Lemma 89.19.11. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal{C}_\Lambda $ satisfying (RS). Let $x' \to x$ be a morphism lying over a surjective ring map $A' \to A$ with kernel $I$ annihilated by $\mathfrak m_{A'}$. Let $x_0$ be a pushforward of $x$ to $\mathcal{F}(k)$. Then $\text{Inf}(x'/x)$ has a free and transitive action by $T_{\text{id}_{x_0}} \mathit{Aut}(x') \otimes _ k I = \text{Inf}_{x_0}(\mathcal{F}) \otimes _ k I$.
Proof. This is just the analogue of Lemma 89.17.5 in the setting of automorphism sheaves. To be precise, we apply Remark 89.6.4 to the functor $\mathit{Aut}(x') : \mathcal{C}_{A'} \to \textit{Sets}$ and the element $\text{id}_{x_0} \in \mathit{Aut}(x)(k)$ to get a predeformation functor $F = \mathit{Aut}(x')_{\text{id}_{x_0}}$. By Lemmas 89.19.6 and 89.16.11 $F$ is a deformation functor. Hence Lemma 89.17.5 gives a free and transitive action of $TF \otimes _ k I$ on $\text{Lift}(\text{id}_ x, A')$, because as $\text{Lift}(\text{id}_ x, A')$ is a group it is always nonempty. Note that we have equalities of vector spaces
\[ TF = T_{\text{id}_{x_0}} \mathit{Aut}(x') \otimes _ k I = \text{Inf}_{x_0}(\mathcal{F}) \otimes _ k I \]
by Lemma 89.19.7. The equality $\text{Inf}(x'/x) = \text{Lift}(\text{id}_ x, A')$ of Remark 89.19.8 finishes the proof. $\square$
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