Lemma 89.17.5. Let $\mathcal{F}$ be a deformation category. Let $A' \to A$ be a surjective ring map in $\mathcal{C}_\Lambda$ whose kernel $I$ is annihilated by $\mathfrak m_{A'}$. Let $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(A))$. If $\text{Lift}(x, A')$ is nonempty, then there is a free and transitive action of $T\mathcal{F} \otimes _ k I$ on $\text{Lift}(x, A')$.

Proof. Consider the ring map $g : A' \times _ A A' \to k[I]$ defined by the rule $g(a_1, a_2) = \overline{a_1} \oplus a_2 - a_1$ (compare with Lemma 89.10.8). There is an isomorphism

$A' \times _ A A' \xrightarrow {\sim } A' \times _ k k[I]$

given by $(a_1, a_2) \mapsto (a_1, g(a_1, a_2))$. This isomorphism commutes with the projections to $A'$ on the first factor, and hence with the projections of $A' \times _ A A'$ and $A' \times _ k k[I]$ to $A$. Thus there is a bijection

89.17.5.1
\begin{equation} \label{formal-defos-equation-one} \text{Lift}(x, A' \times _ A A') \longrightarrow \text{Lift}(x, A' \times _ k k[I]) \end{equation}

By Remark 89.17.4 there is a bijection

89.17.5.2
\begin{equation} \label{formal-defos-equation-two} \text{Lift}(x, A') \times \text{Lift}(x, A') \longrightarrow \text{Lift}(x, A' \times _ A A') \end{equation}

There is a commutative diagram

$\xymatrix{ A' \times _ k k[I] \ar[r] \ar[d] & A \times _ k k[I] \ar[r] \ar[d] & k[I] \ar[d] \\ A' \ar[r] & A \ar[r] & k. }$

Thus if we choose a pushforward $x \to x_0$ of $x$ along $A \to k$, we obtain by the end of Remark 89.17.4 a bijection

89.17.5.3
\begin{equation} \label{formal-defos-equation-three} \text{Lift}(x, A' \times _ k k[I]) \longrightarrow \text{Lift}(x, A') \times \text{Lift}(x_0, k[I]) \end{equation}

Composing (89.17.5.2), (89.17.5.1), and (89.17.5.3) we get a bijection

$\Phi : \text{Lift}(x, A') \times \text{Lift}(x, A') \longrightarrow \text{Lift}(x, A') \times \text{Lift}(x_0, k[I]).$

This bijection commutes with the projections on the first factors. By Remark 89.17.3 we see that $\text{Lift}(x_0, k[I]) = T\mathcal{F} \otimes _ k I$. If $\text{pr}_2$ is the second projection of $\text{Lift}(x, A') \times \text{Lift}(x, A')$, then we get a map

$a = \text{pr}_2 \circ \Phi ^{-1} : \text{Lift}(x, A') \times (T\mathcal{F} \otimes _ k I) \longrightarrow \text{Lift}(x, A').$

Unwinding all the above we see that $a(x' \to x, \theta )$ is the unique lift $x'' \to x$ such that $g_*(x', x'') = \theta$ in $\text{Lift}(x_0, k[I]) = T\mathcal{F} \otimes _ k I$. To see this is an action of $T\mathcal{F} \otimes _ k I$ on $\text{Lift}(x, A')$ we have to show the following: if $x', x'', x'''$ are lifts of $x$ and $g_*(x', x'') = \theta$, $g_*(x'', x''') = \theta '$, then $g_*(x', x''') = \theta + \theta '$. This follows from the commutative diagram

$\xymatrix{ A' \times _ A A' \times _ A A' \ar[rrrrr]_-{(a_1, a_2, a_3) \mapsto (g(a_1, a_2), g(a_2, a_3))} \ar[rrrrrd]_{(a_1, a_2, a_3) \mapsto g(a_1, a_3)} & & & & & k[I] \times _ k k[I] = k[I \times I] \ar[d]^{+} \\ & & & & & k[I] }$

The action is free and transitive because $\Phi$ is bijective. $\square$

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