homeomorphism between Ring (Annulus) and Cylinder How to prove that a Ring (Annulus) in $R2$ is homeomorphic to a Cylinder in $R3$?
I understand that I need to find a function that maps every point in the ring to a point on the cylinder.
 A: Note first that it is easy to scale and translate both the ring and the cylinder, so we may choose the ones that fit us best. 
If you consider a closed ring and a finite closed cylinder, you may describe them by 
$$
R=\{(r\cos t, r\sin t),\ 1\leq r\leq 2,\ 0\leq t<2\pi\},\ \ \ \ 
C=\{(\cos t, \sin t, z),\ 0\leq z\leq 1\}.
$$
Then take $\gamma:R\to C$ given by $$\gamma(r\cos t, r\sin t)=(\cos t, \sin t, r-1).$$
To confirm that this is well-defined, this is 
$$
\gamma(x,y)=\left(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}},-1+\sqrt{x^2+y^2}\right)
$$
If the cylinder is infinite, the ring cannot be closed (as it would be compact). So if 
$$
R=\{(r\cos t, r\sin t),\ 1< r< 2,\ 0\leq t<2\pi\},\ \ \ \ 
C=\{(\cos t, \sin t, z),\ z\in\mathbb R\},
$$
you can take $$\gamma(r\cos t,r\sin t)=(\cos t,\sin t, \delta(r)),$$ where $\delta:(1,2)\to\mathbb R$ is a homeomorphism. 
A: If $A = \{z \in \mathbf C : 1<|z| < 2\}$ is your ring, $B = \mathbf S^1\times ]1,2[$ is your cylinder, then you can take
$z\mapsto (z/|z|, |z|)$
It's continous and so is it's inverse.
But try to see it geometrically!
A: Intuitively: if you have a hollow cylinder (like a pipe, or a toilet roll core) and take a picture down the center of it, what you get on that picture is actually an annulus . The only reason you think of it as a cylinder is your brain ticking you into thinking that the image you see is three-dimensional.
Example image from google image search: 
(source: alamy.com) 
