm{gr}},\end{aligned}$ in which the first map is composition with the functoriality map, $r$ is composition with $D\to L^{\mathrm{gr}}_{P/R'}$ and $\partial(c)=\tilde c[1]\circ\delta$, is exact. Indeed, $\operatorname{Hom}(-,M[1])$ carries the triangle and its rotation $L^{\mathrm{gr}}_{P/R}[-1]\xrightarrow{-\delta[-1]}D\to L^{\mathrm{gr}}_{P/R'}\to L^{\mathrm{gr}}_{P/R}$ to exact sequences [Tags 0145, 0149]; the term before the first group is $\operatorname{Hom}(D,M)_{\mathrm{gr}}=0$; and composition with $-\delta[-1]$ has the same kernel as $\partial$, the shift identifying $\operatorname{Hom}(L^{\mathrm{gr}}_{P/R}[-1],M[1])$ with $\operatorname{Ext}^2(L^{\mathrm{gr}}_{P/R},M)_{\mathrm{gr}}$.

By Theorem 4.5 of the lesson on graded modules, applied to $R'\to P$, the middle group is $\operatorname{Exal}_{R'}(P,M)_{\mathrm{gr}}$: an extension $X$ corresponds to the composite $\kappa_M(X)\colon L^{\mathrm{gr}}_{P/R'}\to L^{\mathrm{gr}}_{P/X}\to M[1]$. The composite $D\to L^{\mathrm{gr}}_{P/R'}\to L^{\mathrm{gr}}_{P/X}$ is the $P$-linear extension of the map $L^{\mathrm{gr}}_{R/R'}\to L^{\mathrm{gr}}_{P/X}$ induced by the commutative square formed by $R'\to X$ and $R\to P$, functoriality maps being compatible with composition. By the naturality in Proposition 4.1 there, on $H^{-1}$ it is the $P$-linear extension of the map $c_X\colon N\to M$ of kernels that $R'\to X$ defines. So $r(X)=c_X$, the solutions are the extensions $X$ with $r(X)=c$, and isomorphisms of solutions are isomorphisms of extensions. By exactness, there is such an $X$ exactly when $\partial(c)=0$, and their classes then form a coset of the image of the injective map from $\operatorname{Ext}^1(L^{\mathrm{gr}}_{P/R},M)_{\mathrm{gr}}$. This proves (1) and the torsor statement.

By Theorem 4.5 there, the automorphisms of a solution $X$ are the maps $x\mapsto x+\theta(\bar x)$ for the graded $R'$-derivations $\theta$ of $P$ into $M$. These vanish on the image of $R'$, which is that of $R$, $R'\to R$ being surjective; so they are the graded $R$-derivations, and $\operatorname{Der}_R(P,M)_{\mathrm{gr}}=$ $\operatorname{Hom}(\Omega_{P/R},M)_{\mathrm{gr}}=$ $\operatorname{Ext}^0(L^{\mathrm{gr}}_{P/R},M)_{\mathrm{gr}}$ (Lemma 3.1 (1) and Theorem 3.3 (2) there, and [Tag 06XS]). (3) The maps $\widetilde{\varphi c}$ and $\varphi[1]\circ\tilde c$ induce the same map on $H^{-1}(D)$, so they are equal, and $\operatorname{ob}(\varphi c)=\varphi[2]\circ\tilde c[1]\circ\delta$. $\square$

Lemma 5.2 (dual numbers over dual numbers). Let $f_o\colon E_o\to F_o$ be a map of $B_o$-modules, $R=B_o\oplus E_o$ and $P=B_o\oplus F_o$, with $P$ an $R$-algebra by $\mathrm{id}\oplus f_o$. Let $C(f_o)$ be the cone of $f_o$, the complex $E_o\xrightarrow{f_o}F_o$ with $F_o$ in degree $0$ [Tag 014E], and $j_o\colon F_o\to C(f_o)$ the inclusion.

Let $L$ be $L^{\mathrm{gr}}_{P/R}$ or $L^{\mathrm{gr}}_{P/B_o}$, and let $K$ and $Q$ be $B_o$-modules. For every $i$, $g\mapsto k^1(g)$ is an isomorphism from $\operatorname{Ext}^i_P(L,K\langle1\rangle)_{\mathrm{gr}}$ to $\operatorname{Ext}^i_{B_o}(k^1(L),K)$, and $\operatorname{Ext}^i_P(L,Q\langle0\rangle)_{\mathrm{gr}}=0$.
There is a canonical isomorphism $\gamma\colon k^1(L^{\mathrm{gr}}_{P/R})\to C(f_o)$ in $D(B_o)$ with $\gamma\circ k^1(\psi)=j_o\circ\beta$, where $\psi\colon L^{\mathrm{gr}}_{P/B_o}\to L^{\mathrm{gr}}_{P/R}$ is the functoriality map and $\beta\colon k^1(L^{\mathrm{gr}}_{P/B_o})\to F_o$ is $k^1$ of the augmentation followed by Lemma 3.1 (4) of the lesson on graded modules, an isomorphism by Theorem 3.3 (5) there.
Let $\alpha\colon k^1(L^{\mathrm{gr}}_{R/B_o}\otimes^{\mathbf L}_RP)\to E_o$ be $k^1$ of the augmentation $L^{\mathrm{gr}}_{R/B_o}\otimes_RP\to\Omega_{R/B_o}\otimes_RP$ followed by $k^1(\Omega_{R/B_o}\otimes_RP)=k^1(\Omega_{R/B_o})\cong E_o$ (Lemma 3.1 (4) there). Then $(\alpha,\beta,\gamma)$ is an isomorphism of triangles from $k^1$ of the transitivity triangle of $B_o\to R\to P$ to the triangle $(E_o,F_o,C(f_o),f_o,j_o,-p)$, $p\colon C(f_o)\to E_o[1]$ the projection, which is distinguished [Tags 014E, 014L].

