  A left-order on a group $G$ is a total order $<$ on $G$ such that for any
$f$, $g$ and $h$ in $G$ we have $f < g \Leftrightarrow hf < hg$. We construct a
finitely generated subgroup $G$ of $\operatorname{Homeo} (I^2;\delta I^2)$, the
group of those homeomorphisms of the disc that fix the boundary pointwise, and
show $G$ does not admit a left-order. Since any left-order on
$\operatorname{Homeo} (I^2;\delta I^2)$ would restrict to a left-order on $G$
this shows that $\operatorname{Homeo} (I^2;\delta I^2)$ does not admit a
left-order. Since $\operatorname{Homeo} (I;\delta I)$ admits a left-order it
follows that neither $G$ nor $\operatorname{Homeo} (I^2;\delta I^2)$ embed in
$\operatorname{Homeo} (I;\delta I)$.
