TITLE: When is a continous $\epsilon$-isometry of the sphere surjective?
QUESTION [9 upvotes]: Equip $\mathbb S^n$ with the standard round metric. Let $f : \mathbb S^n \to \mathbb S^n$ be a continous map satisfying $\vert d(f(x),f(y)) - d(x,y)\vert \leq \epsilon$. 
Is $f$ is surjective for all $0 \leq \epsilon < \epsilon_0$ for some positive $\epsilon_0$? 
My guess would be that the answer is yes and maybe $\epsilon_0 = \pi$.

REPLY [15 votes]: Your guess seems to be true. If a map $f$ is not surjective then $f$ can be considered as a continuous map from $S^n$ to $R^n$. Hence there exist two opposite points on $S^n$ which maps to the same point by Borsuk-Ulam theorem.