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Prove that the cone over the unit circle is homeomorphic to the closed unit disc.

  1. Jun 2, 2012 #1
    1. The problem statement, all variables and given/known data

    This question comes out of "Introduction to Topology" by Mendelson, from the section on Identification Topologies.

    Let D be the closed unit disc in R^2, so that the boundary, S, is the unit circle. Let [tex] C=S\times [0,1], [/tex] and [tex]
    A=S \times \{1\} \subset C. [/tex] Prove that C/A is homeomorphic to D.

    2. Relevant equations



    3. The attempt at a solution

    I feel as though the map [tex]
    p:C/A \rightarrow D \\
    p(x,y,z)=(x,y) [/tex]
    should define a nice homeomorphism. It has an obvious inverse, but even proving that the forward one is continuous is proving to be a problem for me. This may be for lack of experience working with the identification topology on C/A, or maybe I'm taking the wrong approach here.

    p is continuous iff for every δ open in D, [tex] p^{-1}(δ):=ψ[/tex] is open in C/A. A subset of C/A is open iff in turn,
    [tex]
    f^{-1}(ψ)\subset C
    [/tex]
    is open, where
    [tex]
    f:C\rightarrow C/A
    [/tex] is defined by
    [tex]
    f|_{C-A}=id|_{C-A}\\
    f(A)=(0,0,1).
    [/tex]

    So for the moment, the problem is: prove that [tex]f^{-1}(ψ)[/tex] is open.

    I'm open to more elegant approaches, if you can get me started. I feel as though my approach might be too mechanical, to be honest. There must be a nicer way of doing this.
     
  2. jcsd
  3. Jun 3, 2012 #2
    wait, your title says cone over the unit circle (without the bottom im assuming), but s^1x[0,1]/(s^1x{1}) would be a cylinder without a top?
     
  4. Jun 3, 2012 #3
    To be clear, by C/A, I mean the quotient space obtained by identifying the subset A with a point (in this case, the point (0,0,1) ). Generally, what I've seen is the notation C/A is reserved for a quotient space, whereas C\A is the complement of A in C (also written C-A).

    So in the context of this problem, C/A is a cone without a bottom, and C\A is a hollow, bottomless, topless cylinder which is missing the upper "lip".

    Specifically, by C/A I mean the set [tex]C/A=(C-A)\cup (0,0,1) [/tex]
    which is given a topology as follows: σ is open in C/A if and only if [tex]f^{-1}(σ)[/tex] is open in C, where the function f is defined as in my original post.
     
  5. Jun 4, 2012 #4
  6. Jun 7, 2012 #5
    Thank you for the suggestion, tt. It wasn't exactly the same problem, but it pointed me in the right direction. In the event anybody ever needs help with this problem, I'll post an outline of the solution here.

    The key was defining two auxillary functions; one to take the set C\A (the hollow bottomless, topless cylinder without the top lip) to the hollow bottomless cone without the "tip", and another one to then flatten the image into the punctured (at the origin) closed disc. This can be done with cylindrical coordinates, and a little bit of thinking about the geometry. These functions can both be proven to be homeomorphisms onto their images.

    Finally, build the homeomorphism C/A->D by mapping C\A to the punctured disc by composing the aforementioned homeomorphisms, and sending a*, the image of the the 'quotiented-out' set A (in the space C/A, a*=μ(A), where μ is the quotient map), to the origin, 0, of the disc. Prove stuff.

    Cheers.
     
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