Is it true, the space of embeddings segments is homotopy equivalent to the subspace of all line segments? Consider the space of smooth embeddings of the segment in the plane with the compact-open topology. Denote by X the quotient space obtained from the equivalence relation $a \sim b$ if and only if $\mathrm{Im}~ a = \mathrm{Im}~ b$
Consider the retraction $f: X \to X$ such that $f(s)$ is the line segment connecting the ends of s (it is obvious that the function is continuous and idempotent). Is it true, that $f$ is homotopic to the $id$? if not, can it be otherwise proved that the subspace of all segments is a deformation retract?
 A: Here are some remarks which may be relevant.


*

*First of all it seems to me that the correct topology to use is the Whitney $C^\infty$-topology on the embedding space.

*Let $M$ be an  closed manifold. There is a free action of the group of diffeomorphisms $\text{Diff}(M)$ on the space of embeddings $E(M,\Bbb R^n)$.  The orbit space of the action, call it $S(M,\Bbb R^n)$, is identified with subspace of all submanifolds of $\Bbb R^n$ that are diffeomorphic to $M$.   I remember hearing as a graduate student that $E(M,\Bbb R^n)\to S(M,\Bbb R^n)$ is a Serre fibration. This result may actually be due to Cerf.

*If a version of the result in (2) holds for compact manifolds $M$ with boundary $\partial M$, then the map $E(M,\Bbb R^n)\to S(M,\Bbb R^n)$ is a fibration where the group acting in this case is the diffeomorphisms of $M$ which restrict to the identity on $\partial M$. (However, I do not know if this is true.)

*Assuming the statement in (3) is true, then take $M = I = [0,1]$. The diffeomorphism group $\text{Diff}(I \text{ rel } \partial I)$ in this case is weakly contractible. It follows that the map $E(I,\Bbb R^n) \to S(I,\Bbb R^n)$ is a weak homotopy equivalence. 
