Region between two Jordan curves where one is in the interior of the other Let $C_1,C_2$ be two simple closed curves in the plane (ie Jordan curves) such that $C_1$ lies in the interior of $C_2$. Is it true that the region between the two curves, (i.e., $C_1 \cup C_2 \cup(I_2\cap E_1)$, where $I_2$ is the interior of $C_2$, $E_1$ is the exterior of $C_1$) is homeomorphic to the region between two cocentric circles? Pictorially: 
I am aware of Schoenflies theorem, but I don't know how to generalize it to deal with two curves.
 A: Here is one proof.
Step 1. Prove the same theorem in the PL setting, i.e. if $C_1, C_2$ are simple disjoint piecewise-linear curves in $R^2$, one contained in the region bounded by the other, than the (closed) region $A$ between the two curves is homeomorphic to the annulus.  This is an appication of the classification of surfaces with boundary: You compute the Euler characteristic of $A$ and show that it equals zero. Then argue that every compact connected (orientable) surface with two boundary components and of zero  Euler characteristic, is homeomorphic to the annulus.
Step 2 (the hard one): Quote Theorem 13 from Chapter 10 in
Moise, Edwin E., Geometric topology in dimensions 2 and 3, Graduate Texts in Mathematics. 47. New York - Heidelberg - Berlin: Springer-Verlag. X, 262 p. (1977). ZBL0349.57001.
In this theorem he proves that if $K\subset R^2$ is a compact homeomorphic to a simplicial complex, then there is a homeomorphism $R^2\to R^2$ which sends $K$ to a subcomplex of $R^2$. Now, apply this theorem to $K=C_1\cup C_2$. Lastly, appeal to Step 1.
