# 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?

• Most of the techniques that I've tried to address this question founder on issues with the point set topology on X. The function "arc length" $X \to \Bbb R$ is not continuous - in fact, it is not even bounded above by a continuous function - because, eg, one can converge to a straight line in the compact-open topology with a sequence of very tightly packed helix-shaped curves. – Tyler Lawson Feb 1 at 14:43
• @TylerLawson: It might help to get some clarification from the author. It looks like you are thinking of the $C^0$ version of the compact-open topology. There are versions adapted to the smooth mapping space as well, and it's unclear if this is what the author means. More often it would just be called the weak topology, or the Whitney topology, in that context. – Ryan Budney Feb 1 at 15:43

1. First of all it seems to me that the correct topology to use is the Whitney $$C^\infty$$-topology on the embedding space.
2. 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.
3. 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.)
4. 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.
• It seems to me that the OP's question is: "Is the restriction map $S(I, \Bbb R^2) \to S(S^0,\Bbb R^2)$ an equivalence?" This seems harder to me. – Mike Miller Feb 2 at 5:09
• $E(I,\Bbb R^2)$ is weakly equivalent to $S^1$. Assuming the fibration assertion, it would follow that $S(I,\Bbb R^2)$ is also weakly equivalent to $S^1$. But $S(S^0,\Bbb R^2)$ is identified with $S^1$ as well (it's a configuration space of unordered subsets of $\Bbb R^2$ of cardinality two). So I think the OP's question will follow from what I remarked. – John Klein Feb 2 at 12:48