I'm looking for a statement, if it exists, of the following sort:
Let $X$ be a compact metric space. Let $f:X\to X$ be a function, and suppose that for some (fixed) $\epsilon>0$, for all $x,y\in X$, $$ | d(f(x) , f(y) ) - d(x,y) | < \epsilon . $$ Then $f$ is $\epsilon$-surjective, meaning that for every $y\in X$, $$ \inf_{x\in X} d(y, f(x)) < \epsilon .$$
Can anyone help me, and if possible, provide a reference? Possibly replacing $\epsilon$-surjectivity with $2\epsilon$, or any other similar function of $\epsilon$.
What you want to prove is false.
Let $X = \{x_1,\dots,x_n\}$ with $d(x_2,x_1) = \epsilon$, $d(x_3,x_1),d(x_3,x_2) = 2\epsilon$, $d(x_4,x_1),d(x_4,x_2),d(x_4,x_3) = 3\epsilon$, etc.. Define $f: X \to X$ by $f(x_1) = x_1$ and $f(x_i) = x_{i-1}$ for $2 \le i \le n$. Then $f$ is $\epsilon$-close to being an isometry, but is $(n-1)\epsilon$ away from being surjective.