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Nov 14, 2017 at 8:26 comment added Laurent Moret-Bailly @Sasha: the result is false for general $Y$ (I'll edit my answer later); it might be true if $Y$ is normal.
Nov 13, 2017 at 21:01 comment added Sasha @Friedrich Knop: Thank you! Just to clarify for myself, the proof below by Moret-Bailly will work too for establishing the fact I wanted, right?
Nov 13, 2017 at 20:59 comment added Sasha @Laurent Moret-Bailly: It is a well-known theorem that this quotient is isomorphic to an affine space. But I guess that this doesn't really matter if one knows how to work with analytic spaces well enough, your proof below will carry?
Nov 13, 2017 at 14:24 comment added Laurent Moret-Bailly Concerning the group-theoretic motivation, why is $W\backslash\mathfrak{h}$ smooth?
Nov 13, 2017 at 8:42 comment added Friedrich Knop For a reference to your invariant theoretic application see my answer to mathoverflow.net/questions/237666/….
Nov 12, 2017 at 18:57 vote accept Sasha
Nov 12, 2017 at 18:42 comment added Sándor Kovács @Sasha: Right, it would just mash everything into a wedge. I knew it was too simple...
Nov 12, 2017 at 17:43 answer added Laurent Moret-Bailly timeline score: 9
Nov 12, 2017 at 2:49 comment added Sasha @Sándor Kovács: root $\circ$ power is not identity in general, that's your mistake if I am not mistaken?
Nov 11, 2017 at 17:19 comment added nfdc23 @abx: I had the same initial gut reaction, but it is unnecessary: since $f$ is a finite surjection, hence topologically a quotient map, a function on $Y$ whose composition with $f$ is continuous must itself be continuous.
Nov 11, 2017 at 17:16 comment added abx You should at least assume that your function is continuous, otherwise there are obvious counter-examples.
Nov 11, 2017 at 16:48 history asked Sasha CC BY-SA 3.0