The question of quasi-isometry

Question

I cannot show the following quetion.

"Let X and Y be metric spaces , and f: X → Y be a quasi-isometry.Show that there exists a finite metric space Z and a map F: X × Z → Y which is also a quasi-isometry and is onto."

If f is onto , this question is clear. So, the problem is the case that f is not onto.

Please advice to me

Answer 1

Here is the proof in the case where "finite metric space" means "bounded metric space", as in your comments.

Let $f: X\to Y$ be a quasi-isometry. Define the metric space $Z$ to be the set $Y$ equipped with the discrete metric taking only the values $0$ and $1$. Since $f$ is a quasi-isometry, there exists a constant $R$ so that $$ \bigcup_{x\in X} B(f(x), R)= Y, $$ where $B(y,R)$ is the (closed) metric $R$-ball in $Y$ centered at $y$. Now, for each $y\in Y$ there exists a surjective map $g_y: Z\to B(y, R)$, since cardinality of $Z$ is greater than (or equal to) cardinality of $B(y, R)$. Then define the map $$ F: X\times Z\to Y $$ by sending each $(x, z)$ to $g_{f(x)}(z)$. By construction, $F$ is surjective.

Let us check that $F$ is a quasi-isometry. Note that the map $F$ is within distance $R$ from the map $$ f\circ p : X\times Z\to Y $$ where $p(x,z)=x$. Since both $p$ and $f$ are quasi-isometries, so is $F$. qed

Answer 2

This is not true, at least for the definition of quasi-isometry that I'm familiar with. The inclusion map $\mathbb Z\to\mathbb R$ is a quasi-isometry. Multiplying $\mathbb Z$ by a finite (or countable) metric space gives another countable space, which of course cannot be mapped onto $\mathbb R$.