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May 21, 2010 at 13:31 comment added Tyler Lawson @Noah: Corrected. The set notation was wrong and I had a sign error. My apologies.
May 21, 2010 at 13:30 history edited Tyler Lawson CC BY-SA 2.5
more corrections because I am an idiot
May 21, 2010 at 13:15 comment added Noah Stein Apparently I can't edit my own comments. By "instead" I meant "instead of $\{y<0\}$".
May 21, 2010 at 13:14 comment added S. Carnahan Yes, it is a bit like real points of $y^2 = x^2(x+a)$ for $a$ very very small.
May 21, 2010 at 13:13 comment added Noah Stein I'm confused about this set $R^2\setminus\{y<0\}$. If it is to be homeomorphic to $R\times (0,\infty)$ it should be open. Do you mean $\{y\leq 0,x=0\}$ instead? Also, I'm not sure about that $g$ being an inverse, because as written $g(0,u)=(0,0)$ for all $u$.
May 21, 2010 at 13:07 comment added Pete L. Clark In fact, this is is somewhat reminiscent of the universal cover of the nodal cubic that we were talking about some months ago, no?
May 21, 2010 at 13:05 comment added Pete L. Clark Never mind -- I see it now. Thanks.
May 21, 2010 at 12:58 comment added Pete L. Clark @Scott: I can't quite see it yet. Could you be a little more more explicit?
May 21, 2010 at 12:45 comment added S. Carnahan @Pete: No, it is not Hausdorff. You glue a bunch of real lines along open rays.
May 21, 2010 at 12:43 comment added Tyler Lawson @Chris: The "usual" coordinate charts make it look like a quasifibration. I've added some details about a change of coordinates that makes it look more like an actual fibration.
May 21, 2010 at 12:42 history edited Tyler Lawson CC BY-SA 2.5
added details
May 21, 2010 at 12:39 comment added Pete L. Clark Weird. Is the universal cover just the usual real line?
May 21, 2010 at 12:32 comment added Chris Schommer-Pries Is this a fibration or only a quasi-fibtration? (a quasi-fibration is still good enough for your argument).
May 21, 2010 at 12:20 history answered Tyler Lawson CC BY-SA 2.5