Annulus Theorem I'm trying to read Rolfsen's "Knots and Links" and I'm a little discouraged that I can't do one of the first and seemingly more important exercises.  The question is 

Use the Schoenflies theorem (every topological imbedding $S^1 \rightarrow \mathbb{R}^2$ is the boundary of a 2-disk) to prove the annulus theorem: Given two disjoint imbeddings of $S^1$ in $\mathbb{R}^2$, where one is "inside" the other (I.e. one is in the bounded component of the complement of the other), then the space "between" the two curves is homemorphic to $S^1 \times [0,1]$.  

There is a hint that says "connect the two curves with arcs".  Unfortunately I can't say much about "what I've tried." At this point in the book, we know only that all simple closed curves in $\mathbb{R}^2$ or $S^2$ are "equivalent" in the sense that there exists a homeomorphism of the ambient space taking one to the other.  This allows us to assume one of the curves is the standardly imbedded $S^1$. After that I'm totally unsure of what to do. I would appreciate any hints or advice.  Thank you for reading my question. 
 A: I haven't thought this through carefully, but just following the hint:
choose a point $a$ on the outer cirlce and a point $b$ on the inner circle.  Join $a$ and $b$ with an arc that doesn't cross the two circles.  Now "thicken up" this arc very slightly (and thicken up each of $a$ and $b$ to a very short piece of circular arc on their respective circles) so that now we have to the two circles with a very thing rectangle sitting between them.  (I think that making this "thickened" picture rigorous is probably part of what is involved in translating the hint into a full argument.)
Now the consider the boundary of the union of the two circles and this rectangle.  If you are picturing what I want you to picture, you will see that it is a circle.  (Just to be sure: if you take an annulus and snip out a small rectangle joining the inner and outer annulus, you get a kind of "sliced" annulus which is homeomorphic to a disk with circle as boundary.  But you can verify that the boundary of the region I am describing is a circle without knowing that the region inbetween the two circles is an annulus just follow the outer circle around from the top left corner of the rectangle all the way around to the top right corner, then move along the right boundary of the rectangle to its bottom right corner, now go back around the inner circle to the bottom left corner of the rectangle, and now go back up the left edge of the rectangle to get back where you started: you've described a circle.)  So it bounds a disk.
So the region between the two circles, with a thin rectangle removed, is a disk.  Gluing back in the rectangle should make into an annulus, as required.
