5
$\begingroup$

Let $M$ be a smooth manifold and $N$ a closed embedded submanifold. Assume that they have the same dimension. In this case are they equal?

EDIT: M is connected.

$\endgroup$
1
  • $\begingroup$ You can use open-closed argument here. $\endgroup$
    – user99914
    Jul 2, 2014 at 14:08

2 Answers 2

8
$\begingroup$

Yes. If $N\subseteq M$ is an embedded submanifold (without boundary) of the same dimension as $M$, then the inclusion map $N\hookrightarrow M$ is a smooth embedding, which means that its differential at each point is bijective. It follows from the inverse function theorem that its image is open in $M$. Since $N$ is both open and closed, it is all of $M$.

$\endgroup$
2
  • $\begingroup$ Do we really need to require that $N$ do not have a boundart? I am sorry to ask, but my geometry is not that great... $\endgroup$
    – J.L
    Jul 3, 2014 at 18:07
  • $\begingroup$ Yes, we really do. If $N$ has a nonempty boundary, then it need not be open in $M$. Just think of the case in which $M=\mathbb R^n$ and $N$ is the closed unit ball. $\endgroup$
    – Jack Lee
    Jul 3, 2014 at 19:43
0
$\begingroup$

As the original poster added later, you require connectedness:

As Jack argued, the inclusion is an embedding. Since the manifolds have same dimension, it is an open map. Thus the image is open. By assumption it is also closed. Therefore, the image is a connected component of $M$---but it need not be $M$ itself, unless $M$ has only one component.

$\endgroup$

You must log in to answer this question.

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