I am using the following definitions:
Let $X,Y$ be metric spaces.
We say that a map $f:X\to Y$ is coarse if:
$1.$ the pre-image of any bounded set in $Y$ under $f$ is bounded in $X$; and
$2.$ for any $R>0$ there exists $S>0$ such that for all $x,y\in X$ if $d_X(x,y)<R$ then $d_Y(f(x),f(y))<S$.
We say that a coarse map $f:X\to Y$ is a coarse equivalence if there exists a coarse map $g:Y\to X$ such that $d_Y(f\circ g(y),y)$ and $d_X(g\circ f(x),x)$ are (uniformly) bounded. That is, $f\circ g$ and $g\circ f$ are close to the identity maps.
$X$ and $Y$ are coarsely equivalent if such a coarse equivalence exists.
Now, I want to show that if $X$ and $Y$ are coarsely equivalent then $asdim(x)=asdim(Y)$.
My approach:
Let $f:X\to Y$ and $g:Y\to X$ be as above. Let $R>0$ be given.
Let $S>0$ be such that if $d_X(x,y)<R$ then $d_Y(f(x),f(y))<S$.
Let $\mathcal{U}=\{U_i\}_{i\in I}$ be a uniformly bounded cover for $Y$ with $S$-multiplicity $asdimY+1$.
Let $\mathcal{V}=\{f^{-1}(U_i)\}_{i\in I}$ be a uniformly bounded (since $f$ pulls back bounded sets to bounded sets) cover of $X$.
By construction $B_R(x)$ intersects as many elements of $\mathcal{V}$ as the number of elements of $\mathcal{U}$ intersecting $f(B_R(x))$. But $f(B_R(x))\subseteq B_S(f(x))$.
Since $B_S(f(x))$ intersecting at most $asdimY+1$ elements of $\mathcal{U}$, we obtain $asdimX\leq asdimY$.
Now, use coarsity of $g$ to show $asdimY\leq asdimX$.
The problem is that I actually didn't use the fact that the compositions are close to the identity maps. So, I've shown that whenever $f:X\to Y$ is a coarse map, $asdimX\leq asdimY$, which I suspect to be true.
Thank you for any help.