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Homotopic maps of the n-sphere

  1. Jun 14, 2012 #1
    "Let f,g:Sn→Sn be maps so that f(x) and g(x) are not antipodal for any x. Show that f and g are homotopic."

    Here's my initial approach: I figured it would be easier to work in In instead, so I note that Sn is the quotient of the n-cube with its boundary. Therefore, each map Sn→Sn can be lifted to a map In→Sn which is constant on the boundary in Rn.

    It may also be possible to use a covering space to lift the other end of the map, getting In→Rn. I played around with this in the n=1 case (the only case where I'm certain it'd work) and it didn't seem to help.

    In any case, In is contractible, so we have a homotopy F between f' and g' (our lifted maps). I just can't figure out how to use the fact that f and g are never antipodal to prove that F can be deformed into a homotopy which is constant on the boundary at ever stage. I understand intuitively that that restriction prevents one map from ever "wrapping around" more than the other, but I'm not certain how to apply it.
     
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  3. Jun 14, 2012 #2

    lavinia

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    Not sure but if the two maps are never anitpodal then there is a unique length minimizing great circle connecting them.
     
  4. Jun 14, 2012 #3

    Bacle2

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    Consider a homotopy between the two points in R^n and then rescale so that the

    homotopy is moved to Sn
     
  5. Jun 14, 2012 #4
    A homotopy between which two points? Having a homotopy between f(x) and g(x) for each fixed x doesn't give me a homotopy between f and g, since those homotopies need not vary continuously with x.
     
  6. Jun 14, 2012 #5
    If homotopies between corresponding points in paths constituted a homotopy of those paths, then every path-connected space would be simply connected.
     
  7. Jun 14, 2012 #6

    lavinia

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    You can move along the great circles by moving along the line segment in Euclidean space connecting f(x) to g(x) then projecting onto the sphere. The line segment will not pass through the origin because f(s) and g(x) are not antipodal.
     
  8. Jun 14, 2012 #7

    Bacle2

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    I meant a homotopy between f(x) and g(x) in general; the (arguably) simplest homotopy.
     
  9. Jun 14, 2012 #8
    That worked perfectly:

    [itex]F(z,t) = \frac{tf(z) + (1-t)g(z)}{|tf(z) + (1-t)g(z)|}[/itex]

    is my homotopy. Thanks a lot!
     
  10. Jun 15, 2012 #9

    Bacle2

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    Glad it helped.
     
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