# 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.

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.

• What definition of degree are you using? Nov 23, 2014 at 15:41
• for $f:X\to Y$, $dim(X)=dim(Y)$ then the intersection number $I_2(f,{y})$ are the same for all $y\in Y$, we call this common value $deg_2(f)$ Nov 23, 2014 at 15:59

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 $\beta$?

• $f((-1,0))=-f((1,0))$ and $g$ is the angle of $f$ and the range of $g$ is from $0$ to $\pi$, right? Nov 23, 2014 at 23:23
• No, but what is $g(\pi)-g(0)$? Nov 24, 2014 at 0:18
• $g(\pi)-g(0)=0$ , so $g(\pi)= g(-\pi)$, which make $g$ is even function, but how this can help me? Nov 24, 2014 at 1:10
• if $\cos\beta=-\cos\alpha$ and $\sin\beta=-\sin\alpha$, then $\alpha = \beta + k\pi$. From your first hint, this implies $g(t+\pi)= g(t)+ k\pi$ for some odd integer $k$, because if $k$ is even then the equation in your first hint will not hold. Am I correct? Nov 24, 2014 at 22:29
• Thank you very much, I think I understand better about antipodal maps now. Nov 24, 2014 at 22:31

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 $$deg_2(f) = 1$$

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.

• Diane, what do you mean by $f(x)=xe^{i\theta}$? (The OP is using $2\pi$ where you are using $1$.) So, the crucial thing to argue is: Because $f$ is odd, $g(t+\pi) = g(t)+k\pi$ for some odd integer $k$. You didn't justify this claim. Nov 24, 2014 at 21:50
• I'm not trying to give him/her a full solution, just trying to give him another view on what his professor told him. I guess I didn't do a good job then. Nov 24, 2014 at 21:55