# Inverse Function Theorem in Algebraic Geometry

Suppose that $X$ and $Y$ are smooth complex algebraic varieties, and that $f:X\rightarrow Y$ is an etale morphism in the sense that $d_xf:T_xX\rightarrow T_{f(x)}Y$ is an isomorphism for all $x\in X$. Must $f$ be (at least locally) an open immersion? Certainly, difficulties can arise if the differential fails to be an isomorphism at a single point. Consider the map $\mathbb{A}^1\rightarrow\mathbb{A}^1$, $z\mapsto z^2$.

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You can only say $f$ is a local biholomorphism around $x$. It is not, in general, a local isomorphism of algebraic varieties around $x$: just take a covering of curves of different genus minus the ramification points. Or, even simplier, take your example $\mathbb{A}^1\setminus \{ 0 \} \to \mathbb{A}^1\setminus \{ 0 \}$ $z\mapsto z^2$. The inverse function theorem however holds with respect to the étale topology for étale morphisms. – Qfwfq May 25 '13 at 23:17
What about something of the form: "every smooth map is locally split over an étale cover"? I vaguely remember reading this somewhere, can't be sure it's true. – David Roberts May 25 '13 at 23:43
If $X=Y=\mathbb{A}^n$, your question is the famous Jacobian conjecture. – Jérémy Blanc May 26 '13 at 6:52
In some trivial but profound sense every etale map is locally an isomorphism. Let "locally" mean "locally in the etale topology". – Tom Goodwillie May 26 '13 at 20:55

## 1 Answer

Maybe this "answer" is too far away from schemes to be what is desired. But here it goes anyway.

I believe in general that for complex affine varieties $X,Y$, that a morphism $f:X\to Y$ is étale iff it is a local analytic isomorphism in the analytic topology. When $X,Y$ are smooth, it is enough to just check that the tangent spaces are isomorphic at every point.

For example, if a finite group $\Gamma$ acts on a connected and normal variety $X$ algebraically, then the mapping $X\to X//\Gamma$ is étale if and only if $\Gamma$ acts freely (i.e. stabilizers are trivial).

One can find projections like that who are not homeomorphic to their image; a necessary requirement for open immersions. For example, consider the étale map $\mathrm{SL}_2(\mathbb{C})\to \mathrm{SL}_2(\mathbb{C})/\mathbb{Z}_2\cong \mathrm{SO}(3,\mathbb{C})$.

Conversely however, open immersions are always étale.

EDIT: I added some detail to the general statement about finite quotients, and replaced my original example since it was not correct. In particular, the tangent map to $\mathbb{C}^* \to \mathbb{C}^*//\mathbb{Z}_2 \cong \mathbb{C}$ (where $\mathbb{Z}_2$ acts by $z\mapsto 1/z$) is not an isomorphism at $\pm 1$, and so it is not étale at those points (it is at all other points though).

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