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Identification Space

  1. Feb 23, 2010 #1
    Is this a cone after Identifaction?

    Let A = (I × I)/J
    where J = (I × {1}) ∪ ({0; 1} × I) ⊂ I × I
     
  2. jcsd
  3. Feb 23, 2010 #2

    quasar987

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    It's a closed disk. But if by "cone" you mean baseless cone (i.e. a child's birthday party hat : http://www.utterwonder.com/archives/images/happy_birthday_party_hat-thumb.jpg [Broken]), then one can also say that A is a cone, since the two are homeomorphic.
     
    Last edited by a moderator: May 4, 2017
  4. Feb 23, 2010 #3
    how would you construct that homeomorphism
     
  5. Feb 23, 2010 #4

    quasar987

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    Take C the cone with base a unit circle in the xy plane and vertex at say (0,0,1). Let also D be the unit disk centered at (0,0,0). Then the projection down on the xy-plane C-->D²:(x,y,z)-->(x,y,0) is the desired homeomorphism.
     
  6. Feb 24, 2010 #5
    Could I do it using the circles
    Ct = {(x; y) ∈ D2 | (x − t)2 + y2 = (1 − t)2} (t ∈ I)
    to construct a homeomorphism f : A → D2
     
  7. Feb 24, 2010 #6

    quasar987

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    What do you mean?

    Also, for "x quared", put x^2 instead of x2 please.
     
  8. Feb 24, 2010 #7
    Sorry about that I constructed it as follows is this ok?

    A -> D^2

    (s,t) in I x I -> (cos(2pi(1-t)s),sin(2pi(1-t)s)) in D^2
     
  9. Feb 24, 2010 #8

    quasar987

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    This is a map from I x I to D^2. I thought you were trying to construct a homeomorphism between C and D^2. I don't see the link.
     
  10. Feb 24, 2010 #9
    I was trying to create a homeomorphism between the identified I x I and D^2 such that Ct is satisfied.
     
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