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Let $f:S^1→S^1$ be any smooth map. There exists a smooth map $g:\mathbb R \to \mathbb R$ such that $f(\cos(t),\sin(t) )=(\cos(g(t)),\sin(g(t) )$ and satisfying $g(2π)=g(0)+2πq$ for some integers $q$. Prove that $\deg_2 (f) \equiv q \mod 2$

Mod 2 degree of $f$ $(\deg_2(f)) $ theorem: if $f:X\to Y$ is smooth from a cmpact manifold $X$ into a connected manifold $Y$ and $\dim X=\dim Y$ then $I_2(f,{y})$ is the same for all $y\in Y$. This common value is call the mod 2 degree of $f$, denoted $\deg_2 (f)$

From the hypothesis, I have $f:S^1→S^1$ is smooth and $\dim S^1 =\dim S^1$ ,ofcourse. So clearly I can use this theorem an got $I_2(f,{S^1})$ is the same. but how can I show that $\deg_2 (f) \equiv q \mod 2$

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The loop $f$ is homotopic to the loop $\tilde{f}(e^{it})=e^{iqt}$ (well-defined for every $t\in\mathbb{R}$) through the homotopy $$ H:S^1\times [0,1]\rightarrow S^1\quad;\quad H(e^{it},s)=e^{i[sg(t)+(1-s)qt]}$$ $H$ is well-defined because $g(t+2\pi)=g(t)+2\pi q$ for every $t\in\mathbb{R}$: the smooth function $h(t)=g(t+2\pi)-g(t)$ assumes discrete values in $2\pi\mathbb{Z}\subseteq\mathbb{R}$, for connectedness then $h(t)=h(0)=2\pi q$. Since homotopic maps have the same mod 2 degree you have $$ \#f^{-1}(y)\equiv\#\tilde{f}^{-1}(y)\equiv q\pmod 2$$ where $y\in S^1$ is a regular value for both $f,\tilde{f}$

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  • $\begingroup$ what is $h(t)$, is it how $H$ defined? $\endgroup$ Oct 21, 2014 at 22:04
  • $\begingroup$ $h:\mathbb{R}\rightarrow\mathbb{R}$ is the smooth map defined by $t\rightarrow g(t+2\pi)-g(t)$, I used the fact that it is constant ($h(\mathbb{R})\subseteq 2\pi\mathbb{Z}$ is connected so it consists of a single point) to prove that the homotopy $H:S^1\times[0,1]\rightarrow S^1$ between $f$ and $\tilde{f}$ defined by $H(e^{it},s)=e^{i(sg(t)+(1-s)qt)}$ is well-defined or, in other words, $H(e^{it},s)$ doesn't depend on the choice of $t\in\mathbb{R}$. $\endgroup$
    – Gabriele
    Oct 22, 2014 at 16:31

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