TITLE: Retraction of a Riemannian manifold with boundary to its cut locus
QUESTION [11 upvotes]: This question is edited following the comment of Joseph. He pointed out that the main object of the first version of this question is the cut locus.
Recall that the cut locus of a set $S$ in a geodesic space $X$
is the closure of the set of all points $p \in X$ that have two or more distinct shortest paths in $X$ from $S$ to $p$.
http://en.wikipedia.org/wiki/Cut_locus
A simple lemma shows that, for a disk $D^2$ with a Riemannian metric and piecewise smooth generic boundary, the cut locus of $D^2$ with respect to its boundary is a tree. 
A picture of such tree can be found on page 542, figure 17 of the article of Thurston "Shapes of polyhedra". The tree is white.
http://arxiv.org/PS_cache/math/pdf/9801/9801088v2.pdf
For an ellipse on the 2-plane, the tree is the segment that joins its focal points.
More generically for a Riemannian manifold $M^n$ with boundary, the cut locus of $\partial M$ should be a deformation retract of $M$.  (I guess it is a $CW$ complex of dimension less than $n$.) To prove this lemma, notice that $M^n\setminus\operatorname{cut-locus}(\partial M^n)$ is canonically foliated by geodesic segments that join $X$ with $\partial M$.
I wonder if this lemma has a name or maybe is contained in some textbook on Riemannian geometry?

REPLY [6 votes]: The 'simple lemma' certainly isn't always true.  Let's consider just open disks in the Euclidean plane with a flat metric. Fremlin (Proc. LMS, 1997) calls the set of points without a unique nearest boundary point the "skeleton".  He gives an example where the skeleton is somewhere dense (that is, its closure, which you call the "cut locus", has nonempty interior).  The skeleton will always be an R-tree.