Timeline for A question about connectedness in Euclidean space

7 events

when toggle format	what		by	license	comment
Nov 8, 2011 at 5:19	vote	accept	Kwong
Nov 8, 2011 at 5:19	comment	added	Kwong		Thanks for the neat answer. So we can replace $\mathbb{R}^n$ by a connected space with vanishing $H_1$, and $K$ to be closed subset. Nice.
Nov 7, 2011 at 14:37	comment	added	Richard Rast		This answer has the advantage of answering his extension question (that is, on what manifolds this is still true), so bravo
Nov 7, 2011 at 11:49	history	edited	Pierre	CC BY-SA 3.0	added 83 characters in body
Nov 7, 2011 at 11:43	comment	added	Pierre		you also have $H_1(\mathbb{R}, \mathbb{R} -K)=0$ using that $H_1(\mathbb{R})=0$ (crucially!), so by excision $H_1(U, U-K)=0$, too. The last part of my answer was incorrect, let me edit (sorry i answered too quickly).
Nov 7, 2011 at 10:24	comment	added	Kwong		I am not very familiar with relative homology, but I think we need more than that. I can only see that if $H_0(U, U-K)=0$, the inclusion induces a surjective map from $H_0(U-K)$ to $H_0(U)$, by the long exact sequence of relative homology. (Correct me if I am wrong. ) Also, if your argument is correct, doesn't it imply that if $K$ is the equator of the torus $T^2$, and $U$ is a tubular neighborhood of $K$, then $U-K$ is connected?
Nov 7, 2011 at 8:17	history	answered	Pierre	CC BY-SA 3.0