

Title: Curve Homotopic to a Point Contour Post by JP05 on Apr 9^{th}, 2006, 1:26pm Show there exists at least one closed curve C in the complex plane that is homotopic to the point contour z = 0. 

Title: Re: Curve Homotopic to a Point Contour Post by Icarus on Apr 9^{th}, 2006, 6:41pm z=sin(t) 

Title: Re: Curve Homotopic to a Point Contour Post by JP05 on Apr 10^{th}, 2006, 11:58am Care to elaborate? 

Title: Re: Curve Homotopic to a Point Contour Post by Obob on Apr 10^{th}, 2006, 1:52pm C is simply connected. Therefore every closed curve is homotopic to the point contour z=0. 

Title: Re: Curve Homotopic to a Point Contour Post by Icarus on Apr 10^{th}, 2006, 3:42pm I assumed he meant the curve itself has to be homotopic to a point, not that there is a homotopy of the ambient space that carries it to a point. As for elaborating: z(r,t)=sin(rt). 

Title: Re: Curve Homotopic to a Point Contour Post by Michael_Dagg on Apr 10^{th}, 2006, 4:24pm The first example Icarus gave works: the homotopy between z = sin(t) and z = 0 is X(s,t) = (1  s)w(t), where w(t) = sin(t), s, t in [0,1]. 

Title: Re: Curve Homotopic to a Point Contour Post by JP05 on Apr 10^{th}, 2006, 6:09pm Yeah, I was just asking how so to get more. That result with (1s)sin(t) wrote over [0,1] is in Whitehead's book but it talks about reverse path elimination. But, sin(rt) works too. 

Title: Re: Curve Homotopic to a Point Contour Post by Icarus on Apr 11^{th}, 2006, 4:29pm Contours are homotopic to a point if they do not encircle any points in their complement. I.e., the complement has a single component. Effectively, this means they "start" at some point z_{0}, traverse a path to some other point z_{1}, where they reverse direction and retrace the same path back to z_{0} to complete the loop (I am of course oversimplifying  they can turn around many times). Any path that does this is homotopic to z_{0}, for instance, by "pulling z_{1} in"  that is, by turning around earlier than z_{1}. z = sin(t) was simply the easiest example I could come up with. Its image consists of the unit interval [1, 1], and can be considered a contour by limiting it to [0, 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif]. My z = sin(rt) answer was supposed to be a homotopy of contours, but alas, it falls short, as Michael was too kind to say. However, his trick of X(s,t) = (1  s)w(t) will work for any radial contour. 

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