Proof. (1) Write $L=L^{\mathrm{gr}}_{P/A}$ with $A=R$ or $A=B_o$; in both cases $A^0=P^0=B_o$. As a simplicial graded $P$-module, $L$ has free terms $P^{(X_n)}$ with $X_n\in T^{\mathrm{gr}\geq0}$ (Definition 3.2 there), which live in degrees $\geq0$. For a graded $P$-module $Y$ in degrees $\geq0$, $(P^+Y)^0=0$, and $(P^+Y)^1$ is the image of the multiplication $F_o\otimes_{B_o}Y^0\to Y^1$, which is injective when $Y$ is free. So the quotient map $L\to B_o\otimes_PL=L/P^+L$ induces an isomorphism $k^0(L)\to k^0(B_o\otimes_PL)$ and an exact sequence of simplicial $B_o$-modules $\begin{aligned}&0\to F_o\otimes_{B_o}k^0(L)\\&\quad\to k^1(L)\xrightarrow{\ \eta\ }k^1(B_o\otimes_PL)\to0.\end{aligned}$ By Theorem 3.3 (4) there, $k^0(L)=L_{B_o/B_o}$, which is acyclic by Theorem 3.3 (3), $B_o$ being the polynomial algebra over itself on no variables. Its terms are free, so $F_o\otimes_{B_o}k^0(L)$ is acyclic (Lemma 2.1 of the lesson on derived tensor products, applied to the quasi-isomorphism $k^0(L)\to0$), and $\eta$ is a quasi-isomorphism. As $B_o\otimes_PL$ computes $B_o\otimes^{\mathbf L}_PL$ (Proposition 1.6 (3) there), Proposition 1.6 (2) there gives isomorphisms $\begin{gathered}\operatorname{Ext}^i_P(L,K\langle1\rangle)_{\mathrm{gr}}\\ \cong\operatorname{Ext}^i_{B_o}(k^1(B_o\otimes_PL),K),\\ \operatorname{Ext}^i_P(L,Q\langle0\rangle)_{\mathrm{gr}}\\ \cong\operatorname{Ext}^i_{B_o}(k^0(B_o\otimes_PL),Q)=0.\end{gathered}$ The first is the derived adjunction [Tag 0DVC]: it sends $g$ to $\varepsilon\circ G(g)$, where $G=k^1(B_o\otimes^{\mathbf L}_P-)$ and $\varepsilon\colon G(K\langle1\rangle[i])\to K[i]$ is induced by the counit. The quotient maps $Y\to B_o\otimes_PY$ give a natural transformation from $k^1$ to $G$ on bounded above complexes of flat graded modules, hence on $D^-(P)_{\mathrm{gr}}$. It is $\eta$ at $L$, and at $K\langle1\rangle[i]$ its composite with $\varepsilon$ is the identity of $k^1(K\langle1\rangle[i])=K[i]$, because $B_o\otimes_PK\langle1\rangle=K\langle1\rangle$. By naturality, $\varepsilon\circ G(g)\circ\eta=k^1(g)$. As $\eta$ is an isomorphism, $g\mapsto k^1(g)$ is an isomorphism.

(2) Let $f=\mathrm{id}\oplus f_o\colon R\to P$ and $C'=R\times P$, graded degree by degree, an $R$-algebra by the injective map $(\mathrm{id},f)$. By the proof of Theorem 3.4 of the lesson on graded modules, the transitivity triangle of $B_o\to R\to P$ is the triangle of the exact sequence of simplicial graded $P$-modules $\begin{aligned}&0\to L^{\mathrm{gr}}_{R/B_o}\otimes_RP\\&\quad\xrightarrow{\ u\ }L^{\mathrm{gr}}_{C'/B_o}\otimes_{C'}P\\&\quad\xrightarrow{\ v\ }W\to0,\end{aligned} \tag{5.1}$ obtained from the sequence (3.1) there for $B_o\to R\to C'$ by $-\otimes_{C'}P$. Its middle term is identified with $L^{\mathrm{gr}}_{P/B_o}$ by the quasi-isomorphism $\pi$ induced by the projection $C'\to P$, and $W$ with $L^{\mathrm{gr}}_{P/R}$ by an isomorphism $\iota\colon W\to L^{\mathrm{gr}}_{P/R}$ of $D(P)_{\mathrm{gr}}$ given by quasi-isomorphisms; under these identifications $\pi\circ u$ is the functoriality map of $f$ and $\iota\circ v=\psi\circ\pi$.

Normalize and apply $k^1$, which is exact: this gives an exact sequence of complexes of $B_o$-modules $0\to X\xrightarrow{k^1u}Y\xrightarrow{k^1v}Z\to0$. Let $a=\alpha\colon X\to E_o$ be the map of (3) (here $k^0(\Omega_{R/B_o})=0$), and $b=\beta\circ k^1(\pi)\colon Y\to F_o$. Both are quasi-isomorphisms. For $b$ this is Theorem 3.3 (5) there. For $a$: by Theorem 3.3 (5) there, $\theta^+$ gives a quasi-isomorphism $U\to L^{\mathrm{gr}}_{R/B_o}$ from a complex $U$ of free graded modules generated in degrees $\geq1$, whose $k^1$, followed by the augmentation and Lemma 3.1 (4), is a quasi-isomorphism $k^1(U)\to E_o$. The map $U\otimes_RP\to L^{\mathrm{gr}}_{R/B_o}\otimes_RP$ is a quasi-isomorphism (Proposition 1.6 (3) there), $x\mapsto x\otimes1$ identifies $k^1(U)$ with $k^1(U\otimes_RP)$, and $a$ composed with $k^1$ of this map is that quasi-isomorphism. By the naturality of the augmentations and of Lemma 3.1 (4) there, $b\circ k^1u=f_o\circ a$.

