Let $f: [0,1]^2 \to [0,1]^2$ be a homeomorphism of the square. By the n-grid I mean the collection of smaller squares $$[a/2^n,a/2^n + 1/2^n] \times [b/2^n,b/2^n + 1/2^n] \text{ for } a,b = 0,\ldots, 2^{n}-1$$ for some natural number $n$. I would like to take some grid fine enough that $f$ can be approximated by permuting the grid elements.
Formally, let $\epsilon >0$ be given. Does there exist an $n$-grid and permutation $\sigma$ of the squares in the grid such that for each $x \in [0,1]^2$ we have $d(f(x),\sigma (A)) < \epsilon$ for every grid element $A$ with $x \in A$?
This looks a little like simplicial approximation but I can't find anything about simplicial approximations of homeomorphisms being bijective.
No. Take $\epsilon = \frac{1}{100}$. Take any homeomorphism mapping $[0,\frac{1}{2}]\times[\frac{1}{2},1]$ into $[0,\frac{1}{2^{10}}]\times[1-\frac{1}{2^{10}},1]$. For $n=1$, we must send $[0,\frac{1}{2}]\times[\frac{1}{2},1]$ to itself, and so the inverse of the point $(\frac{1}{2^9},1-\frac{1}{2^9})$ (say) must be far from its corresponding cube. And for any $n \ge 2$, some cubes in $[0,\frac{1}{2}]\times[\frac{1}{2},1]$ will be sent far away from $[0,\frac{1}{2^{10}}]\times[1-\frac{1}{2^{10}},1]$.