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Let $a(t)=e^{2\pi it}$ be the generator of $\pi_1(S^1,1)$ and we define degree of $f$ this way:

$$[\omega^{-1}]\cdot f_*([a])\cdot [\omega]=\deg(f)[a]$$

where $\omega$ is any path from $f(a)$ to $1$.

Prove: if $\deg(f)=\deg(g) \iff f\simeq g$.

I think from $f\simeq g$ to $\deg(f)=\deg(g)$ is easy. But I have difficulty in the opposite direction, that is, $\deg(f)=\deg(g) \Rightarrow f\simeq g$.

I know $f\simeq g \Rightarrow [\omega^{-1}]\cdot f_*([a])\cdot [\omega]=g_*([a])$. Is the opposite direction still correct?

Balarka Sen
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Tom
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  • See here http://math.stackexchange.com/questions/550216/degree-of-continuous-maps-from-s1-to-s1-two-equivalent-properties/550254#550254 – Haha Dec 08 '13 at 17:14
  • There is "it turns out that the degree map gives a bijection from $[S^1,S^1]$ to the integers." I am not clear about this statement. – Tom Dec 08 '13 at 17:27

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