How to find an inverse of a multivariable function?

Question

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.

Answer 1

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.

Answer 2

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:

$$ f^{-1}(\tilde{x},\tilde{y})= \left(\frac{\tilde{x}}{\|(\tilde{x},\tilde{y})\|},\frac{\tilde{y}}{\|(\tilde{x},\tilde{y})\|},\ln(\|(\tilde{x},\tilde{y})\|) \right) $$

Answer 3

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}