Let $C$ be the cone of $k^1u$, with $C^n=Y^n\oplus X^{n+1}$ and $d(y,x)=(dy+k^1u(x),-dx)$ [Tag 014E]. The map $q\colon C\to Z$, $(y,x)\mapsto k^1v(y)$, is a quasi-isomorphism [Tag 0152]. The map $(y,x)\mapsto(b(y),a(x))$ from $C$ to $C(f_o)$ is a map of complexes because $b\circ k^1u=f_o\circ a$, and a quasi-isomorphism because $a$ and $b$ are (the five lemma for the sequences $0\to Y\to C\to X[1]\to0$ and $0\to F_o\to C(f_o)\to E_o[1]\to0$). Let $\gamma$ be the composite of $k^1(\iota)^{-1}$, $q^{-1}$ and this map; no choice enters. The inclusion of $Y$ into $C$ composed with $q$ is $k^1v$, and composed with the map to $C(f_o)$ it is $j_o\circ b$. Hence $\gamma\circ k^1(\psi)\circ k^1(\pi)=$ $\gamma\circ k^1(\iota)\circ k^1v=$ $j_o\circ b=$ $j_o\circ\beta\circ k^1(\pi)$, and $k^1(\pi)$ is invertible.

(3) The first square commutes because $b\circ k^1u=f_o\circ a$, with $b=\beta\circ k^1(\pi)$ and $\pi\circ u$ the first map of the transitivity triangle; the second is (2). By construction the transitivity triangle is the triangle of (5.1) through $\pi$ and $\iota$, so its third map $\delta$ satisfies $\delta\circ\iota=\delta_{(5.1)}$, and $k^1$ of the latter is $-\mathrm{pr}\circ q^{-1}$, with $\mathrm{pr}\colon C\to X[1]$ the projection [Tag 0152]. As $-p\circ(b,a[1])=a[1]\circ(-\mathrm{pr})$ on $C$, $-p\circ\gamma=a[1]\circ(-\mathrm{pr})\circ q^{-1}\circ k^1(\iota)^{-1}=$ $\alpha[1]\circ k^1(\delta)$. $\square$

Theorem 5.3 (the single obstruction). In the situation of Proposition 4.2, let $R$, $P$, $C(f_o)$ and $\gamma$ be as in Lemma 5.2, and let $M$ be the graded $P$-module with $M^0=I$ and $M^1=K$, on which $p\in F_o$ acts by $I\to K$, $x\mapsto u_o(x\otimes p)$.

For every $i$, $\Theta(h)=k^1(h)\circ\gamma^{-1}$ defines an isomorphism $\Theta$ from $\operatorname{Ext}^i_P(L^{\mathrm{gr}}_{P/R},M)_{\mathrm{gr}}$ to $\operatorname{Ext}^i_{B_o}(C(f_o),K)$.
There is a class $\omega(B,E,f_o,f_1,u_o)$ in $\operatorname{Ext}^2_{B_o}(C(f_o),K)$ that vanishes if and only if a solution exists.
If it vanishes, the isomorphism classes of solutions form a torsor under $\operatorname{Ext}^1_{B_o}(C(f_o),K)$, and the automorphism group of a solution is $\operatorname{Ext}^0_{B_o}(C(f_o),K)=$ $\operatorname{Hom}_{B_o}(\operatorname{coker}f_o,K)$.

Proof. $M$ is a graded $P$-module because $F_o\cdot F_o=0$ in $P$ and $M^2=0$. Its submodule $K\langle1\rangle$ has quotient $I\langle0\rangle$. In the long exact sequence of $\operatorname{Ext}(L^{\mathrm{gr}}_{P/R},-)_{\mathrm{gr}}$ the groups with coefficients $I\langle0\rangle$ vanish (Lemma 5.2 (1)), so composition with the inclusion $K\langle1\rangle\subseteq M$ is an isomorphism $\operatorname{Ext}^i(L^{\mathrm{gr}}_{P/R},K\langle1\rangle)_{\mathrm{gr}}\to\operatorname{Ext}^i(L^{\mathrm{gr}}_{P/R},M)_{\mathrm{gr}}$. As $k^1$ of this inclusion is the identity of $K$, (1) follows from Lemma 5.2.

Let $R'=B\oplus E$. The map $R'\to R$ is surjective with kernel $N=I\oplus J$, of square zero: $I^2=0$, $I$ annihilates $J$, and $E\cdot E=0$; the product of $x\in I$ and $e\in E$ is $u(E)(x\otimes\bar e)\in J$. Let $c\colon N\to M$ be the identity on $I$ and $f_1$ on $J$. It is $R$-linear: for $e\in E_o$ and $x\in I$, $c(e\cdot x)=$ $f_1(u(E)(x\otimes e))=$ $u_o(x\otimes f_o(e))=$ $f_o(e)\cdot c(x)$, and for $x\in J$ both sides vanish, $N$ and $M$ having nothing in degree $2$.

The solutions of Proposition 4.2 and their isomorphisms correspond to the solutions of Lemma 5.1 for $(R',R,P,M,c)$ and their isomorphisms. Given $(F,f)$, let $X=B\oplus F$, with $X\to P$ given by $B\to B_o$ and $F\to F_o$, and $R'\to X$ by the identity of $B$ and $f$. The kernel of $X\to P$ is $I\oplus K$, of square zero, and $x\in I$ times a lift of $p\in F_o$ is $u(F)(x\otimes p)=u_o(x\otimes p)$, so the kernel is $M$ as a graded $P$-module; and $R'\to X$ carries $I$ identically and $J$ by $f_1$, that is, $N$ by $c$. Conversely, let $X$ be a solution. As $P$ and $M$ vanish in degrees $\geq2$, so does $X$. Its degree-zero part $X^0$ is a $B$-algebra and an extension of $B_o$ by $I$, the map $B\to X^0$ inducing the identity on $I$ and on $B_o$, so $B\to X^0$ is an isomorphism by the five lemma. Then $F=X^1$ is a $B$-module and an extension of $F_o$ by $K$, with $u(F)=u_o$ by the module structure of $M$; the degree-one part $f\colon E\to F$ of $R'\to X$ is $B$-linear, induces $f_o$ since $R'\to X\to P$ is $R'\to R\to P$, and induces $f_1$ on $J$. The two constructions are inverse to each other up to the isomorphism $B\cong X^0$, and they match isomorphisms and automorphisms, which are determined by their degree-one parts.

