This is the exercise 6 in Chapter 2, Section A (2.A.6) in Rolfsen.
I know that two such arcs with common endpoints and disjoint interiors are homotopic via straight line homotopy. I am looking for extending this homotopy to the plane. By Alexander Lemma, it is possible if the homotopy is defined from circle, not from just an arc. So that I can at least this homotopy to unit disk.
Any idea to help?
My copy of Rolfsen has the sentence "Moreover, the [ambient] isotopy may be taken to be fixed on any neighbourhood of the closure of the region bounded by $A \cup B$." in this exercise. This sentence isn't quite right.
Since $A$ and $B$ have disjoint interiors, the region bounded by $A \cup B$ is (homeomorphic to) an open disk, $D$. Then $A \cup B \cup D$ is the closure of the region bounded by $A \cup B$. Any ambient isotopy taking $A$ to $B$ drags $A$ through every point of $D$, so is not fixed on any of $A \cup B \cup D$. The correct sentence would be "Moreover, the isotopy may be taken to be fixed on any neighbourhood of the complement of a neighborhood of the closure of the region bounded by $A \cup B$." That is, the ambient isotopy only needs to distort a small fringe outside $A \cup B \cup D$.
Of course, this is also a hint. Your ambient isotopy is permitted to be constant beyond that narrow fringe (... and up to homeomorphism, "narrow" is largely meaningless.) Could you, for instance, find an explicit ambient isotopy of the plane, fixed outside $[-2,2]\times[-2,2]$ which takes the semicircle $\{(x,-\sqrt{1-x}) \mid -1 \leq x \leq 1\}$ to the semicircle $\{(x,\sqrt{1-x}) \mid -1 \leq x \leq 1\}$? (Here, our fringe, which we are allowed to distort, is the region bounded on the interior by the unit circle and on the exterior by the given square.) I predict you could even make a piecewise linear ambient isotopy...