# Antipodal mapping from $S^n$ to $S^{n-1}$

Let $S^n$ denote the unit sphere in $\mathbb{R}^{n+1}$. An antipodal mapping $f:A\rightarrow B$ is a continuous function such that $f(-x)=-f(x)$ for all $x\in A$.

An antipodal mapping $S^n\rightarrow S^{n-1}$ is also a nowhere zero antipodal mapping $S^n\rightarrow \mathbb{R}^n$.

Why is the above sentence true? I realize that $S^{n-1}\subseteq\mathbb{R}^n$, but I don't understand the "nowhere zero" part.

I would presume that "nowhere zero" simply means that the function does not take 0 as a value; i.e., for every $x\in S^n$, we have $f(x)\neq 0$. This is true here since $f(x)\in S^{n-1}$ and $0\notin S^{n-1}$.