Let $\omega(B,E,f_o,f_1,u_o)=\Theta(\operatorname{ob}(c))$. Lemma 5.1 and (1) give (2) and (3), and $\operatorname{Ext}^0(C(f_o),K)=\operatorname{Hom}(H^0(C(f_o)),K)$ [Tag 06XS], with $H^0(C(f_o))=\operatorname{coker}f_o$. $\square$

Remark 5.4 (quotient modules). If $f_o$ is surjective with kernel $N_o$, the inclusion $N_o[1]\to C(f_o)$, by $N_o\subseteq E_o$ in degree $-1$, is a quasi-isomorphism, so $\operatorname{Ext}^i_{B_o}(C(f_o),K)\cong\operatorname{Ext}^{i-1}_{B_o}(N_o,K)$. The obstruction of Theorem 5.3 then lies in $\operatorname{Ext}^1_{B_o}(N_o,K)$, the solutions form a torsor under $\operatorname{Hom}_{B_o}(N_o,K)$, and a solution has no automorphism other than the identity. If moreover $u(E)$ and $u_o$ are surjective, then $f_1$ is determined by $f_o$ through $f_1\circ u(E)=u_o\circ(1\otimes f_o)$, and the morphism $f\colon E\to F$ of every solution is surjective: $F=f(E)+K$ because $f_o$ is surjective, and $K=u_o(I\otimes F_o)=IF=I\,f(E)\subseteq f(E)$. So the solutions are quotients of $E$ lifting the quotient $F_o$ of $E_o$: this is the deformation theory of quotient modules.

Proposition 5.5 (the canonical obstruction). In the situation of Theorem 5.3, let $M_o$ be the pushout of $u(E)\colon I\otimes E_o\to J$ and $1\otimes f_o\colon I\otimes E_o\to I\otimes F_o$, with its maps $\lambda\colon J\to M_o$ and $\mu\colon I\otimes F_o\to M_o$, and let $(f_1,u_o)\colon M_o\to K$ be the map induced by $f_1$ and $u_o$. As $\lambda\circ u(E)=\mu\circ(1\otimes f_o)$, Theorem 5.3 applies to $K=M_o$, $f_1=\lambda$, $u_o=\mu$; let $\omega(B,E,f_o)=\omega(B,E,f_o,\lambda,\mu)$ be its class, in $\operatorname{Ext}^2_{B_o}(C(f_o),M_o)$.

$\omega(B,E,f_o,f_1,u_o)=$ $(f_1,u_o)[2]\circ\omega(B,E,f_o)$. If $u(E)$ is an isomorphism, so is $\mu$.
Let $\delta_A$ be the third map of the transitivity triangle of $A\to R\to P$, and $\operatorname{at}_{B_o/A}(f_o)=k^1(\delta_A)\circ\gamma^{-1}$, a map from $C(f_o)$ to $k^1(L^{\mathrm{gr}}_{R/A}\otimes^{\mathbf L}_RP)[1]$. Let $e\in\operatorname{Ext}^1_R(L^{\mathrm{gr}}_{R/A},N)_{\mathrm{gr}}$ be the class of the graded $A$-extension $R'=B\oplus E$ of $R$ by $N=I\oplus J$ (Theorem 4.5 of the lesson on graded modules). Then $k^1(N\otimes_RP)=M_o$ and $\begin{aligned}\omega(B,E,f_o)&=k^1(\mathrm{aug})[2]\\&\quad\circ k^1(e\otimes^{\mathbf L}_RP)[1]\\&\quad\circ\operatorname{at}_{B_o/A}(f_o),\end{aligned}$ where $\mathrm{aug}\colon N\otimes^{\mathbf L}_RP\to N\otimes_RP$ is the augmentation.

Proof. Let $M_{\mathrm u}$ and $c_{\mathrm u}\colon N\to M_{\mathrm u}$ be the module and the map of the proof of Theorem 5.3 for the data $(M_o,\lambda,\mu)$: $M_{\mathrm u}^0=I$, $M_{\mathrm u}^1=M_o$, and $c_{\mathrm u}$ is the identity on $I$ and $\lambda$ on $J$.

(1) The graded map $\varphi\colon M_{\mathrm u}\to M$ that is the identity on $I$ and $(f_1,u_o)$ on $M_o$ is $P$-linear, because $(f_1,u_o)\circ\mu=u_o$, and $\varphi c_{\mathrm u}=c$. By Lemma 5.1 (3), $\operatorname{ob}(c)=\varphi[2]\circ\operatorname{ob}(c_{\mathrm u})$. Applying $k^1$ and composing with $\gamma^{-1}$ gives the formula, as $k^1(\varphi)=(f_1,u_o)$. A pushout of an isomorphism is an isomorphism.

(2) In degree $0$, $N\otimes_RP$ is $I\otimes_{B_o}B_o=I$. In degree $1$ it is generated by $J\otimes_{B_o}B_o=J$ and $I\otimes_{B_o}F_o$, subject to the relations $(e\cdot x)\otimes1=x\otimes f_o(e)$ for $x\in I$ and $e\in E_o$, where $e\cdot x=u(E)(x\otimes e)$; so it is $M_o$. Let $\mathrm q\colon N\otimes_RP\to M_{\mathrm u}$ be the $P$-linear extension of $c_{\mathrm u}$; it is the identity in degrees $0$ and $1$, so $k^1(\mathrm q)=\mathrm{id}_{M_o}$. The maps $\tilde c_{\mathrm u}$ and $\mathrm q[1]\circ\mathrm{aug}[1]\circ(\tau\otimes^{\mathbf L}_RP)$ from $D$ to $M_{\mathrm u}[1]$, where $\tau\colon L^{\mathrm{gr}}_{R/R'}\to N[1]$ is the truncation of Proposition 4.1 there, both induce $\mathrm q$ on $H^{-1}(D)=N\otimes_RP$, so they are equal (Lemma 5.1). The structure map $A\to R'$ and the identities of $R$ and $P$ form a map from the diagram $A\to R\to P$ to the diagram $R'\to R\to P$; by the naturality of transitivity triangles (Theorem 3.4 there), $\delta=(\ell\otimes^{\mathbf L}_RP)[1]\circ\delta_A$, with $\ell\colon L^{\mathrm{gr}}_{R/A}\to L^{\mathrm{gr}}_{R/R'}$ the functoriality map. As $e=\tau\circ\ell$ (Theorem 4.5 there), $\begin{aligned}\operatorname{ob}(c_{\mathrm u})&=\mathrm q[2]\circ\mathrm{aug}[2]\\&\quad\circ(e\otimes^{\mathbf L}_RP)[1]\circ\delta_A.\end{aligned}$ Apply $k^1$ and compose with $\gamma^{-1}$. $\square$

By Exercise 6.4, the object $k^1(L^{\mathrm{gr}}_{R/A}\otimes^{\mathbf L}_RP)$ in the target of $\operatorname{at}_{B_o/A}(f_o)$ sits in a distinguished triangle with $L_{B_o/A}\otimes_{B_o}F_o$ and $E_o$, whose third map is $(1\otimes f_o)[1]\circ\operatorname{at}_{B_o/A}(E_o)$.

