Understanding Manifolds: Questions & Answers

[Fellowroot]

I'm just trying to understand something about manifolds.

What is meant when a manifold doesn't have boundary? I thought the boundary was where the manifold "ends" so to speak. Like a boundary point, something where you take a small nhbd (neighborhood) and you get something inside the set and then something outside the set.

But then there are objects that arn't manifolds apparently but they do have boundary.

Another thing. Excuse the simplicity of this question, but why are we so concerned about making maps and charts and atlases and such. What is the whole purpose of doing this? And why do we care about these maps being smooth?

Also what is the purpose of restricting things like here with local diffeomorphisms.

http://en.wikipedia.org/wiki/Local_diffeomorphism

As seen in the link they restrict F which is the map to U. Why do this? What exactly do they mean by this?

Thanks.

 

[Cruz Martinez]

Another thing. Excuse the simplicity of this question, but why are we so concerned about making maps and charts and atlases and such. What is the whole purpose of doing this? And why do we care about these maps being smooth?Thanks.

One of the reasons charts are used is because you can do calculus on R^n, and so by using the charts you can do calculus on the manifold.

With the second question, do you mean to ask why the charts have to be compatible?

 

[WWGD]

The reason why you use local homeos. is because a manifold is a space that is not (globally*) quite like Euclidean space, but
it is locally like Euclidean space. And this statement means in a more rigorous way that every point has a neighborhood that is
homeomorphic to R^n , since homeomorphisms preserve basic topological properties of a space. Hope this is what you were asking.

* Not necessarily so, but sometimes so.

 

[lavinia]

Have you read the definition of the boundary of a manifold?

 

[Bacle2]

A boundary is just a sort of "kink" in the manifold, so that near the boundary point the manifold is not locally like R^n. But I agree with lavinia; to get more out of an answer, it helps if you read the def first and ask something more specific.