Proposition 9.10.41A7FEC3124D0

Proof. Let $x\in \mathcal{X}(\mathbb {C})$. Then we claim that $\mathcal{X}$ is unibranch at $x$ if and only if $\mathcal{X}^{\mathrm{An}}$ is unibranch at $x$. By GAGA, there is a natural morphism of ringed spaces $(\mathcal{X}^{\mathrm{An}},\mathcal{O}_{\mathcal{X}^{\mathrm{An}}})\rightarrow (\mathcal{X},\mathcal{O}_{\mathcal{X}})$. Then since $\mathcal{O}_{\mathcal{X},x}$ is excellent, $\mathcal{X}$ is unibranch at $x$ if and only if $\widehat{\mathcal{O}_{\mathcal{X},x}}=\widehat{\mathcal{O}_{\mathcal{X}^{\mathrm{An}},x}}$ is integral ([Théorème 18.9.1, EGAIV4]). The latter implies that $\mathcal{O}_{\mathcal{X}^{\mathrm{An}},x}$ is unibranch by [Chapitre 5, Exercise 2.8, AC]. Running the same argument the other way round, we conclude that $\mathcal{O}_{\mathcal{X},x}$ is unibranch if and only if $\mathcal{O}_{\mathcal{X}^{\mathrm{An}},x}$ is. $\square$