Proposition 5.6 (comparison with Section 2). Let $F_o$ and $K$ be $B_o$-modules and $u_o\colon I\otimes_{B_o}F_o\to K$, and apply Theorem 5.3 to $E=0$, $f_o=0$ and $f_1=0$. Then $C(f_o)=F_o$, $\gamma=\beta$ (Lemma 5.2 (2), $\psi$ being the identity), and the solutions are the extensions $F$ of $F_o$ by $K$ over $B$ with $u(F)=u_o$, those of Theorem 2.3. Write $C_o=B_o\oplus F_o$.

$\omega(B,0,0,0,u_o)=\omega(B,u_o)$, the class of Definition 2.2.
Let $M'$ be an extension of $F_o$ by $K$ over $B_o$, $\delta_{M'}\colon F_o\to K[1]$ the third map of its triangle [Tag 0152], and $Y$ the graded $B_o$-extension of $C_o$ by $M$ obtained from $B_o\oplus M'$ by pushout along $K\langle1\rangle\subseteq M$. Then $\Theta(\kappa_M(Y))=-\delta_{M'}$, $\kappa_M$ as in Theorem 4.5 of the lesson on graded modules.
In Theorem 5.3 (3), $\xi\in\operatorname{Ext}^1_{B_o}(F_o,K)$ acts on the solutions by Baer sum with an extension of class $-\xi$: this torsor structure is the inverse of that of Theorem 2.3 (2).

Proof. (1) Here the pushout of Proposition 5.5 is $I\otimes F_o$, with $\mu$ the identity and $(f_1,u_o)=u_o$; and $R=B_o$, $R'=B$, $N=I$, $P=C_o$. The class $e$ of the $A$-extension $B$ of $B_o$ by $I$ is $e(B)=\tau_{B_o/B}\circ c$ of the Conventions. For a complex $X$ of $B_o$-modules, $k^1(X\otimes^{\mathbf L}_{B_o}C_o)=X\otimes^{\mathbf L}_{B_o}F_o$; so $k^1(e\otimes^{\mathbf L}_{B_o}C_o)=e(B)\otimes^{\mathbf L}1_{F_o}$, and $k^1(\mathrm{aug})$ is the augmentation $I\otimes^{\mathbf L}_{B_o}F_o\to I\otimes_{B_o}F_o$. By Theorem 2.4 of the lesson on the Atiyah class, $k^1(\delta_A)\circ\beta^{-1}=\operatorname{at}_{B_o/A}(F_o)$. Proposition 5.5 (2) and Theorem 3.2 give $\omega(B,0,0)=\omega(B,F_o)$, and Proposition 5.5 (1) gives $\omega(B,0,0,0,u_o)=$ $u_o[2]\circ\omega(B,F_o)=$ $\omega(B,u_o)$.

(2) Put $X=B_o\oplus M'$. By the naturality of $\kappa$ in the coefficients, $\kappa_M(Y)$ is $\kappa_{K\langle1\rangle}(X)$ followed by the inclusion, so $\Theta(\kappa_M(Y))=k^1(\kappa_{K\langle1\rangle}(X))\circ\beta^{-1}$. Apply Lemma 5.2 to the projection $p'\colon M'\to F_o$ in place of $f_o$, so that $X$ plays the part of $R$ and $C_o$ that of $P$; let $\gamma'\colon k^1(L^{\mathrm{gr}}_{C_o/X})\to C(p')$, $\alpha'$, $\psi'\colon L^{\mathrm{gr}}_{C_o/B_o}\to L^{\mathrm{gr}}_{C_o/X}$ and $j'\colon F_o\to C(p')$ be the corresponding maps. As $X\to C_o$ is surjective with kernel $K\langle1\rangle$, $\kappa_{K\langle1\rangle}(X)=\tau\circ\psi'$, with $\tau\colon L^{\mathrm{gr}}_{C_o/X}\to K\langle1\rangle[1]$ the truncation of Proposition 4.1 there. By Lemma 5.2 (2), $k^1(\psi')\circ\beta^{-1}=\gamma'^{-1}\circ j'$, so $\Theta(\kappa_M(Y))=k^1(\tau)\circ\gamma'^{-1}\circ j'$.

The complex $C(p')=[M'\xrightarrow{p'}F_o]$ has $H^0=0$ and $H^{-1}=K$, so the inclusion $\nu\colon K[1]\to C(p')$ is a quasi-isomorphism, and a map $C(p')\to K[1]$ is determined by its effect on $H^{-1}$ [Tag 06XS]. We show $k^1(\tau)\circ\gamma'^{-1}=-\nu^{-1}$. Let $\delta_X$ be the third map of the transitivity triangle of $B_o\to X\to C_o$. Its image under $\Phi$ is the third map of the ungraded triangle (Theorem 3.4 there), and the identifications of $H^{-1}(L^{\mathrm{gr}}_{C_o/X})$ with $K$ (Proposition 4.1 there) and of $H^0(L^{\mathrm{gr}}_{X/B_o}\otimes C_o)$ with $\Omega_{X/B_o}\otimes_XC_o$ (Theorem 3.3 (2) there) are given by the same formulas as the ungraded ones; so by Lemma 3.9, $H^{-1}(\delta_X)$ sends $k\in K$ to $\mathrm dk\otimes1$, and $\alpha'\circ H^{-1}(k^1\delta_X)$ is the inclusion $K\subseteq M'$ composed with $H^{-1}(k^1\tau)$. By Lemma 5.2 (3), $-p\circ\gamma'=\alpha'[1]\circ k^1(\delta_X)$, where the projection $p\colon C(p')\to M'[1]$ induces the inclusion $K\subseteq M'$ on $H^{-1}$. Hence $H^{-1}(\gamma')=-H^{-1}(k^1\tau)$, so $k^1(\tau)\circ\gamma'^{-1}$ induces $-\mathrm{id}_K$ on $H^{-1}$, as $-\nu^{-1}$ does.

