0
$\begingroup$

Is every nonempty open set $U\subseteq\mathbb{R}^n$ diffeomorphic to $\mathbb{R}^n$?

I think this is false; perhaps I can take $U$ to be the disjoint union of two open balls. But how can I prove that this is not diffeomorphic to $\mathbb{R}^n$?

More generally, do diffeomorphisms preserve connectedness, or number of connected components?

$\endgroup$

2 Answers 2

7
$\begingroup$

For two sets to be diffeomorphic, they must be homeomorphic. Connectedness (and number of connected compotents) is a topological property and is therefore preserved under homeomorphism. Therefore $\mathbb{R}^n$ is not homeomorphic to any disconnected open subset of itself.

$\endgroup$
6
  • $\begingroup$ and if we assume $U$ connected? $\endgroup$ Dec 6, 2017 at 19:23
  • 6
    $\begingroup$ Still not. Take the annulus $\endgroup$
    – eepperly16
    Dec 6, 2017 at 19:25
  • $\begingroup$ @VeridianDynamics The statement may hold if your set is open and convex or perhaps open and star-like. $\endgroup$
    – eepperly16
    Dec 6, 2017 at 19:29
  • 3
    $\begingroup$ @Arthur Unfortunately, no. There are rather surprising counterexamples. But even better; there are contractible open subsets of $\Bbb R^3$ which are not even homeomorphic to $\Bbb R^3$: eg, the Whitehead manifold $\endgroup$ Dec 6, 2017 at 20:46
  • 2
    $\begingroup$ @eepperly16 This is indeed true for open star-like subsets, but is rather hard to prove. See, eg, this MO post $\endgroup$ Dec 6, 2017 at 20:51
6
$\begingroup$

Diffeomorphism saves connectedness. But here $\mathbb R^n$ is connected and union of two disjoint open sets is not connected.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .