The space of regular curves deformation retracts onto the space of arclength parameterized curves Let X denote the space of smooth maps from the circle into R^3 which have no zero derivatives. This is an open submanifold of the Frechet space of smooth maps from the circle into R^3. Let Y denote the space of such smooth maps which are arc-length parametrized. I believe that this is a submanifold of X. 
Is Y a (strong) deformation retract of X? Is there an easy proof of this? If not, is there a reference which addresses this, and/or similar types of problems?
 A: Any two regular parameterizations of a closed curve differ by a diffeomorphism of the unit circle. If we had a unique arclength parameterization then we would be done, but different arclength parameterizations differ by the circle rotation. Now, the claim, I think, follows from the fact that the subgroup $O(2)\subset Diff(S^1)$ is a deformation retract (see for instance here: Ghys does it for homeomorphisms but the proof is the same in the smooth category). 
To finish the proof, consider the space of regular closed curves (in a fixed target manifold) $C$ and its subspace $A$ of arc-length parameterizaed curves; take the quotient $C/O(2)$, where $O(2)$ acts by precomposition. In view of what we just saw, this quotient strongly retracts to the space of arc-length parameterizaed curves $A/O(2)$ (since such parameterizations are unique up to rotation). I think, this retraction lifts to a retraction $C\to A$.  
A: Could you do something like this?  Given a curve $\gamma:[0,1]\to \mathbb R^3$, let $u(s,t)$ be the solution to the heat equation $u_t = u_{ss}$ with Dirichlet boundary conditions, and $u(s,0) = \theta(s) := \int_0^s |\gamma'(s')| ds'$.  Then let
$\Gamma:[0,1]\times[0,1] \to \mathbb R^3$ be given by
$$ \Gamma(s,t) = \gamma(\theta^{-1}(u(s,t/(1-t)))) ?$$
So $\Gamma(s,0) = \gamma(s)$, and $\tilde\gamma(s) := \Gamma(s,1) = \gamma(\theta^{-1}(s))$ is arclength parametrized?
So I have done it for maps from $[0,1]$, and I am probably assuming that my curves all satisfy $\theta(s) = 1$.  But I think that is the sort of thing I would try.
