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Let $\alpha:[0,1]\rightarrow S^1:t\mapsto e^{2\pi it}$ be a path.

Let $f:S^1\rightarrow S^1$ be a continuous map such that $-f(x)=f(-x)$ on $S^1$.

How do I show that the winding number $Wnd(f\circ \alpha,0)$ is not zero?

I have proven this by proving $f$ is not null-homotopic, but I'm now trying to show to prove this directly.

Here's how I tried.

Note that $-(f\circ \alpha)(t)=(f\circ \alpha)(t+1/2)$.

Set $f\circ \alpha = r e^{i\theta}$

If you draw a circle, then it's intuitively clear that $\theta(0)\neq \theta(1)$.

I think this is due to connectedness of $[0,1]$, but I don't know how to prove this rigorously.

Please help :)

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Divide [0,1] into two parts [0,1/2] and [1/2,1] and note that $r=1$

Now, let's decompose $\theta$ into two parts so that $\theta=\theta_1 \ast \theta_2$ and set $g_1=e^{i\theta_1} , g_2=e^{i\theta_2}$

Note that $Wnd(f\circ \alpha,0)=Wnd(g_1,0) + Wnd(g_2,0)$ and $g_2=-g_1$

Hence, $e^{\theta_1 - \theta_2} = -1$. Since the range of this is discrete, $\theta_1 = \theta_2 + (2n+1)\pi$

Of course $Wnd(g_1,0)$ is not 0 since $\theta_1(0)\neq \theta_1(1)$

Hence $Wnd(f\circ \alpha,0)= 2 Wnd(g_1,0)$ and thus not 0.

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  • $\begingroup$ You need to use braces, e^{\theta_1} to get $e^{\theta_1}$ instead of $e^\theta_1$. In your argument, you wrote $\theta_2 = \theta_1 + \pi/2$. You probably meant $\pi$, since $e^{i\pi} = -1$ and $e^{i\pi/2} = i$. But that need not be, consider $f(z) = z^3$. However, the key idea is right, and only a small fix needed. (I'm not saying more now because I think you learn more by finding the flaw and fixing it yourself. Ping me when you have fixed it, or when you don't find the fix and need more guidance.) $\endgroup$ – Daniel Fischer Dec 17 '14 at 11:13
  • $\begingroup$ @DanielFischer Correted. Is it correct now? $\endgroup$ – cococomi Dec 17 '14 at 11:42
  • $\begingroup$ That's unfair to edit again while I type my comment and thus making it obsolete ;) Yes, now it is correct. $\endgroup$ – Daniel Fischer Dec 17 '14 at 11:49
  • $\begingroup$ Oh.. I didn't notice that :S Thank you! $\endgroup$ – cococomi Dec 17 '14 at 11:50
  • $\begingroup$ No problem. Out of curiosity, did you come up with the argument on your own, or did you take an argument from the literature and modify it to fit this problem? $\endgroup$ – Daniel Fischer Dec 17 '14 at 11:58

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