Lemma 89.18.1. Let $F, G: \mathcal{C}_\Lambda \to \textit{Sets}$ be deformation functors. Let $\varphi : F \to G$ be a smooth morphism which induces an isomorphism $d\varphi : TF \to TG$ of tangent spaces. Then $\varphi $ is an isomorphism.

**Proof.**
We prove $F(A) \to G(A)$ is a bijection for all $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_\Lambda )$ by induction on $\text{length}_ A(A)$. For $A = k$ the statement follows from the assumption that $F$ and $G$ are deformation functors. Suppose that the statement holds for rings of length less than $n$ and let $A'$ be a ring of length $n$. Choose a small extension $f : A' \to A$. We have a commutative diagram

where the map $F(A) \to G(A)$ is a bijection. By smoothness of $F \to G$, $F(A') \to G(A')$ is surjective (Lemma 89.8.8). Thus we can check bijectivity by checking it on fibers $F(f)^{-1}(x) \to G(f)^{-1}(\varphi (x))$ for $x \in F(A)$ such that $F(f)^{-1}(x)$ is nonempty. These fibers are precisely $\text{Lift}(x, A')$ and $\text{Lift}(\varphi (x), A')$ and by assumption we have an isomorphism $d\varphi \otimes \text{id} : TF \otimes _ k \mathop{\mathrm{Ker}}(f) \to TG \otimes _ k \mathop{\mathrm{Ker}}(f)$. Thus, by Lemma 89.17.5 and Remark 89.17.6, for $x \in F(A)$ such that $F(f)^{-1}(x)$ is nonempty the map $F(f)^{-1}(x) \to G(f)^{-1}(\varphi (x))$ is a map of sets commuting with free transitive actions by $TF \otimes _ k \mathop{\mathrm{Ker}}(f)$. Hence it is bijective. $\square$

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