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Homotopy extension property

  1. Apr 11, 2007 #1
    Hi there,

    I've been asked to prove that (R^2, S^1) has the homtopy extension property and then extend it to the general case: (R^n, S^(n-1))
    here's where i've got so far,
    for (R^2, S^1) Let S^1=A, R^2=X
    well S^1 is contained in R^2, so by a theorem, if A contained in X, has a mapping cylinder neighbourhood N, then (X,A) has the homotopy extension property.

    so, let N= {x ; 0.5 <=||x||<= 1.5}, N is closed, A is contained in N. Let b_1={x ; 0.5=||x||} and b_2= {x ; 1.5=||x||} , let B= b_1 union b_2. then B=the boundary of N, B closed implies N\B is open, and A is contained in N\B, so N is a mapping cylinder neighbourhood.

    Now, let f: B->A

    M_f (mapping cylinder) = [(BxI) disjoint union A]/~ where, for x belonging to B, (x,1)~f(x)
    = [(b_2 x [0,0.5] union b_1 x [-0.5,0]) disjoint union A]/~
    isomorphic to Ax[-1,1].

    Define a homeomorphism
    h: Ax[-1,1] -> N by h(e^itheta, t) =(t+1)e^itheta for t belonging to [-1,1] and theta belonging to [0, 2Pi]

    Then Ax[-1,1] is isomorphic to N and h restricted to A union B = Identity.

    Hence N is a mapping cylinder neighbourhood of A and therefore, (X,A) has the Homotopy extension Property.

    is this correct, particularly the bold bit, and do i just have to set A= S^(n-1) and X= R^n, for it to work in the cases for other dimensons?

    Thank You
  2. jcsd
  3. Apr 11, 2007 #2


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    No, the bold part is where your mistake is. You need to define a map f, and then the mapping cylinder is determined by this as you wrote. You seem to have arbitrarily identified some set that you want to be the mapping cylinder without using the definition. Once you have the mapping cylinder (which is its own space) you need to show it's homeomorphic to N, taking B to B and A to A. Here's a hint: f won't be injective.
  4. Apr 11, 2007 #3


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    for my benefit would you please state in detail the property you want to prove?
  5. Apr 12, 2007 #4
    I was firstly asked to prove that (R^2, S^1) has the homotopy extension property.

    Definition of Homotopy extension property: suppose one is given a map f_0:X->y, for A contained in X. and there exists a homotopy f_t: A->y of f_0 restricted to A that one would like to extend to a homotopy f_t:X->y of f_0.
    If (X,A) is such that this extension problem can always be solved, then (X,A) has the homotopy extension property.
  6. Apr 15, 2007 #5


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    you want to show any map defined on the union of the plane and a cylindder passing through the plane perpendicularly, extends to three space.

    That seems obvious. \\

    here is a simple example. consider the product of R^1 with the unit interval [0,1].

    then in there, take the subset R^1 x {0} union {0} x [0,1] union {1} x [0,1].

    you have a map defiend on the second sett hat you want to extend to the fiorst set. just retract the first set ointo the seconmd one.

    say project the set [0,1] x [0,1] back onto {0}x [0,1] union [0,1] x {0} union {1} x [0,1], by projecting from the point (1/2, 1).

    then project the rest of the first set back onto the second set by projecting from say the points (-1, 1) and (2,1).

    then composing the retraction with the map of the subset gives the extension.

    notice i have just done the case of the pair (R^1, S^0).
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