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Let $f:S^1\rightarrow S^1$ is a continuous map with the property that $f(-x)=-f(x)$ for all $x\in X$.

Question is to show that degree of $f$ is odd.

We have $(\exp \circ g)(t)=f(e^{2\pi it})$ for all $t\in \mathbb{R}$.

Now, $f(-x)=(\exp \circ g)(-t)=e^{2\pi i g(-t)}$ and $f(x)=e^{2\pi i g(t)}$

As $f(-x)=-f(x)$ we have $e^{2\pi i g(-t)}=-e^{2\pi i g(t)}$ i.e., $e^{2\pi i (g(-t)-g(t))}=-1=e^{(2n+1)\pi i}$ for all $n\in \mathbb{N}$.

So, for some $m\in \mathbb{N}$ we have $$g(-t)-g(t)=\frac{2n+1}{2}+m$$

I am stuck here and could not proceed further...

I am looking for order of $f$ i.e., $g(1)$...

1 Answers1

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As you say, there exists a continous function $g:\mathbb R\rightarrow \mathbb R$ such that $$\forall t\in \mathbb R, \ f(e^{2i\pi t})=e^{2i\pi g(t)}.$$ Since $f(-z)=-f(z)$, it means that $\forall t\in \mathbb R$, $$e^{2i\pi g(t+\frac{1}{2})}=f(-e^{2i\pi t})=-f(e^{2i\pi t})=-e^{2i\pi g(t)}=e^{2i\pi\left(g(t)+\frac{1}{2}\right)}.$$

Hence (and this is where you made a mistake), there exists $m\in\mathbb Z$ such that $$g\left(t+\frac{1}{2}\right) = g(t)+\frac{1}{2}+m$$ Finally, $$g(t+1)-g(t)=\left[ g(t+1)-g\left(t+\frac{1}{2}\right)\right]+\left[g\left(t+\frac{1}{2}\right)-g(t)\right]=2m+1 $$ and the degree of $f$ is odd.

Bebop
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  • In fourth line why is it $g(t+\frac{1}{2})$ and not $g(-t)$?? –  Feb 27 '15 at 10:33
  • What do you get if you compute $e^{2i\pi g(t+\frac{1}{2})}$ ? – Bebop Feb 27 '15 at 10:37
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    Oh.... I got it... That was my mistake... I was writing $e^{2\pi i g(-t)}$ for $f(-e^{2\pi i t})$... we have $e^{2\pi i g(t+\frac{1}{2})}=f(e^{2\pi i (t+\frac{1}{2})})=f(e^{2\pi i t}\cdot e^{\pi i})=f(-e^{2\pi i t})$... Fine... –  Feb 27 '15 at 10:42