Homotopy of Jordan Curves? A Jordan curve is an injective continuous map from $S^1$ to $\mathbb{R}^2$.
If $\gamma_1,\gamma_2,\gamma_3,\gamma_4$ are four counter clockwise Jordan curves, such that $\gamma_{i+1}$ is contained in the exterior region of $\gamma_i$ for each $i=1,2,3$. Let $A$ be the intersection of the of exterior region of $\gamma_1$ and the interior region of $\gamma_4$, is it true that there is a homotopy from $\gamma_2$ to $\gamma_3$ in $A$?
Here homotopy means homotopy as maps from $S^1$ to $\mathbb{R}^2$.
I thought this seems reasonable, but I couldn't come up with a proof myself, although I do think we may assume that $\gamma_i$'s are all polygonal curves, but it is still hard after that.
Maybe it is possible to use the Jordan Schoenflies theorem, generalized to four curves?
 A: Indeed such a homotopy always exists. More strongly, given any finite set of nested Jordan curves, there is a homeomorphism of the plane which sends all of them to concentric circles. Using this result, the homotopy you desire can be found as a simple radial expansion of one circle to the next. To prove our stronger claim, we use a stronger version of the Schoenflies theorem: every positively oriented Jordan curve can be extended into a compact support homeomorphism (homeomorphism which is identity outside of some compact set). This follows as a quick corollary of most of the usual proofs of the Shoenflies theorem.
We can improve this a bit more: suppose $\gamma_1$ and $\gamma_2$ are both positively oriented and interior to the unit circle, then there's a homeomorphism $H$ turning $\gamma_1$ into $\gamma_2$ which is identity exterior to the unit circle. To prove this, apply the strengthened Shoenflies theorem to extend each $\gamma_i$ into the compact support homeomorphisms $H_i$, and then find $H'=H_2\circ H_1^{-1}$, which sends $\gamma_1$ to $\gamma_2$. Now find a sufficiently large $r<1$ such that both $\gamma_1$ and $\gamma_2$ are still interior to the circle with radius $r$. To get the desired homeomorphism $H$, use Alexander's trick so that $H$ agrees with $H'$ interior to the $r$ circle, but $H$ is identity outside the unit circle. Supposing $H'$ is identity outside the circle of radius $R>1$, the construction looks like this.
$$A(p)=\begin{cases}p &: |p|\leq r \\ \frac{p}{|p|}\left(r+(|p|-r)\frac{R-r}{1-r}\right) &: r\leq |p|\end{cases}$$
$$H = A^{-1}\circ H' \circ A$$
Now we have all the tools to prove our primary claim: given pairwise disjoint Jordan curves $\gamma_1,\cdots,\gamma_n$, arranged so that generally $\operatorname{int}(\gamma_k)\subset\operatorname{int}(\gamma_{k+1})$, we can find a homeomorphism sending them all to concentric circles. The base case $n=1$ is handled by the Schoenflies theorem. For the inductive case, apply inductive hypothesis to find a homeomorphism $H_0$ which sends each of $\gamma_2,\cdots,\gamma_n$ to concentric circles; if necessary, compose this with a translation and magnification so that $H_0$ sends $\gamma_2$ to the unit circle. Now let $C$ be a circle centered at the origin and having radius $\frac{1}{2}$. Notice that both $C$ and $\gamma_1$ are positively oriented Jordan curves within the unit circle. Using our earlier result, find a homeomorphism $H_1$ which sends $\gamma_1$ to $C$ but is identity outside the unit circle. Now we can let $H=H_1\circ H_0$. We see that $H$ agrees with $H_0$ exterior to the unit circle while also sending $\gamma_1$ to $C$, which is concentric with the unit circle. It follows that $H$ sends all the curves $\gamma_1,\cdots,\gamma_n$ to concentric circles.

This is a weaker version of a more general result: given any finite set of pairwise disjoint Jordan curves, there's a homeomorphism that turns all of them into circles. In other words, sets of disjoint loops are topologically indistinguishable from sets of disjoint circles. Your question is solved by simple geometry when all your loops are disjoint circles, so by homeomorphism, it holds in generality.
A: By "homotopy from $\gamma_2$ to $\gamma_3$ in $A$" do you mean a homotopy between maps $\gamma_2 : S^1 \to A$ and $\gamma_3 : S^1 \to A$ where each is a homeomorphism onto its image? If so, the answer is "no" but only for trivial reasons.
Using the convention $S^1 = \{z \in \mathbb{C} : \lvert z \rvert = 1\}$, we can define the maps $\gamma_i : S^1 \to \mathbb{R}^2$ by
$$\gamma_1(e^{it}) = (\cos(t), \sin(t))$$
$$\gamma_2(e^{it}) = (-2\cos(t),-2\sin(t))$$
$$\gamma_3(e^{it}) = (3\cos(t),3\sin(t))$$
$$\gamma_4(e^{it}) = (4\cos(t),4\sin(t))$$
The region $A$ in this case is an annulus, so it's clear that $\gamma_2$ and $\gamma_3$ are not homotopic: they represent different elements of $\pi_1(A) \cong \mathbb{Z}$!
Edit: you might also mean some kind of isotopy, e.g. a homotopy $H$ from $\operatorname{id}_A$ to some map $A \to A$ such that $H(t,{-}) : A \to A$ is a homeomorphism for all $t$ – I'm not sure what's possible in this interpretation.

Edit: Assuming $\gamma_2$ and $\gamma_3$ have the same orientation, the answer ought to be yes (I think, I have an unjustified step). First, homotope $\gamma_2$ and $\gamma_3$ inside $A$ to polygons $p_1$ and $p_2$ (respectively) so that they stay in the interior/exterior regions of each other the entire time (I'm not 100% sure this is possible?). Let $A'$ be the region between these polygons. It's clear that $A'$ deformation retracts onto either of its polygonal boundary curves, which induces isomorphisms $f_i : \pi_1(A) \to \pi_1(B_i) \cong \mathbb{Z}$, where $B_1$ is the "interior boundary" and $B_2$ is the "exterior boundary". Since $[p_1]$ generates $\pi_1(B_1)$ and $[p_2]$ generates $\pi_1(B_2)$, $f_1^{-1}([p_1]) = \pm f_1^{-1}([p_2])$ in $\pi_1(A') \cong \mathbb{Z}$. But $p_1$ and $p_2$ are both oriented counterclockwise, so we conclude that their homotopy classes are equal. Now we have a chain of homotopies $\gamma_2 \to p_1 \to p_2 \to \gamma_3$, all of which stay inside $A$.
