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May 26, 2013 at 20:55 comment added Tom Goodwillie In some trivial but profound sense every etale map is locally an isomorphism. Let "locally" mean "locally in the etale topology".
May 26, 2013 at 18:46 vote accept Peter Crooks
May 26, 2013 at 18:10 answer added Sean Lawton timeline score: 2
May 26, 2013 at 6:52 comment added Jérémy Blanc If $X=Y=\mathbb{A}^n$, your question is the famous Jacobian conjecture.
May 25, 2013 at 23:43 comment added David Roberts 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.
May 25, 2013 at 23:17 comment added Qfwfq 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.
May 25, 2013 at 22:41 history asked Peter Crooks CC BY-SA 3.0