I'm working on a Topology assignment, and we were asked to show that the annulus $A$ of inner radius $1$ and outer radius $2$ is homeomorphic to a cylinder $C$ of radius $1$ and height $1$. Now, I already have the two homeomorphisms, but I thought to myself, I really should be able to just invert the first function. The problem is I'm a little more than confused on how to do this if we are considering the function $f:\mathbb{R^3}\rightarrow\mathbb{R^2}$.
Here is the homeomorphism for both directions: $$f:A\rightarrow C\,\,\,\,\text{where}\,\,\,\,f(x,y,z)=(x(z+1),y(z+1))$$ $$g:C\rightarrow A\,\,\,\,\text{where}\,\,\,\,f(x,y)=\bigg(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}},\sqrt{x^2+y^2}-1 \bigg)$$
A quick computation can be done to verify these are inverse functions. How might I go about finding the other mapping once I have one of these continuous functions?
Set $(x(z+1),y(z+1))=(A,B).$ Then, $A/x=z+1=B/y.\ $ Now, $x^2+y^2=1$ so
$A/x=B/\sqrt{1-x^2}\Rightarrow A^2(1-x^2)=B^2x^2\Rightarrow (A^2+B^2)x^2=A^2\Rightarrow x=\pm\frac{A}{\sqrt{A^2+B^2}}.$
Similarly, $y=\pm\frac{B}{\sqrt{A^2+B^2}}$ and then
$z+1=\pm A(\sqrt{A^2+B^2})/A=\pm\sqrt{A^2+B^2}\Rightarrow z=\pm\sqrt{A^2+B^2}-1.$
I'll leave it to you to determine the signs, and invert the other function.