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Why the figure of eight is not a manifold?

  1. Mar 18, 2013 #1
    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 [itex]\mathbb{R}^2[/itex] it's still connected.

    Is there any other proof without removing the crossing point?
     
  2. jcsd
  3. Mar 18, 2013 #2

    jgens

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    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.

    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.
     
  4. Mar 18, 2013 #3

    WannabeNewton

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    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?
     
  5. Mar 19, 2013 #4
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