# Why is a continuous, degree zero and odd function from $S^n$ an antipodal map?

https://terrytao.wordpress.com/2008/11/27/the-kakeya-conjecture-and-the-ham-sandwich-theorem/

In this blog post, Terry tao states that "It is easy to see that $f$ is continuous, homogeneous of degree zero, and odd, and so its restriction to $S^n$ is an antipodal map."

Why is this the case? It may only be the case that it is true in the specific case of this $f$, as defined in the post, but I can't see why these conditions would lead to this conclusion of antipodality.

• Since $f$ is odd we have that $f(-x) =-f(x)$ which is what you need for it to be the antipodal map. I don't see why the other conditions are required. – R_D Jan 25 '16 at 15:53
• @Rise it need not be the antipodal map. It need not even be continuous. – Matt Samuel Jan 25 '16 at 15:59
• @MattSamuel, I did not know there can be other antipodal maps. What would an example? Thank you. – R_D Jan 26 '16 at 4:28
• @Rise I didn't either, but I'm trusting Tao. In any case, $f(x)=ix$ on $S^3$ considered as the group of unit quaternions is continuous and odd, and can be extended to $\mathbb R^4$ as homogenous of degree 0, but is not "the" antipodal map. – Matt Samuel Jan 26 '16 at 4:32
• @MattSamuel, good point. Thank you. – R_D Jan 26 '16 at 4:34