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Here"analytically isomorphic"means that the completion of the local rings of two points in some complex spaces are isomorphic. The smooth case is trivial so the only interesting case is that these two points are singular points. I think analytially isomorphism of points does not imply local isomorphism for complex spaces,but I can’t find a counterexample. Or maybe complex spaces associated to algebraic varieties over C may behave better,I don't have any intuition for this problem.

asked Oct 14 at 8:10
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    $\begingroup$ Your definition does not even make sense. You can say that there is a bijection $f: X\to Y$ of two analytical spaces such that for every $x\in X,ドル there exists an isomorphism of completed local rings at $x$ and $f(x)$. But with this definition, for instance, any two smooth complex manifolds of the same (positive) dimension will be isomorphic. Such a definition would be quite useless. $\endgroup$ Commented Oct 14 at 8:19
  • $\begingroup$ The definition has nothing to do with morphism,it does not even say the isomorphism is induced from a local homomorphism,and it's quoted from hartshorne chapter1.5,below thm 5.5A@Moishe Kohan $\endgroup$ Commented Oct 14 at 9:08

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