Why is a branched line in R2 not a topological manifold?

[CSteiner]

Is there a topologist out there that wants to explain why exactly a branched line in R2 is not not a topological manifold? I know it's because there doesn't exist a chart at the point of branching, but I don't understand why not. I'm just starting to self study this, so go easy on me :).

 

[Orodruin]

but I don't understand why not.

How do you suggest such a chart is constructed?

 

[CSteiner]

Well why don't we form a chart (U, x) where U is an element of the subset topology of the standard topology on R2. Then U can be a subset of of the branch containing the point of branching with open endpoints on each of the three branches. x can then map each of these points to a point in R2. This mapping seems to be invertable, and continuous in both directions.

 

[Orodruin]

But it is not a mapping from an open subset of R^2.

 

[CSteiner]

But it is not a mapping from an open subset of R^2.

Why not? Sorry if this seems like a stupid question, but I'm only just now learning (and trying to understand) the definitions of these things. Open means that it belongs the standard topology on R2, right? Why doesn't it?

 

[Orodruin]

Because it is not a union of balls. This is the very definition of the standard topology.

 

[micromass]

Is there a topologist out there that wants to explain why exactly a branched line in R2 is not not a topological manifold? I know it's because there doesn't exist a chart at the point of branching, but I don't understand why not. I'm just starting to self study this, so go easy on me :).

Do you know the concept of "connected"?

 

[CSteiner]

Do you know the concept of "connected"?

Not rigorously.

 

[micromass]

You should learn about that then. Learn about "cut points" too. With that you can prove your OP.

 

[CSteiner]

You should learn about that then. Learn about "cut points" too. With that you can prove your OP.

Will do, thanks for pointing me in the right direction.