Lemma 89.27.3. Let $U: \mathcal{C}_\Lambda \to \textit{Sets}$ be a prorepresentable functor. Let $\varphi : U \to U$ be a morphism such that $d\varphi : TU \to TU$ is an isomorphism. Then $\varphi$ is an isomorphism.

Proof. If $U \cong \underline{R}|_{\mathcal{C}_\Lambda }$ for some $R \in \mathop{\mathrm{Ob}}\nolimits (\widehat{\mathcal{C}}_\Lambda )$, then completing $\varphi$ gives a morphism $\underline{R} \to \underline{R}$. If $f: R \to R$ is the corresponding morphism in $\widehat{\mathcal{C}}_\Lambda$, then $f$ induces an isomorphism $\text{Der}_\Lambda (R, k) \to \text{Der}_\Lambda (R, k)$, see Example 89.11.14. In particular $f$ is a surjection by Lemma 89.4.6. As a surjective endomorphism of a Noetherian ring is an isomorphism (see Algebra, Lemma 10.31.10) we conclude $f$, hence $\underline{R} \to \underline{R}$, hence $\varphi : U \to U$ is an isomorphism. $\square$

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