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Homeomorphisms in Topology

  1. Oct 17, 2005 #1
    One question I've had lately in my independent study of topology is the problem of how to show two sets are homeomorphic to each other. I am not sure how I would go about doing this in a general, or even specific case. One problem that wants me to demonstrate this is in Mendelson:

    Prove that an open interval (a,b) considered as a subspace of the real line is homeomorphic to the real line.

    Now, I know that a homeomorphism is a map where both it and its inverse are continuous, so I'm wondering if all that is needed to show two sets are homeomorphic is to simply define a mapping between the two. However, I don't know if this is right. Can someone please clarify?

  2. jcsd
  3. Oct 17, 2005 #2
    Yes, that is pretty much it. The mapping has to be one-to-one so that an inverse can exist. The mapping has to be continuous and the inverse has to be continuous. So here the task is to define a mapping from the interval (a,b) to the entire real line [itex](-\infty,+\infty)[/itex].
  4. Oct 18, 2005 #3
    How would you do that? I saw someone once parametrize to make it work, but I guess I just don't know how one would go about doing that.
  5. Oct 18, 2005 #4

    George Jones

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    What's with the LaTeX preview feature? It never seems to work for me.

    Back to the matter at hand.

    Hint: think tan(x) restricted to the domain (-pi/2 , pi/2).

  6. Oct 18, 2005 #5

    George Jones

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    If you don't want to muck around mapping (a , b) onto (-pi/2 , pi/2), an easier example that I think works is f: (a , b) -> R given by

    f(x) =1/(x -a) + 1/(x - b).

  7. Oct 18, 2005 #6
    Great, thanks a lot you guys. That was a big help. I guess it just depends on being creative in your map-defining.
  8. Oct 18, 2005 #7
    How about showing that two spaces are not homeomorphic.

    To do that you would have to show that one has a topological property that the other does not right?

    By topological property I guess I mean any property of a topological space that is always preserved by homeomorphism...

    Maybe its even harder to prove that any propety is always preserved by homeomorphisms... (I don't immediately see how to prove such a statement).

    maybe I've just confused myself.

    I guess a better way to prove two spaces are not homeomorphic would be to just assume they are homeomorphic and show that the existence of a homeomorphism between them leads to a contradiction.

    Am I right? or does that first method work after all?
  9. Oct 19, 2005 #8

    matt grime

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    well, trivially, the answer is "yes" since "is a member of some equivalence class of topological spaces modulo homeomorphism" is a topological property. That is to say determining homeomorphism type and showing they are not equivalent would obviously do.

    no, it is usually easy to prove certain properties are always preserved by homeomorphism. What is hard is generating a complete set of data that uniquely characterizes the space and is not as difficult as working out every possible homeomorphic image.

    your "two methods" appear to be the same: show that they have some different invariants under homeomorphism (that would be the contradiction). there are some easy to calculate invariants: homotopy groups, homology groups etc.

    In the case of smooth (connected compact) 2-folds then the genus and orientability determine the space up to homeomorphism. That is every compact connected surface is homeomorphic to either a torus with n holes (n in N) or a torus with n holes and a mobius cap stitched in. (equivalently the fundamental group determines it).

    in the case of 3-folds (things that look locally like R^3) then we do not know what data is sufficient to characterize them. see eg the poincare and geometrization conjectures.
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