TITLE: Effective contraction of a loop. Reference or a simple proof?
QUESTION [12 upvotes]: Let $M$ be a compact simply connected R. manifold. Let $x$ be a base point and let $\gamma$ be a smooth loop in $M$ starting and ending at $x$. 

Is there a base point preserving retraction of $\gamma$ to $x$  such that every point on $\gamma$ travels a distance at most, say, $2diam(M)$?

I think the answer is yes. The scheme I have in mind is the following
Consider the cut locus $CL(x)$ of $x$. By perturbing the metric and the curve we can assume that $CL(x)$ is triangulable and $\gamma$ intersects it at finitely many points. Then slide these points off the $CL(x)$. Then the new curve does not intersect $CL(x)$ and can be contracted to $x$ following the geodesics.
This is probably too complicated...

Is there a simple proof? A reference?

Remark: 1) I would be happy with any estimate (not necessarily $2diam(M)$) which is independent of $\gamma$.
2) If there is a good curve shortening procedure that is not base-point-preserving please share it as well.

REPLY [11 votes]: Yes there is a bound independent of $\gamma$. Fix a fine triangulation of $M$ (say, with simplices 10 times smaller than the injectivity radius of the metric). For each vertex $q$ of the triangulation, fix a shortest path $s_q$ connecting $q$ to the marked point $p$. It is easy to deform any loop $\gamma$, via a short homotopy, into a path in the 1-skeleton of the triangulation. Then, for every vertex occuring on this path, "pull" a small piece of the loop to $p$ along the corresponding $s_q$. Now the loop is a product of "elementary" loops of the form $s_q^{-1}[qq']s_{q'}$ where $[qq']$ is an edge of the triangulation. Contract these "elementary" loops one-by-one. Since there are only finitely many different elementary loops, and the same elementary loops work for all $\gamma$'s, we have a uniform upper bound (depending on $M$) for the length of the homotopy.
On the other hand, there is no bound for the homotopy length in terms of the diameter. In addition to the great Bill Thurston's answer, let me mention another construction. For every $\varepsilon>0$ there is a metric on $S^3$ with sectional curvature $\le 1$ and diameter $\le\varepsilon$ (it was first constructed in Gromov "Almost flat manifolds, see also Buser and Gromoll "On the almost negatively curved $3$-sphere"). In this metric, take a geodesic starting from $p$ of length $2\varepsilon$ and connect its endpoint back to $p$ by a shortest path. Since the curvature is bounded above by 1, any contraction of this curve will have a trajectory longer than $\pi$ (I think this fact is called Klingenberg's lemma but I am not sure).