2
$\begingroup$

Does all smooth embedding $\iota: S\to M$ is proper if $\dim S < \dim M$? I thought then the slice criterion implies $S$ is locally closed, so closed in $M$. But this is equivalent to saying that the embedding is proper. Is it correct? Also what can happen in the topological category?

$\endgroup$

1 Answer 1

1
$\begingroup$

No: the embedding $(-1,1) \to \Bbb R^2$ given by $t \mapsto (t,0)$ is smooth, but not proper, since the preimage of the compact subset $[-1,1]\times\{0\}$ is not compact.

$\endgroup$
2
  • $\begingroup$ Note that locally closed does not imply closed. Any open subset of $\Bbb R^n$ is locally closed. $\endgroup$
    – Didier
    Commented Jul 10 at 10:04
  • 1
    $\begingroup$ Yes, I was trying to use the fact that if we have an open cover and every intersection is closed then closed, but I missed that $M\setminus S$ is not open so we cannot make open cover $\endgroup$ Commented Jul 10 at 10:06

You must log in to answer this question.

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