The Stacks project

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

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. $\square$