0
$\begingroup$

Let $X$ be a topological n-manifold and $U, V \subseteq X$ two open subsets homeomorphic to $\mathbb{R}^n$. Suppose $\emptyset \neq U \cap V$ connected. Is true that $U \cup V$ is an open subset of $X$ homeomorphic to $\mathbb{R}^n$? If it is not in general, is it true for $n = 1$?

$\endgroup$
  • 1
    $\begingroup$ Well, the deleted answer only needed to up the dimension of their example by one. Take a sphere and two charts given by a little more than a hemisphere. Their intersection is connected, homeomorphic to an annulus. $\endgroup$ – user565560 May 30 '18 at 11:38
  • $\begingroup$ @elmer: Go ahead, post that. $\endgroup$ – quasi May 30 '18 at 11:39
  • $\begingroup$ You are right, what a stupid question.. And for n = 1? $\endgroup$ – Marco All-in Nervo May 30 '18 at 11:42
  • $\begingroup$ Slight cheating, but consider $U, V$ being disjoint open intervals on $X = \mathbb{R}$. $\endgroup$ – Adayah May 30 '18 at 11:51
  • 1
    $\begingroup$ Any connected $1$-manifold without boundary is homeomorphic to either $\mathbb{R}$, or $S^1$. Based on that, it's easily verified that, for $n=1$, no counterexample is possible (assuming $U\cap V\ne\varnothing$). $\endgroup$ – quasi May 30 '18 at 11:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.