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TOPOLOGY: homeomorphism between quotient spaces

  1. Mar 23, 2008 #1
    Show the following spaces are homeomorphic: [tex]\mathbb{R}^2, \mathbb{R}^2/I, \mathbb{R}^2/D^2[/tex].

    Note: [tex]D^2[/tex] is the closed ball of radius 1 centered at the origin. [tex]I[/tex] is the closed interval [0,1] in [tex]\mathbb{R}[/tex].

    THEOREM:
    It is enough to find a surjective, continuous map [tex]f:X\rightarrow Y[/tex] to show that [tex]X/S(f)\approx Y[/tex].

    I can easily show [tex]R^2/D^2 \approx R^2[/tex] by sending every point in [tex]D^2[/tex] to the origin and, for every point outside [tex]D^2[/tex], letting [tex]f(r,\theta)=(r-1,\theta)[/tex].

    I am having difficulty showing that [tex]\mathbb{R}^2/I[/tex] is homeomorphic to either [tex]\mathbb{R}^2[/tex] or [tex]\mathbb{R}^2/D^2[/tex]. First, I do not understand the notation [tex]\mathbb{R}^2/I[/tex] since I is traditionally a subset of R, not R^2. Does [tex]\mathbb{R}^2/I[/tex] mean:

    (1) points on the x-axis are equivalent iff the x coordinate falls in [0,1] and all other points (any point x,y where y=/=0) are not equivalent, OR
    (2) (x,y)~(w,z) iff y=z and [tex]x,w\in[0,1][/tex], OR
    (3) (x,y)~(w,z) iff [tex]x,w\in[0,1][/tex] regardless of y and z?

    If it means (2), then I simply take f so that f(x,y)=(0,y) for x in [0,1], f(x,y)=(x-1,y) for x>1 and f(x,y)=(x,y) for x<0.

    If it is (1), I can find a surjective (but not continous!) map f that takes [0,1]x{0} to (0,0), (x,0) to (x-1,0) for x>1, and leaves all other points where they are.

    For (3), I can find a continuous surjective map from R^2 to R^2/I where f(x,y)=(x,y) for [tex]x\leq0[/tex] and f(x,y)=(x+1,y) for x>0.

    Can someone clarify what [tex]\mathbb{R}^2/I[/tex] means? If it means (1), any hints?

    Thanks. :)
     
  2. jcsd
  3. Mar 23, 2008 #2
    Asked a friend - I think, we're using definition (2) for the problem, in which case I've got a solution.

    Thanks. :)
     
  4. Nov 29, 2010 #3
    Sorry if this is too late for your needs.

    A general result that will help you here is that if A is contractible in X, then

    X/A is homeo to X , which kind of makes sense, since you may consider A as

    a single point in X, and then X/{pt.} is a trivial quotient. In this case, D^2 is

    contractible in R^2 , and I is ( I am.?) also contractible, both in R, and in R^2.
     
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