Bernard

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	228 reputation	bio	website		visits	member for	8 months
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May 1	comment	Function that cannot be extended continuously out of sphere Thanks. I wasn't actually interested in S^n itself but had in my mind a general compact subset hence I did not give much thought to the (obvious) solution above.
May 1	accepted	Function that cannot be extended continuously out of sphere
May 1	asked	Function that cannot be extended continuously out of sphere
Mar 18	accepted	Showing an analytic map has closed irreducible image
Mar 17	comment	Showing an analytic map has closed irreducible image If we define variety as integral separated scheme of finite type (here over $\mathbb C$) isn't it Hausdorff then?
Mar 17	comment	Showing an analytic map has closed irreducible image Isn't it true that $Y$ is necessarily Hausdorff since it is a variety? I think this follows from the requirement that it should be separated, but am not 100% sure. Do you have a reference for why this is true if $Y$ is Hausdorff?
Mar 17	revised	Showing an analytic map has closed irreducible image added 136 characters in body
Mar 16	revised	Showing an analytic map has closed irreducible image added 236 characters in body
Mar 16	comment	Showing an analytic map has closed irreducible image Wait a minute. Sorry to "unaccept" the answer, but your map $g$ hence $\tilde g$ is $analytic$, and your argument seems to treat it like an algebraic map, unless I am hugely missing something here...
Mar 16	comment	Showing an analytic map has closed irreducible image Ok, that makes sense. Thanks a lot!
Mar 15	comment	Showing an analytic map has closed irreducible image That $X$ may be covered by affine open sets $Spec(B)$ such that each $\pi^{-1}(SpecB)$ is covered by finitely many affine open sets $Spec(A)$ where $A$ is finitely generated $B$-algebra.
Mar 15	comment	Showing an analytic map has closed irreducible image You claim $\pi$ is quasi-finite; ie it has finite fibers (true by hypothesis) and is of finite type. Why is the latter true?
Mar 15	comment	Showing an analytic map has closed irreducible image Don't you need $\pi$ to be finite for it to be quasi-finite? And also, why is the dimension the same?
Mar 15	revised	Showing an analytic map has closed irreducible image added 289 characters in body
Mar 15	revised	Showing an analytic map has closed irreducible image added 106 characters in body
Mar 15	comment	Showing an analytic map has closed irreducible image Thanks for fixing it.
Mar 14	asked	Showing an analytic map has closed irreducible image
Mar 10	comment	Application of Hirzebruch Riemann Roch Ok, sorry I'll keep that in mind.
Mar 10	asked	Application of Hirzebruch Riemann Roch
Feb 15	comment	Is there some geometrical intuition behind separable and or purely inseparable extensions? Ok, thanks, I'll look these up!