Next, $\nu^{-1}\circ j'=\delta_{M'}$. Let $C(i)$ be the cone of the inclusion $i\colon K\to M'$, with the quasi-isomorphism $q\colon C(i)\to F_o$ and the projection $\mathrm{pr}\colon C(i)\to K[1]$, so that $\delta_{M'}=-\mathrm{pr}\circ q^{-1}$ [Tag 0152]. The map of complexes $C(i)\to C(p')$ equal to $i$ in degree $-1$ and to $p'$ in degree $0$ is $j'\circ q+\nu\circ\mathrm{pr}$, and it is null-homotopic, by the identity of $M'$ from degree $0$ of $C(i)$ to degree $-1$ of $C(p')$. So $j'=-\nu\circ\mathrm{pr}\circ q^{-1}=\nu\circ\delta_{M'}$. Together, $\Theta(\kappa_M(Y))=-\nu^{-1}\circ j'=-\delta_{M'}$.

(3) Every graded $B_o$-extension of $C_o$ by $K\langle1\rangle$ is $B_o\oplus M'$ for an extension $M'$ of $F_o$ by $K$ over $B_o$, its degree-zero part being $B_o$. So, by Theorem 4.5 there, the bijectivity of $\Theta$ and (2), every $\xi$ is $\Theta(\kappa_M(Y))$ for the $Y$ built from an extension $M'$ of class $-\xi$. By Lemma 5.1 and the proof of Theorem 5.3, $\xi$ acts by adding, in $\operatorname{Exal}_B(C_o,M)_{\mathrm{gr}}$, the class of $Y$ viewed over $B$ (composition with the functoriality map $L^{\mathrm{gr}}_{C_o/B}\to L^{\mathrm{gr}}_{C_o/B_o}$). The Baer sum of $B\oplus F$ and $Y$ is formed degree by degree: in degree $0$ it is the Baer sum of $B$ with the trivial extension $B_o\oplus I$, which is $B$, and in degree $1$ it is the Baer sum of $F$ and $M'$. $\square$

So, for $E=0$, the graded transitivity triangle of $B\to B_o\to B_o\oplus F_o$, with the coefficients $I\oplus K\langle1\rangle$ on which $F_o$ acts through $u_o$, gives Theorem 2.3 (1)–(3) for each $u_o$, with the same obstruction class. The exact sequence of the proof of Theorem 2.3, which lets $u_o$ vary, comes from the derived tensor product $F_o\otimes^{\mathbf L}_BB_o$.

6. Exercises

Exercise 6.1. Let $k$ be a ring of $T$, $B=k[t]/(t^3)$, $B_o=k[t]/(t^2)$, $I=(t^2)$, and $M_o=k=B_o/tB_o$. Show directly that there is no $B$-module extension $M$ of $k$ by $I\otimes k$ with $u(M)=\mathrm{id}$, so that $\omega(B,k)\neq0$ when $k\neq0$. Compute $t_{[-1}(k\otimes^{\mathbf L}_BB_o)$ as a two-term complex of $B_o$-modules.

Exercise 6.2. In Theorem 2.3, let $M_o=B_o^n$ be free and $J=I\otimes M_o=I^n$. Show that $\omega(B,M_o)=0$, and that the extensions with $u=\mathrm{id}$ are all isomorphic to $B^n$ when $H^1(T,I)=0$.

Exercise 6.3. Let $Y\to S$ be a smooth morphism of schemes, $Y\subseteq Y'$ a first-order thickening over $S$ with ideal $\mathcal I$, and $L$ an invertible $\mathcal O_Y$-module. Show that the local connections on $L$ relative to $S$ form a torsor under $\Omega_{Y/S}$, obtained from the $\mathcal O_Y^\times$-torsor of local bases of $L$ by $\mathrm d\log\colon f\mapsto f^{-1}\mathrm df$. Deduce that if the class of $L$ in $H^1(Y,\mathcal O_Y^\times)$ maps to zero in $H^1(Y,\Omega_{Y/S})$, then $L$ extends to an invertible $\mathcal O_{Y'}$-module.

Exercise 6.4. In the situation of Proposition 5.5, show that there is a distinguished triangle of $D(B_o)$ $\begin{aligned}&L_{B_o/A}\otimes_{B_o}F_o\to k^1(L^{\mathrm{gr}}_{R/A}\otimes^{\mathbf L}_RP)\\&\quad\to E_o\xrightarrow{\ w\ }(L_{B_o/A}\otimes_{B_o}F_o)[1],\\&w=(1\otimes f_o)[1]\circ\operatorname{at}_{B_o/A}(E_o),\end{aligned}$ whose first map is $k^1$ of the functoriality map tensored with $P$, and whose second map is $k^1$ of the functoriality map $L^{\mathrm{gr}}_{R/A}\otimes_RP\to L^{\mathrm{gr}}_{R/B_o}\otimes_RP$ followed by the quasi-isomorphism $\alpha$ of Lemma 5.2 (3). (Apply $-\otimes^{\mathbf L}_RP$ and $k^1$ to the transitivity triangle of $A\to B_o\to R$, and use Theorem 2.4 of the lesson on the Atiyah class.)

