Lemma 90.27.3. 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 90.11.14. In particular f is a surjection by Lemma 90.4.6. 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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