Why the figure of eight is not a manifold?

[maxverywell]

Why the figure of eight is not a manifold?

I have read somewhere that if we remove the crossing point than the the figure of eight becomes disconnected, but by removing one point in \mathbb{R}^2 it's still connected.

Is there any other proof without removing the crossing point?

 

[jgens]

I have read somewhere that if we remove the crossing point than the the figure of eight becomes disconnected, but by removing one point in \mathbb{R}^2 it's still connected.

This does show that the figure eight cannot be a 2-manifold, but it does not show that it has no manifold structure at all. You need some additional arguments for that. For example to show that the figure eight is not a 1-manifold, you could note in a sufficiently small neighborhood of the crossing point, if you remove this point then you end up with four connected components. The higher-dimensional cases are pretty easy too.

Is there any other proof without removing the crossing point?

Sure. You could use something like the classification of 1-manifolds to show that the figure eight is not a manifold, but doing that is like killing a fly with a sledgehammer. The connectedness argument is nice because it is simple and does not rely on a bunch of heavy machinery.

 

[WannabeNewton]

I'm curious as to why you want a different way anyways? A very similar example is the wedge sum of two copies of the real line which happens to be homeomorphic to the disjoint union of the x and y axes which can be shown to fail to be a manifold in the same exact method alluded to by Jgens: by looking at the 4 connected path components one gets by removing the origin; higher dimensional cases are yet again rather easy. Why make it more involved than it needs to be?

 

[maxverywell]

Thanks!