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.

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 $\beta$?