Prove that any map $f\colon S^1 \to S^1$ mapping antipodal point to antipodal point has $\deg_2(f)=1$ by a direct computation.

Question

Without using the Borsuk-Ulam theorem. Prove that any map $f\colon S^1 \to S^1$ mapping antipodal point to antipodal point has $\deg_2(f)=1$ by a direct computation.

I know that $f$ map anitpodal point to antipodal point, meaning $f$ is an odd function.

From one of my previous exercise, I have shown that for any smooth map $f\colon S^1 \to S^1$ there exist a smooth map $g\colon R\to R$ such that $f(\cos t, \sin t)=(\cos g(t), \sin g(t))$ and satisfying $g(t+2\pi)=g(t)+2\pi q$ for some integer $q$ and $\deg_2(f)= q \mod 2$

Since $f$ is an odd function, $f(-x)=-f(x)$ My professor says I need to use this fact to find a $g\colon R\to R$ such that $g(s+\pi)=g(s)+\pi q$ where $q$ is odd. But I have no clue how to do that.

Answer 1

Use the $g$ you already have from the previous exercise. What do you know about $f((-1,0))$ compared to $f((1,0))$? What does this tell you about $g$?

EDIT Further hint: First, \begin{align*} (\cos(g(t+\pi)),\sin(g(t+\pi)))&=f(\cos(t+\pi),\sin(t+\pi)) = f(-\cos t,-\sin t) = -f(\cos t,\sin t)\\ &= -(\cos(g(t)),\sin(g(t))). \end{align*} Next, if $\cos\beta=-\cos\alpha$ and $\sin\beta=-\sin\alpha$, what must be the relation between $\alpha$ and $\beta$?

Answer 2

Here you have a complete answer.

We know that there exist a map $g: \mathbb{R} \to \mathbb{R}$ and a $q \in \mathbb{Z}$ such that \begin{align} & g(t + 2\pi ) = g(t) + 2 \pi q \ \forall \ t \in \mathbb{R} \tag{1}\\ & f(\cos t, \sin t) = (\cos g(t), \sin g(t)) \tag{2}\\ & deg_2(f) = q \pmod 2 \tag{3} \end{align} By $(2)$ and because $f$ map anitpodal points to antipodal points we must have \begin{align*} (\cos g(t+\pi), \sin g(t+\pi)) & = f(\cos (t+\pi), \sin (t+\pi) )\\ & = f(-\cos t, -\sin t ) \\ & = - f(\cos t, \sin t ) = - (\cos g(t), \sin g(t) ) \end{align*} that is $\cos g(t+\pi)= -\cos g(t)$ and $\sin g(t+\pi)= -\sin g(t)$, so it must exist a $q' \in \mathbb{Z}$ such \begin{equation} g(t + \pi) = g(t) + q'\pi \tag{4} \end{equation} then $$ g(t + 2\pi) = g(t + \pi + \pi)= g(t+\pi)+ q'\pi = g(t) + 2\pi q' $$ but by $(1)$ we deduce that $q'=q$, so in (4) we can replace $q'$ for $q$. Finally because $f(1,0)=(-1,0)$, whith $(2)$ we obtain that $\cos g(0) = -1$ and then that $g(0)=k_o\pi$ where $k_o$ is odd. Analogously since $f(-1,0)=(1,0)$ we get $g(\pi) = k_e \pi$ where $k_e$ is even. Then by (4) $$ g(\pi) = g(0) + q\pi \ \Longrightarrow k_e \pi = k_o\pi + q\pi $$ so $q=k_e-k_o$, then we must have that $q$ is an odd integer. Now by (3) we conclude that indeed $$ deg_2(f) = 1 $$

Answer 3

Correct me if I'm wrong but here is how I understand this problem.

First you know that $f$ is an odd function. $f:S^1 \to S^1$ can be defined as $f(x)=xe^{i\theta}$

Then you lift $f$ to a map $g: R\to R$ such that $g(x+1)=g(x)+ deg(f)$. I think of $2\pi$ as one circle, you come back to where you begin, so I just call it $1$.

Since $f$ is odd $g(x+\frac1 2)=g(x)+ \frac k 2$ for some odd integer $k$. It follow that $deg(f)=k$ must be odd

One more thing, I'm not sure if you are allowed to use this theorem : for $f:S^n \to S^n $ is continuous, the antipodal map of $f$ has degree $(-1)^{n+1}$ for $n$ is the dimension of the sphere.