Is there a compact metric space $X$ of covering dimension $2$ without a Lipschitz surjection on $[0,1]^2$?


For a space $X$ with Hausdorff dimension greater than $2$, we have a negative answer (see Theorem 1.1 in article <https://arxiv.org/pdf/1203.0686.pdf>).



Lipschitz mappings, covering dimension
Let $\ n\in\mathbb N.\ $ It's convenient to consider the injective metric
in $\ \mathbb R^n,\ $ it's Lipschitz equivalent to the Euclidean metrics.


By dimension, let's mean the *topological dimension* dim (say, *covering* -- for metric compact spaces, *topological* has only one standard meaning).




---


**Theorem**   For every metric compact space $\ X,\ $ with a topologically agreeing metrics $\ d,\ $ such that $\ \dim(X)\ge n,\ $
there exists a Lipschitz surjection
$\ f:X\to[-1;1]^n.$




---


**Proof**   Due to the dimension $\ \ge n\ $ there are
$\ (F\_k:\ k=\pm1 \ldots \pm n)\,\ $ closed subsets of $\ X\ $ such that


$$ \forall\_{k=1}^n\quad F\_{-k}\cap F\_k\ =\ \emptyset $$
and
$$ \bigcap\_{k=1}^n\,S\_k\ \ne\ \emptyset $$
for arbitrary closed separators S\_k between $\ F\_{-k}\ $ and $\ F\_k,\ $
where $\ 1\le k\le n.$


Let $\ g\_k:F\_{-k}\cup F\_k\to[-1;1]\ $ be defined by $\ g\_k(x)=-1\ $
for every $\ x\in F\_{-k},\ $ and $\ g\_k(x)=1\ $ for every $\ x\in F\_k,\ $
whenever $\ 1\le k\le n.\ $ These maps are obviously Lipschitz. Since $\ [-1;1]\ $ is metrically injective (is a **metric** AR), there are
Lipschitz maps $\ f\_k:X\to [-1;1]\ $ that extend the respective
$\ g\_k,\ $, hence


$$ f\ :=\ \triangle\_{k=1}^n f\_k:X\to[-1;1]^n $$
is a Lipschitz map.


Such continuous maps that start with the non-separable collections
$\ F\_{\pm k}\ $ are well-known and easily seen to be surjective
(or even *universal*).


**End of Proof**


