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Let $X$ and $Y$ be two complex reduced affine algebraic or analytic varieties, possibly singular. Take a regular proper function $$f\colon X \to Y $$ and assume that it is bijective at the level of $\mathbb{C}$-points. Moreover, assume that for every point $x$ of $X$ the differential $$df(x)\colon T_xX \to T_{f(y)}Y $$ is an isomorphism, where the tangent spaces are Zariski tangent spaces.

Can we conclude that $X$ and Y are isomorphic? Do we need any assumption on the singularities? In particular, should $Y$ be normal?

In other words, I am asking under which conditions on the singularities the implicit function theorem holds. I remember some notes by Kollar, where this issue was related with Canonical singularities, but I could not find them anymore.

Any reference or example is welcome.

thanks

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    $\begingroup$ Simple counterexample --- $X = Spec (k[t]/t^2)$, $Y = Spec (k[t]/t^3)$, the map is natural. $\endgroup$
    – Sasha
    Commented Feb 25, 2015 at 20:07
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    $\begingroup$ So I need at least $X$ and $Y$ reduced $\endgroup$
    – Giulio
    Commented Feb 25, 2015 at 20:11
  • $\begingroup$ thanks Sasha, I edited my question and added reduced in the hypothesis. $\endgroup$
    – Giulio
    Commented Feb 25, 2015 at 21:03
  • $\begingroup$ W. S. Loud, Some Singular Cases of the Implicit Function Theorem The American Mathematical Monthly Vol. 68, No. 10 (Dec., 1961), pp. 965-977 $\endgroup$
    – user21574
    Commented Jul 24, 2017 at 22:39
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    $\begingroup$ There is a chaper about singular version of implicit function theorem link.springer.com/book/10.1007%2F978-1-4614-5981-1 $\endgroup$
    – user21574
    Commented Jul 24, 2017 at 22:43

2 Answers 2

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In Joe Harris' book, Algebraic Geometry, this is theorem 14.9.

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At least you should assume that $f$ is a homeomorphism (equivalently: proper, or finite).

Otherwise here is a counterexample: let $X=X_1\coprod X_2$ where $X_1$ is the union of the coordinate axes in $\mathbb{C}^2$, and $X_2=\mathbb{C}^*$. Let $Y\subset \mathbb{C}^2$ be the union of $X_1$ and the diagonal, and $f$ be the obvious map (identity on $X_1$, ant $t\mapsto (t,t)$ on $X_2$). It is clearly bijective, and an isomorphism if you remove the origin from $X_1$ and $Y$. At the origin, the Zariski tangent spaces of $X_1$ and $Y$ are the same, namely $\mathbb{C}^2$.

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  • $\begingroup$ $f$ is bijective, so finite $\endgroup$
    – Giulio
    Commented Feb 25, 2015 at 21:03
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    $\begingroup$ Dear Giulio, in Algebraic Geometry the property of finiteness has a specific meaning which implies properness. The example of Laurent is bijective but not proper since the image of the closed subset $X_2$ is not closed in $Y$. $\endgroup$
    – Matthieu Romagny
    Commented Feb 25, 2015 at 21:20
  • $\begingroup$ Ok, I will edit the answer, and assume that $f$ is proper. Thanks $\endgroup$
    – Giulio
    Commented Feb 26, 2015 at 22:01

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