The Stacks project

Remark 90.19.4. Assume $\mathcal{F}$ is a predeformation category. Then

1. for $x_0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(k))$ the automorphism group $\text{Aut}_ k(x_0)$ is trivial and hence $\text{Inf}_{x_0}(\mathcal{F}) = \text{Aut}_{k[\epsilon ]}(x'_0)$, and

2. for $x_0, y_0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(k))$ there is a unique isomorphism $x_0 \to y_0$ and hence a canonical identification $\text{Inf}_{x_0}(\mathcal{F}) = \text{Inf}_{y_0}(\mathcal{F})$.

Since $\mathcal{F}(k)$ is nonempty, choosing $x_0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(k))$ and setting

we get a well defined group of infinitesimal automorphisms of $\mathcal{F}$. With this notation we have $\text{Inf}(\mathcal{F}_{x_0}) = \text{Inf}_{x_0}(\mathcal{F})$. Please compare with the equality $T\mathcal{F}_{x_0} = T_{x_0}\mathcal{F}$ in Remark 90.12.5.