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Thread: Constructing homotopies of functions to the n-sphere

  1. #1
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    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 $\displaystyle f,g:X \to \mathbb{S}^n$ have the property that for all $\displaystyle x\in X$, $\displaystyle \| f(x)-g(x)\| < 2 $, then f is (continuously) homotopic to g.

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

    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)

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

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

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

    But the problem with both these functions is that they don't nessicarily satisfy $\displaystyle \| F(x,t)\| = 1$, 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?

    Thanks.
    Last edited by cduston; Oct 13th 2008 at 05:55 AM.
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  2. #2
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    Notice that the condition $\displaystyle \|f(x)-g(x)\|<2$ is just another way of saying $\displaystyle g(x)\ne-f(x)$ (because the only way that $\displaystyle \|f(x)-g(x)\|$ 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 $\displaystyle F(x,t)
    = \frac{tg(x) + (1-t)f(x)}{\|tg(x) + (1-t)f(x)\|}$. That's the required homotopy.
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  3. #3
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    Looks good to me, thanks a ton!
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