# Homeomorphism exchanging two homotopic paths

Let $X$ be a compact simply connected space and $\gamma_1, \gamma_2 : [0,1] \to X$ be two (homotopic) simple paths between two different points $x,y \in X$. Does there exist a homeomorphism $\varphi : X \to X$ such that $\varphi \circ \gamma_1= \gamma_2$?

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The paths are free homotopic or based homotopic? I guess the first option. –  Sigur Jan 26 '13 at 18:01
@Sigur: I don't know the difference... The same homotopy used to define the fundamental group? –  Seirios Jan 26 '13 at 18:33
No. If you have two paths with the same endpoints, then you can ask for homotopies preserving these endpoints. But also you can consider free homotopies, i.e. it is not necessary to preserve them. –  Sigur Jan 26 '13 at 18:35

Let $X$ be a closed disk, let $x$ and $y$ be two points on its boundary, let $\gamma_1$ be a path from $x$ to $y$ along the boundary, and let $\gamma_2$ be a path from $x$ to $y$ through the interior of the disk. A homeomorphism from the disk onto itself takes the boundary to the boundary, so it can't send $\gamma_1$ to $\gamma_2$.

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I don't think so. Let $X = S^2\vee S^3$, the one point union of $S^2$ and $S^3$. Since both $S^2$ and $S^3$ are compact and simply connected, so is $X$. Let $\gamma_1$ be a path in $S^2$ (starting at the wedge point, if you want) and $\gamma_2$ a path in $S^3$.

I claim there is no homeomorphism $\phi$ moving $\gamma_1$ to $\gamma_2$. The point is that any homeomorphism of $X$ must send the wedge point to itself, and then this implies the $S^2$ must be sent to itself and likewise the $S^3$ must be sent to itself.

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I have forgotten to mention that $\gamma_1$ and $\gamma_2$ have the same extremities; I edited my question. –  Seirios Jan 26 '13 at 18:26
@Seirios The same example still works with minor modifications: Let the paths start and end at the joining point. If you insist on simple paths, join the spheres along a segment instead of only a point. Just make sure that one path enters only the $2$-sphere and the other one only the $3$-sphere. Invariance of dimension excludes the possibility of having a homeomorphism carrying one path into the other. –  Martin Jan 26 '13 at 18:35

A positive answer is given by the isotopy extension theorem for smooth manifolds without boundary: if two smooth embeddings of an interval are isotopic, then there is a diffeomorphism of $X$ that interchanges them. If the dimension of $M$ is $\geq 4$, then two embeddings of an interval are isotopic iff they are homotopic.

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Very interesting. Do you have a reference? –  Seirios Jan 28 '13 at 8:52
Yes, it's in Kosinski's book . In this answer you have the reference and a comment on the case of topological manifolds (math.stackexchange.com/questions/157038/…) –  user17786 Jan 28 '13 at 20:56
Do you happen to know what happens for smooth manifolds (without boundary) of dimension $2$ or $3$? –  Jason DeVito Jan 31 '13 at 14:09
@JasonDeVito, The isotopy extension theorem holds for any dimension, and the comparison between isotopies and homotopies is the main theorem of this article: eudml.org/doc/139228. –  user17786 Feb 1 '13 at 10:34
@JasonDeVito, I'm not sure I understood your question. If you're asking whether two homotopic embeddings of a circle in a $3$-manifold $M$ are ambient isotopic, then $M = \mathbb R^3$ gives a counterexample. On the other hand, it is a theorem that any two homotopic embeddings of a circle into a surface are isotopic. –  user17786 Feb 1 '13 at 10:47