Let $S$ be a subset of $\mathbb{R}^n$. I would like to call $S$


1. a Lipschitz(1) hypersurface if for every $x\in S$ there is a hyperplane $H$ so that the orthogonal projection onto $H$ is a bi-Lipschitz map from a neighbourhood of $x$ in $S$ onto an open subset of $H$, and
2. a Lipschitz(2) hypersurface if for every $x\in S$ there is a bi-Lipschitz map $\psi$ from $B\times(-1,1)$ onto a neighbourhood of $x$ in $\mathbb{R}^n$ so that $\psi^{-1}(S)=B\times\{0\}$, where $B$ is an open subset of $\mathbb{R}^{n-1}$.



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> It seems clear enough (\*) that Lipschitz(1) implies Lipschitz(2). But is the converse true? And if not, what is a simple counterexample?
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I have come across the notion of regions with Lipschitz boundaries in a number of papers on boundary value problems for PDEs. But every such paper seems to take the notion of Lipschitz-ness for granted.


(\*) If $S$ is Lipschitz(1), then after a rotation of the axes, it locally looks like the graph of a Lipschitz function $\gamma\colon\mathbb{R}^{n-1}\to\mathbb{R}$. Put $\psi(x,t)=(x,\gamma(x)-t)$ to obtain the Lipschitz(2) property.



Are these two notions of Lipschitz hypersurface equivalent?
Counterexample in $\mathbb R^3$: Let $C$ be the cube $max(|x|,|y|,|z|)\le 1$. Let $S$ be the intersection of $C$ with $z=0$. Choose a piecewise linear (PL) homeomorphism from the boundary of $C$ to itself such that the boundary of $S$ goes to a very zigzaggy set. Extend to a PL (therefore Lipschitz) homeomorphism from $C$ to itself by coning to the center point $(0,0,0)$. The image of $S$ (minus its boundary) will be Lipschitz(2) but not Lipschitz(1); there will be no good direction such that the projection to a hyperplane is one to one near the center.


