All quasi-isometry is coarsely surjective.

Question

Definition. $(X,d_X)$, $(Y,d_Y)$ metric spaces. $f:X\to Y$ is quasi-isometry if

$d_Y(f(x),f(x'))\leq Ld_X(x,x')+C$

exists coarse inverse, i.e. $\overline{f}:Y\to X$ such that

$d_X(\overline{f}f(x),x)\leq C$

$d_Y(f\overline{f}(y),y)\leq C$

and $d_X(\overline{f}(y),\overline{f}(y'))\leq Ld_Y(y,y')+C$

for all $x,x'\in X$, for all $y,y'\in Y$

How prove that $f:X\to Y$ ($L-C$) quasi-isometry then $f$ is coarsely surjective? i.e. $\forall y\in Y, \exists f(x)\in f(X)$ such that $d_Y(y,f(x))\leq C$

Answer 1

Put $x=\bar f(y)$.

..........

Given $y\in Y$ we need to find $x$ satisfying given property. But the definition of a quasi-isometry trivially implies that $x=\bar f(y)$ fits.