Nearest-point projection is the tubular retraction after shrinking

Write $x=E(p,v)=p+v$ in the tubular coordinates from [L1]. Because $v$ is orthogonal to $T_{p}S$, the function $q\mapsto \mathrm{\parallel }x-q{\mathrm{\parallel }}^2$ has vanishing first derivative at $q=p$. Its Hessian on the tangent directions equals the Euclidean metric plus terms that go to zero with $\mathrm{\parallel }v\mathrm{\parallel }$. Therefore, after shrinking the radius if necessary, $q=p$ is a strict local minimizer on each normal fibre.

On each compact piece of $S$, the radius can be shrunk once more so that this local minimizer is the only point of $S$ at the same or smaller distance from $x$. Applying this on a locally finite cover yields a still smaller tubular neighbourhood on which every point has a unique nearest point in $S$.

In the tubular coordinates, that unique nearest point is exactly the base point $p$ of the normal vector $v$. But [L2] defines the tubular retraction by sending $x=E(p,v)$ to $p$. Hence the nearest-point projection and the tubular retraction agree on the shrunken tube.