How to find an inverse of a multivariable function? Let $ X = \mathbb{R}^2 - \{(0,0)\}$ the punctuated plane.
Let $ Y = \{(x,y,z) \in \mathbb{R}^3 | x^2 + y^2 =1\}$ the circular cylinder.
Prove that they are homeomorphic to each other.
To prove this, just check that the following function is a homeomorphism.
$ f: Y \to X, \quad f(x,y,z)=(xe^z,ye^z)$
My problem is that I don't know how to reverse this kind of functions.
 A: We need a correct homeomorphism:
$$f:\boxed{Y\to X},f(x,y,z)=(xe^z,ye^z)$$
Now given $(a,b)=(xe^z,ye^z)\in X$ first find $z$:
$$\sqrt{a^2+b^2}=e^z\sqrt{x^2+y^2}=e^z\implies z=\log\sqrt{a^2+b^2}$$
Thus
$$f^{-1}:X\to Y,f(a,b)=\left(\frac a{\sqrt{a^2+b^2}},\frac b{\sqrt{a^2+b^2}},\log\sqrt{a^2+b^2}\right)$$
and $f$ and $f^{-1}$ are in fact diffeomorphisms.
A: Geometrically we can thinks  like this: Your function $f$ maps $(x,y,z) \in Y$ to $(x,y)e^z$ in $X$. As $(x,y,z)$ is in $Y$, we know that $x^2+y^2=1$ so $(x,y)$ has norm 1 and therefore the norm of $(x,y)e^z$ come purely from the $e^z$ factor.
We want a inverse of $f$, i.e. $f^{-1}:X \rightarrow Y$. It's looks that have to be something like this: Given $(\tilde{x},\tilde{y}) \in X$ we know that $(\tilde{x},\tilde{y}) \neq (0,0)$ and therefore $\|(\tilde{x},\tilde{y})\|\neq0$, so we can write
$$(\tilde{x},\tilde{y}) = \frac{(\tilde{x},\tilde{y})}{\|(\tilde{x},\tilde{y})\|} \|(\tilde{x},\tilde{y})\|.
$$
Doing that we have $\frac{(\tilde{x},\tilde{y})}{\|(\tilde{x},\tilde{y})\|}$ has norm $1$, we can define $x=\frac{\tilde{x}}{\|(\tilde{x},\tilde{y})\|}$ and $y=\frac{\tilde{y}}{\|(\tilde{x},\tilde{y})\|}$, i.e. $\|(x,y)\|=1$.
On the other hand, we can calculate $z$ from $\|(\tilde{x},\tilde{y})\|=e^z$, i.e. $z=\ln(\|(\tilde{x},\tilde{y})\|)$. From this, if $(\tilde{x},\tilde{y}) \in X$ then:
$$
f^{-1}(\tilde{x},\tilde{y})= \left(\frac{\tilde{x}}{\|(\tilde{x},\tilde{y})\|},\frac{\tilde{y}}{\|(\tilde{x},\tilde{y})\|},\ln(\|(\tilde{x},\tilde{y})\|) \right)
$$
A: I try to answer Anne's observation.
Injectivity:
Applying the definition, of must prove that:
\begin{equation}
f(x_1, y_1, z_1) = f(x_2, y_2, z_2) \Rightarrow (x_1, y_1, z_1) = (x_2, y_2, z_2)
\end{equation}
In this case:
\begin{equation}
(x_1, y_1)\,e^{z_1} = (x_2, y_2)\,e^{z_2} \Rightarrow 
\begin{cases}
(x_1, y_1) = (x_2, y_2) \\
e^{z^1} = e^{z_2}
\end{cases}
\Rightarrow
\begin{cases}
x_1= x_2\\
y_1 = y_2\\
z_1 = z_2
\end{cases}
\end{equation}
Surjectivity:
Applying the definition, of must prove that:
\begin{equation} 
\forall\,(u,v)\in X \quad \exists \, (x,y,x)\in Y : f(x,y,z) = (u,v)
\end{equation}
In this case:
\begin{equation}
\begin{cases}
xe^z = u \\
ye^z = v
\end{cases}
\Rightarrow
\begin{cases}
x= ue^{-z} \\
y= ve^{-z}
\end{cases}
\end{equation}
Now from:
\begin{equation}
(u^2+v^2)\,e^{-2z} = 1 \Rightarrow z=\ln\sqrt{u^2+v^2})
\end{equation}
So, I get:
\begin{equation}
\begin{cases}
x= \frac{u}{\sqrt{u^2+v^2}} \\
y= \frac{v}{\sqrt{u^2+v^2}} \\
z=\ln(\sqrt{u^2+v^2})
\end{cases}
\end{equation}
Finally, I get the inverse:
\begin{equation}
f^{-1}(u,v) = \Big(\frac{u}{\sqrt{u^2+v^2}},\frac{v}{\sqrt{u^2+v^2}}, \ln\sqrt{u^2+v^2}\Big)
\end{equation}