Exercise 6.5. In the situation of Theorem 5.3, let $p\colon C(f_o)\to E_o[1]$ be the projection, so that $(E_o,F_o,C(f_o),f_o,j_o,-p)$ is a distinguished triangle [Tags 014E, 014L], and write $\omega=\omega(B,E,f_o,f_1,u_o)$. Show that $\omega\circ j_o\in\operatorname{Ext}^2_{B_o}(F_o,K)$ vanishes if and only if there is an extension $F'$ of $F_o$ by $K$ over $B$ with $u(F')=u_o$. Deduce that if there is such an $F'$, then $\omega$ lies in the image of the injective map from the cokernel of $f_o^*\colon\operatorname{Ext}^1_{B_o}(F_o,K)\to\operatorname{Ext}^1_{B_o}(E_o,K)$ to $\operatorname{Ext}^2_{B_o}(C(f_o),K)$ induced by composition with $-p$.

7. Complete solutions

Solution 6.1. $I=t^2B\cong k$ and $I\otimes_{B_o}k\cong k$. Suppose $M$ were such an extension, $m$ a local lift of $1\in k$. Then $tm\in J$ because $t\cdot1=0$ in $k$, so $t^2m=t(tm)=0$, $t$ acting by zero on $J=I\otimes k$. But $u(M)(t^2\otimes1)=t^2m$ must be the generator $t^2\otimes1$, a contradiction unless $k=0$. By Theorem 2.3 (1), $\omega(B,k)\neq0$. For the truncation: $k$ has the periodic flat resolution $\cdots\to B\xrightarrow{t^2}B\xrightarrow{t}B\to k$ over $B$; tensoring with $B_o$ gives $\cdots\to B_o\xrightarrow{0}B_o\xrightarrow{t}B_o$, as $t^2=0$ in $B_o$. So $t_{[-1}(k\otimes^{\mathbf L}_BB_o)=[B_o\xrightarrow{t}B_o]$ in degrees $-1,0$, with $H^0=k$ and $H^{-1}=tB_o\cong k$. By the first part and Theorem 2.3 (1), its truncation class is nonzero, so this complex is not quasi-isomorphic to $k\oplus k[1]$.

Solution 6.2. $B^n$ is an extension of $B_o^n$ by $I^n=I\otimes B_o^n$ over $B$ with $u=\mathrm{id}$, so $\omega(B,B_o^n)=0$ by Theorem 2.3 (1). By Theorem 2.3 (2) the extensions with $u=\mathrm{id}$ form a torsor under $\operatorname{Ext}^1_{B_o}(B_o^n,I^n)=$ $\operatorname{Ext}^1_{B_o}(B_o,I)^{n^2}=$ $H^1(T,I)^{n^2}$, containing $B^n$; if $H^1(T,I)=0$, every such extension is isomorphic to $B^n$.

Solution 6.3. On an open where $L=\mathcal O_Yv$, a connection is $\nabla(fv)=\mathrm df\otimes v+f\eta\otimes v$ with a $1$-form $\eta$ (Exercise 4.2 of the lesson on the Atiyah class, $n=1$). In the basis $gv$, $g$ a unit, $\nabla(gv)=$ $(\mathrm dg+g\eta)\otimes v=$ $(g^{-1}\mathrm dg+\eta)\otimes gv$, so the form becomes $\eta+g^{-1}\mathrm dg$. Hence the torsor of local connections is the pushout of the torsor of local bases (under $\mathcal O_Y^\times$) along $\mathrm d\log$, and its class in $H^1(Y,\Omega_{Y/S})$ is the image of the class of $L$ [Tag 03AJ]. If this class vanishes, $L$ has a connection. As $Y\to S$ is smooth, $L_{Y/S}\simeq\Omega_{Y/S}$ is locally free, and $L$ is flat; by Corollary 3.3 of the lesson on the Atiyah class, $\operatorname{at}_{Y/S}(L)=0$. By Theorem 3.2, $\omega(Y',L)=0$, and by Theorem 2.3 (4) there is a flat $\mathcal O_{Y'}$-module $L'$ with $L'\otimes\mathcal O_Y\cong L$; it is invertible. Locally choose $e\in L'$ lifting a basis of $L$, and let $Q$ and $N$ be the cokernel and the kernel of $\mathcal O_{Y'}\to L'$, $1\mapsto e$. Then $Q\otimes\mathcal O_Y=0$, so $Q=\mathcal IQ=\mathcal I^2Q=0$; and since $L'$ is flat, $0\to N\otimes\mathcal O_Y\to\mathcal O_Y\to L\to0$ is exact with $\mathcal O_Y\to L$ an isomorphism, so $N=\mathcal IN=\mathcal I^2N=0$.

Solution 6.4. By Theorem 3.4 of the lesson on graded modules, the transitivity triangle of $A\to B_o\to R$ is $\begin{aligned}&L_{B_o/A}\otimes_{B_o}R\to L^{\mathrm{gr}}_{R/A}\\&\quad\to L^{\mathrm{gr}}_{R/B_o}\xrightarrow{\ \delta'\ }(L_{B_o/A}\otimes_{B_o}R)[1],\end{aligned}$ the first term computing the derived tensor product because $L_{B_o/A}$ has free terms. Apply $-\otimes^{\mathbf L}_RP$, computed termwise on bounded above complexes of flat graded modules (Proposition 1.6 (3) there), and the exact functor $k^1$; both carry distinguished triangles to distinguished triangles. The first term becomes $k^1(L_{B_o/A}\otimes_{B_o}P)=L_{B_o/A}\otimes_{B_o}F_o$, and the third becomes $k^1(L^{\mathrm{gr}}_{R/B_o}\otimes_RP)$, which $\alpha$ identifies with $E_o$. The maps $x\mapsto x\otimes1$ give a natural transformation $\nu$ from the identity to $-\otimes^{\mathbf L}_RP$, compatible with shifts, so $(\delta'\otimes^{\mathbf L}_RP)\circ\nu=\nu[1]\circ\delta'$. On $k^1$ of the first term, $\nu$ is $1\otimes f_o\colon L_{B_o/A}\otimes_{B_o}E_o\to L_{B_o/A}\otimes_{B_o}F_o$. On the third term, $\alpha\circ k^1(\nu)$ is the isomorphism $\beta_R\colon k^1(L^{\mathrm{gr}}_{R/B_o})\to E_o$ of Theorem 3.3 (5) there, by the naturality of the augmentation, $x\mapsto x\otimes1$ being the identity of $k^1(\Omega_{R/B_o})=k^1(\Omega_{R/B_o}\otimes_RP)$. Hence the third map of the new triangle, read through $\alpha$, is $\begin{aligned}&k^1(\delta'\otimes^{\mathbf L}_RP)\circ\alpha^{-1}\\&\quad=k^1(\delta'\otimes^{\mathbf L}_RP)\circ k^1(\nu)\circ\beta_R^{-1}\\&\quad=(1\otimes f_o)[1]\circ k^1(\delta')\circ\beta_R^{-1}.\end{aligned}$ By Theorem 2.4 of the lesson on the Atiyah class, applied to the dual numbers $R=B_o\oplus E_o$, $k^1(\delta')\circ\beta_R^{-1}=\operatorname{at}_{B_o/A}(E_o)$. So the third map is $w$.

