Timeline for A characterization of convexity

Current License: CC BY-SA 2.5

9 events

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Aug 6, 2010 at 16:06	comment	added	Tom Goodwillie		I didn't know for a fact that this key step was legitimate: weak contractibility of the fibers of $p:Y\to X$ implies that $p$ is a weak homotopy equivalence if $Y\subset X\mathbb R^n$ is open and $p$ is the projection. But it's OK, even if $\mathbb R^n$ is replaced by any space. A student of mine rn across this lemma in another context. The reference is the appendix of Graeme Segal, Classifying Spaces Related to Foliations, Topology vol. 17 pp. 367-382.
Jun 27, 2010 at 1:13	comment	added	Tom Goodwillie		Thanks. I don't swear that it's correct, though.
Jun 24, 2010 at 18:28	comment	added	Cristos A. Ruiz		I'm chosing this one as the correct answer. There was also Matessi's correct counterexample and Pak's reference (by the way, that book looks really good). But I think there's more merit in this one, and it looks correct to me, also I don't know if the case of $X$ open was known, and the method used to solve it is very appealing to me. Thanks very much for your responses!
Jun 24, 2010 at 18:21	vote	accept	Cristos A. Ruiz
Jun 22, 2010 at 0:38	comment	added	Victor Protsak		Yes, you are right, thank you! The vertex of the parabola is below the $xy$-plane, so there are two pieces. Once I drew it, it became obvious.
Jun 21, 2010 at 23:51	comment	added	Tom Goodwillie		Your $X$ intersects the plane $x=z+1$ in a disconnected set.
Jun 21, 2010 at 22:34	comment	added	Victor Protsak		I don't think that Claim 2 is true if $X$ is the interior of my modification of Diego Matessi's example: $X=\{ (x,y,z)\in\mathbb{R}^2\times\mathbb{R}_+: x<0 \text{ or } x^2<y^2+z^2\}.$
Jun 21, 2010 at 20:40	history	edited	Tom Goodwillie	CC BY-SA 2.5	added 6 characters in body
Jun 21, 2010 at 20:12	history	answered	Tom Goodwillie	CC BY-SA 2.5