**Proof.**
We apply Remark 90.6.4 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 90.19.6 and 90.16.11 $F$ is a deformation functor. By definition $T_{\text{id}_{x_0}} \mathit{Aut}(x) = TF = F(k[\epsilon ])$ which has a natural $k$-vector space structure specified by Lemma 90.11.8.

Addition is defined as the composition

\[ F(k[\epsilon ]) \times F(k[\epsilon ]) \longrightarrow F(k[\epsilon ] \times _ k k[\epsilon ]) \longrightarrow F(k[\epsilon ]) \]

where the first map is the inverse of the bijection guaranteed by (RS) and the second is induced by the $k$-algebra map $k[\epsilon ] \times _ k k[\epsilon ] \to k[\epsilon ]$ which maps $(\epsilon , 0)$ and $(0, \epsilon )$ to $\epsilon $. If $A \to B$ is a ring map in $\mathcal{C}_\Lambda $, then $F(A) \to F(B)$ is a homomorphism where $F(A) = \mathit{Aut}(x)_{\text{id}_{x_0}}(A)$ and $F(B) = \mathit{Aut}(x)_{\text{id}_{x_0}}(B)$ are groups under composition. We conclude that $+ : F(k[\epsilon ]) \times F(k[\epsilon ])\to F(k[\epsilon ])$ is a homomorphism where $F(k[\epsilon ])$ is regarded as a group under composition. With $\text{id} \in F(k[\epsilon ])$ the unit element we see that $+(v, \text{id}) = +(\text{id}, v) = v$ for any $v \in F(k[\epsilon ])$ because $(\text{id}, v)$ is the pushforward of $v$ along the ring map $k[\epsilon ] \to k[\epsilon ] \times _ k k[\epsilon ]$ with $\epsilon \mapsto (\epsilon , 0)$. In general, given a group $G$ with multiplication $\circ $ and $+ : G \times G \to G$ is a homomorphism such that $+(g, 1) = +(1, g) = g$, where $1$ is the identity of $G$, then $+ = \circ $. This shows addition in the $k$-vector space structure on $F(k[\epsilon ])$ agrees with composition.

Finally, (2) is a matter of unwinding the definitions. Namely $T_{\text{id}_{x_0}} \mathit{Aut}(x)$ is the set of automorphisms $\alpha $ of the pushforward of $x$ along $A \to k \to k[\epsilon ]$ which are trivial modulo $\epsilon $. On the other hand $T_{\text{id}_{x_0}} \mathit{Aut}(x_0)$ is the set of automorphisms of the pushforward of $x_0$ along $k \to k[\epsilon ]$ which are trivial modulo $\epsilon $. Since $x_0$ is the pushforward of $x$ along $A \to k$ the result is clear.
$\square$

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