Solution 6.5. Keep the notation of the proof of Theorem 5.3, and let $\psi\colon L^{\mathrm{gr}}_{P/B_o}\to L^{\mathrm{gr}}_{P/R}$ be the functoriality map. By Lemma 5.2 (2), $\gamma^{-1}\circ j_o=k^1(\psi)\circ\beta^{-1}$, so $\omega\circ j_o=k^1(\operatorname{ob}(c)\circ\psi)\circ\beta^{-1}.$ The argument of the proof of Theorem 5.3 (1), with Lemma 5.2 (1) for $L^{\mathrm{gr}}_{P/B_o}$, shows that $h\mapsto k^1(h)$ is injective on $\operatorname{Ext}^2_P(L^{\mathrm{gr}}_{P/B_o},M)_{\mathrm{gr}}$. So $\omega\circ j_o=0$ if and only if $\operatorname{ob}(c)\circ\psi=0$.

The inclusions $B\to R'$ and $B_o\to R$ and the identity of $P$ form a map from the diagram $B\to B_o\to P$ to the diagram $R'\to R\to P$, which induces $\psi$ on the third terms. By the naturality of transitivity triangles (Theorem 3.4 of the lesson on graded modules), $\delta\circ\psi=\lambda'[1]\circ\delta_1$, where $\delta_1$ is the third map of the transitivity triangle of $B\to B_o\to P$ and $\lambda'\colon L_{B_o/B}\otimes^{\mathbf L}_{B_o}P\to L^{\mathrm{gr}}_{R/R'}\otimes^{\mathbf L}_RP$ is the $P$-linear extension of the functoriality map $L_{B_o/B}\to L^{\mathrm{gr}}_{R/R'}$. Hence $\operatorname{ob}(c)\circ\psi=(\tilde c\circ\lambda')[1]\circ\delta_1$. By the naturality in Proposition 4.1 there, applied to the map from the surjection $B\to B_o$ to the surjection $R'\to R$, the map $\tilde c\circ\lambda'$ induces on $H^{-1}(L_{B_o/B}\otimes^{\mathbf L}_{B_o}P)=I\otimes_{B_o}P$ (Lemma 5.1 for $B\to B_o$) the $P$-linear extension of the inclusion $c_1\colon I\to M$. So $\operatorname{ob}(c)\circ\psi=\operatorname{ob}(c_1)$, the class of Lemma 5.1 for $(B,B_o,P,M,c_1)$. By Lemma 5.1 (1), it vanishes if and only if there is a graded $B$-algebra $X$ with a surjection $X\to P$ whose kernel is $M$ and with $B\to X$ the identity on $I$. As in the proof of Theorem 5.3, with $E=0$, these $X$ are the $B\oplus F'$ for the extensions $F'$ of $F_o$ by $K$ over $B$ with $u(F')=u_o$. By Theorem 2.3 (1), they exist exactly when $\omega(B,u_o)=0$.

For the second part, apply $\operatorname{Hom}(-,K[2])$ to the distinguished triangle $(F_o,C(f_o),E_o[1],j_o,-p,-f_o[1])$, a rotation of the given one [Tags 0145, 0149]. This gives the exact sequence $\begin{aligned}&\operatorname{Ext}^1(F_o,K)\xrightarrow{\ -f_o^*\ }\operatorname{Ext}^1(E_o,K)\\&\quad\xrightarrow{\ (-p)^*\ }\operatorname{Ext}^2(C(f_o),K)\\&\quad\xrightarrow{\ j_o^*\ }\operatorname{Ext}^2(F_o,K).\end{aligned}$ The image of $-f_o^*$ is that of $f_o^*$, so composition with $-p$ induces an injective map from the cokernel of $f_o^*$ onto the kernel of $j_o^*$. If $F'$ exists, $\omega\circ j_o=0$ by the first part, so $\omega$ lies in this image.

Sources

The deformation theory of modules over a square-zero thickening, with the obstruction in $\operatorname{Ext}^2$, the classification by $\operatorname{Ext}^1$ and the automorphisms $\operatorname{Hom}$, and its expression through the derived tensor product $M_o\otimes^{\mathbf L}_BB_o$ (an argument attributed to Grothendieck), the theorem that the obstruction is the product of the Atiyah class and the class of the thickening, and the treatment of morphisms through graded rings of dual numbers, with the obstruction in $\operatorname{Ext}^2_{B_o}(C(f_o),K)$ and its universal form through an Atiyah class of $f_o$, are due to Illusie, in his work on the cotangent complex, where Theorem 3.10 and the example of the quadric surface are proposed as an exercise. The first Chern class in Hodge cohomology and the cohomology of projective space used in Example 3.11 follow the AI-integrated Stacks project, Chapters De Rham Cohomology and Cohomology of Schemes. The AI-integrated Stacks project, Chapter Deformation Theory, treats the obstruction and classification for modules on ringed topoi with injective coefficients. Huybrechts and Thomas extend the product formula to perfect complexes on schemes with truncated Atiyah and Kodaira–Spencer classes: D. Huybrechts and R. P. Thomas, Deformation-obstruction theory for complexes via Atiyah and Kodaira–Spencer classes, arXiv:0805.3527; see also M. Lieblich, Moduli of complexes on a proper morphism, arXiv:math/0502198, and W. Lowen, Obstruction theory for objects in abelian and derived categories, arXiv:math/0407019.