Lemma 97.24.3. In Situation 97.24.2. Assume furthermore that

1. given a short exact sequence of deformation situations as in Remark 97.21.5 and a lift $x'_2 \in \text{Lift}(x, A_2')$ then $o_ x(A_3') \in H^2(K^\bullet \otimes _ A^\mathbf {L} I_3)$ equals $\partial \theta$ where $\theta \in H^1(K^\bullet \otimes _ A^\mathbf {L} I_1)$ is the element corresponding to $x'_2|_{\mathop{\mathrm{Spec}}(A_1')}$ via $A_1' = A[I_1]$ and the given map $T_ x(-) \to H^1(K^\bullet \otimes _ A^\mathbf {L} -)$.

In this case there exists an element $\xi \in H^1(K^\bullet \otimes _ A^\mathbf {L} \mathop{N\! L}\nolimits _{A/\Lambda })$ such that

1. for every deformation situation $(x, A' \to A)$ we have $\xi _{A'} = o_ x(A')$, and

2. $\xi _{can}$ matches the canonical element of Remark 97.21.8 via the given transformation $T_ x(-) \to H^1(K^\bullet \otimes _ A^\mathbf {L} -)$.

Proof. Choose a $\alpha : \Lambda [x_1, \ldots , x_ n] \to A$ with kernel $J$. Write $P = \Lambda [x_1, \ldots , x_ n]$. In the rest of this proof we work with

$\mathop{N\! L}\nolimits (\alpha ) = (J/J^2 \longrightarrow \bigoplus A \text{d}x_ i)$

which is permissible by Algebra, Lemma 10.134.2 and More on Algebra, Lemma 15.58.2. Consider the element $o_ x(P/J^2) \in H^2(K^\bullet \otimes _ A^\mathbf {L} J/J^2)$ and consider the quotient

$C = (P/J^2 \times \bigoplus A \text{d}x_ i)/(J/J^2)$

where $J/J^2$ is embedded diagonally. Note that $C \to A$ is a surjection with kernel $\bigoplus A\text{d}x_ i$. Moreover there is a section $A \to C$ to $C \to A$ given by mapping the class of $f \in P$ to the class of $(f, \text{d}f)$ in the pushout. For later use, denote $x_ C$ the pullback of $x$ along the corresponding morphism $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(A)$. Thus we see that $o_ x(C) = 0$. We conclude that $o_ x(P/J^2)$ maps to zero in $H^2(K^\bullet \otimes _ A^\mathbf {L} \bigoplus A\text{d}x_ i)$. It follows that there exists some element $\xi \in H^1(K^\bullet \otimes _ A^\mathbf {L} \mathop{N\! L}\nolimits (\alpha ))$ mapping to $o_ x(P/J^2)$.

Note that for any deformation situation $(x, A' \to A)$ there exists a $\Lambda$-algebra map $P/J^2 \to A'$ compatible with the augmentations to $A$. Hence the element $\xi$ satisfies the first property of the lemma by construction and property (ii) of Situation 97.24.2.

Note that our choice of $\xi$ was well defined up to the choice of an element of $H^1(K^\bullet \otimes _ A^\mathbf {L} \bigoplus A\text{d}x_ i)$. We will show that after modifying $\xi$ by an element of the aforementioned group we can arrange it so that the second assertion of the lemma is true. Let $C' \subset C$ be the image of $P/J^2$ under the $\Lambda$-algebra map $P/J^2 \to C$ (inclusion of first factor). Observe that $\mathop{\mathrm{Ker}}(C' \to A) = \mathop{\mathrm{Im}}(J/J^2 \to \bigoplus A\text{d}x_ i)$. Set $\overline{C} = A[\Omega _{A/\Lambda }]$. The map $P/J^2 \times \bigoplus A \text{d}x_ i \to \overline{C}$, $(f, \sum f_ i \text{d}x_ i) \mapsto (f \bmod J, \sum f_ i \text{d}x_ i)$ factors through a surjective map $C \to \overline{C}$. Then

$(x, \overline{C} \to A) \to (x, C \to A) \to (x, C' \to A)$

is a short exact sequence of deformation situations. The associated splitting $\overline{C} = A[\Omega _{A/\Lambda }]$ (from Remark 97.21.5) equals the given splitting above. Moreover, the section $A \to C$ composed with the map $C \to \overline{C}$ is the map $(1, \text{d}) : A \to A[\Omega _{A/\Lambda }]$ of Remark 97.21.8. Thus $x_ C$ restricts to the canonical element $x_{can}$ of $T_ x(\Omega _{A/\Lambda }) = \text{Lift}(x, A[\Omega _{A/\Lambda }])$. By condition (iv) we conclude that $o_ x(P/J^2)$ maps to $\partial x_{can}$ in

$H^1(K^\bullet \otimes _ A^\mathbf {L} \mathop{\mathrm{Im}}(J/J^2 \to \bigoplus A\text{d}x_ i))$

By construction $\xi$ maps to $o_ x(P/J^2)$. It follows that $x_{can}$ and $\xi _{can}$ map to the same element in the displayed group which means (by the long exact cohomology sequence) that they differ by an element of $H^1(K^\bullet \otimes _ A^\mathbf {L} \bigoplus A\text{d}x_ i)$ as desired. $\square$

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