other cases by even simpler methods. So in this case there is an $n-2$-plane $V\subseteq {\mathds{R}}^n$ such that if $x\not\in V$ then $P^d$ is not critical at $x$.

There are two steps needed to complete the proof. First, if $x\in S_2$ then by assumption there exists a point $x_i$ our covering which is not too far from $x$ relative to $r'_{x}$. In particular, since $u$ is close to $P^d$ centered at $x_i$ this is roughly equivalent to the statement that $d(x,V)>r'_x\geq r$. We have already mentioned that $P^d$ is therefore not critical at $x$, and with a little work, since $u$ is close to $P^d$, one can use this to show $u$ is not critical at $x$. More effectively, we even have that $r_x\geq r$, which proves (2) for the covering. Making this precise will turn out to require an effective cone-splitting argument.

Finally, let us consider the projection map $P^V:{\mathds{R}}^n\to V$. Since $u$ is close to $P^d$ for in all balls $B_{s}(x_i)$ with $s\in [r_i,1]$, one can use this to prove the projection map, when restricted to the centers of the balls $\{x_i\}$, is $(1+\epsilon)$-biLipschitz. Slightly more precisely, if $x_i,x_j\in \{x_k\}$ are two centers in the covering, then we 