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Thread: Constructing homotopies of functions to the n-sphere

  1. #1
    Jul 2008

    Question [SOLVED]Constructing homotopies of functions to the n-sphere

    Here is problem statement:

    Show that if X is any topological space, and if two continuous maps f,g:X \to \mathbb{S}^n have the property that for all x\in X, \| f(x)-g(x)\| < 2 , then f is (continuously) homotopic to g.

    And here is a homotopy (in our case)
    A function F(x,t): X \times I \to \mathbb{S}^n such that F(x,0)=f(x) and F(x,1)=g(x).

    I don't have any trouble coming up with a function that satisfies the "deforming" property...for instance, there is a smooth function (I won't prove it, but there is, see Milnor or any other diff. top. book)

    \phi (t)=0 for t\leq 1/3
    \phi (t)=1 for t \geq 2/3

    And so our homotopy could simply be
    F(x,t)=\phi (t) g(x) + (1-\phi (t))f(x)

    For that matter, we could simply chose
    F(x,t)=\sin (t) g(x) + \cos (t) f(x)

    But the problem with both these functions is that they don't nessicarily satisfy \| F(x,t)\| = 1, which is required if we want the range to be the n-sphere. So does anyone have any suggestions on how to impose that condition, or any other approaches to this problem that might prove fruitful?

    Last edited by cduston; October 13th 2008 at 05:55 AM.
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  2. #2
    MHF Contributor
    Opalg's Avatar
    Aug 2007
    Leeds, UK
    Notice that the condition \|f(x)-g(x)\|<2 is just another way of saying g(x)\ne-f(x) (because the only way that \|f(x)-g(x)\| could be equal to 2 would be if f(x) and g(x) are antipodal points on the sphere). So the line joining f(x) and g(x) does not pass through the origin. Therefore we can define F(x,t)<br />
 = \frac{tg(x) + (1-t)f(x)}{\|tg(x) + (1-t)f(x)\|}. That's the required homotopy.
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  3. #3
    Jul 2008
    Looks good to me, thanks a ton!
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