Show two spaces are homeomorphic

Let $$S^1$$ be the unit circle in $$\Bbb R^2$$, with the subspace topology. Let $$X \subset \Bbb R^3$$ be given by $$S^1 × [0, 1]$$, and $$Y \subset \Bbb R^2$$ be $$\{(x, y)|1 \leq x^2 + y^2 \leq 2\}$$. Show that $$X$$ and $$Y$$ are homeomorphic.

I know that to show that they are homeomoprhic, I must find a function from $$X$$ to $$Y$$ which is bijective and continuous and one inverse function that is continuous, but I don't know where to start the proof or find a function. How should I approach the problem?

Edit: Okay, so this is how they are look, however I feel like there's a problem because the empty part of the ring. I can't just make z=0 and map cylinder's insides to start from the circle with the radius 1 or can I? • Have you drawn a picture of each set? Can you describe them informally (something like "One is like an eggshell with three holes drilled in it..." would be fine, although that's not a correct description of either shape in this example). You can edit your question by clicking "edit" beneath the question itself, and fill in your thoughts on this. Once you do this, there's another hint: polar coordinates may help. – John Hughes Feb 24 at 11:48
• I edited with the a picture and my thought. It feels wrong squishing the cylinder to a 2d ring with a hole. Can we do that because there are basically infinitely many points? – 70pr4k Feb 24 at 12:02
• Nice work! It's not clear from the left-hand picture, but the cylinder you get is "hollow". So imagine grabbing the top and stretching until it looks like a funnel....and then press downward to the $xy$-plane to get the annulus. It turns out that @JoseCarlosSantos's answer is exactly a formula for doing that. – John Hughes Feb 24 at 12:06
• Ahhh that makes sense now, since unit circle doesnt consist the points inside! Thank you! – 70pr4k Feb 24 at 12:10

The set $$X$$ is a cylinder and the set $$Y$$ is an annulus. Simply flatten the cylinder:$$\begin{array}{ccc}X&\longrightarrow&Y\\(x,y,z)&\mapsto&(z+1)\cdot(x,y)=((z+1)x, (z+1)y)\end{array}$$
• Its inverse is$$(x,y)\mapsto\left(\frac x{\sqrt{x^2+y^2}},\frac y{\sqrt{x^2+y^2}},\sqrt{x^2+y^2}-1\right).$$ – José Carlos Santos Feb 25 at 16:09