I have to show that $S_1=\{(x,y,z)\in \mathbb{R}^3\lvert x^2+y^2+z^2=1, z\neq \pm 1\}$ and
$S_2=\{(x,y,z)\in \mathbb{R}^3\lvert x^2+y^2=1, -1<z<1\}$ are diffeomorphic.
If there is a diffeomorphism $f:S_1\rightarrow S_2$ such that $f$ is bijective and $f$ and $f^{-1}$ are $C^{\infty}$ the surfaces $S_1$ and $S_2$ are called diffeomorphic.
I would probably be able to show that the function would be bijective and $C^{\infty}$, but how do I even find that function?
I know that the sphere (without North- and Southpole) can be parameterized by: $$ \phi_n: \mathbb{R^3}\rightarrow S_1, \phi_n(x,y)=(\frac{2x}{1+x^2+y^2},\frac{2y}{1+x^2+y^2},\frac{-1+x^2+y^2}{1+x^2+y^2}) $$
How would I find the diffeomorphism $f:S_1\rightarrow S_2$?
Edit: Changed $\phi_n(x,y,z)$ to $\phi_n(x,y)$, deleted a wrong parameterization for the cylinder.
The maps $$ \begin{array}{r|ccc} F_1\colon & S_1 & \longrightarrow & S_2 \\ & (x,y,z) & \longmapsto & \displaystyle\left(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}},z\right) \end{array}$$ and $$ \begin{array}{r|ccc} F_2\colon & S_2 & \longrightarrow & S_1 \\ & (x,y,z) & \longmapsto & \displaystyle\left(\sqrt{1-z^2}x,\sqrt{1-z^2}y,z\right) \end{array} $$ are smooth (as restriction of smooth maps on an open set of $\Bbb R^3$) and are inverse of each other.
Here is how I found these functions. Fixing $z=c$ gives an horizontal slice for $S_1$ and $S_2$, and it turns out (this is clear if you draw a picture) that $S_1\cap \{z=c\}$ is a circle of radius $\sqrt{x^2+y^2} = \sqrt{1-c^2}$, while $S_2\cap\{z=c\}$ is a circle of radius $1$. You just have to find the right dilation between these two circles.