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Math Help - degree of a path, path homotopy.

  1. #1
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    degree of a path, path homotopy.

    Let f and g be loops in the unit circle S^1 with base point (1,0). We show if deg(f)=deg(g), then f and g are path homotopic.

    I know f,g: [0,1]-->S^1 and there must be unique paths f* and g* (lifting paths) s.t. f*(0)=0 and g*(0)=0 and s.t. pof*=f, pog*=g where p:R-->s^1 is defined by p(t)=(cos 2 pi t, sin 2 pi t).

    I also know that deg(f)= f*(1), deg(g)=g*(1) which are integers.

    I was thinking about coming up with a path homotopy F:[0,1]x[0,1]-->S^1 s.t. for each t in [0,1]; F(t,0)=f(t) and F(t,1)=g(t)
    for each s in [0,1]; F(0,s)=(1,0) and F(1,s)=(1,0).

    But I am not sure what path homotopy would work here.
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  2. #2
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    The Covering Homotopy Property: If F: I \times I \rightarrow S^1 is a homotopy such that F(0,0) = 1, then there is a lifting of F to a unique covering homotopy \bar{F}: I \times I \rightarrow R for which \bar{F}(0,0)=0.

    You have loops f and g in S^1 with a base point 1 having the same degree. This implies that the covering paths \bar{f}, \bar{g} beginning at common initial point 0 have common terminal point \bar{f}(1)=\bar{g}(1). The following homotopy H:I \times I \rightarrow R demonstrates the equivalence of \bar{f},\bar{g}.

    H(t,s) = (1-s)\bar{f}(t) + s\bar{g}(t), (t,s) \in I \times I .

    By the Covering Homotopy Property, the homotopy pH is a path homotopy between f and g (p is defined in your question).

    You need to make sure the following cases if pH is indeed a path homotopy between f and g.

    pH(0,s), pH(1,s), pH(t,0), and pH(t,1).
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  3. #3
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    Thanks, this makes a lot of sense. But I am wondering, the Covering Homotopy Property says if is a homotopy such that
    F(0,0) = (1,0), then there is a lifting of F to a unique covering homotopy
    F*:IxI -->R for which F*(0,0)=0 and poF* = F

    But does this imply that if H* is a path homotopy between f* and g* s.t. H*(0,0)= 0, then poH* must be a path homotopy between f and g? (where f* and g* are the liftings of f and g respectively).

    Because it seems that the proof is based on that fact.
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  4. #4
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    Quote Originally Posted by math8 View Post
    Thanks, this makes a lot of sense. But I am wondering, the Covering Homotopy Property says if is a homotopy such that
    F(0,0) = (1,0), then there is a lifting of F to a unique covering homotopy
    F*:IxI -->R for which F*(0,0)=0 and poF* = F

    But does this imply that if H* is a path homotopy between f* and g* s.t. H*(0,0)= 0, then poH* must be a path homotopy between f and g? (where f* and g* are the liftings of f and g respectively).

    Because it seems that the proof is based on that fact.
    To define H shown above, f and g should be loops in S^1 with the base point (1,0) having the same degree. Otherwise, the covering paths \bar{f}, \bar{g} do not have the common terminal point.

    By defintion,

    deg(f) = \bar{f}(1) and deg(g) = \bar{g}(1).

    So the degree of f and g should be the same to define H shown above.

    The Covering Homotopy Property simply ensures the existence of a unique covering homotopy \bar{F} of F. If f and g are equivalent loops in S^1, the degree of f and g are the same and we can define H shown above.

    deg(f)=\bar{f}(1)=\bar{F}(1,0)=\bar{F}(1,1)=\bar{g  }(1)=deg(g). (F is defined in your post.)

    \bar{F} and H are basically the same for the above case. The latter is intended to represent the homotopy starting off with \bar{f} and \bar{g}.

    Now, we can conclude that

    For loops f and g in S^1 with base point (1,0), [f]=[g] iff deg(f) = deg(g).
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  5. #5
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    Thanks, this really helps.
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