Manifold path components are open

Question

Can someone explain why the path components of a manifold are open? I'm a little confused at how to demonstrate this fact and it would obviously help me understand manifolds better.

Is path component of manifold related to the disconnected component of manifold? — annie marie cœur, Apr 04 '21 at 19:34
And is the set of path component of manifold, equivalent to the set of disconnected component of manifold? — annie marie cœur, Apr 04 '21 at 19:35

Alex Youcis · Answer 1 · 2013-04-10T03:05:23.203

1

Hint: Manifolds are locally homeomorphic to Euclidean balls, and so simply connected and path connected, and pretty much whatever you want.

EDIT: Thanks to Andreas Blass--I misread your question, my previous response would have been if you asked whether or not the connected components of a manifold are open.

edited Apr 10 '13 at 03:05

answered Apr 10 '13 at 02:47

Alex Youcis

54,059

score 1 · Answer 2 · answered Apr 10 '13 at 02:53

1

Alex Youcis's answer is true, but I think "locally pathwise connected" is more useful for this question.

answered Apr 10 '13 at 02:53

Andreas Blass

71,833

score -1 · Answer 3 · answered Jul 15 '18 at 10:54

-1

A manifold is locally euclidean implies it is locally path connected. Hence components and path components are the same and they are open.

answered Jul 15 '18 at 10:54

Tcm

29

Manifold path components are open

3 Answers3