If $\varphi(x,y)=(x,F(x,y))$ invertible, why it's inverse is of the form $\psi(u,v)=(u, G(u,v))$?

Question

Let $\varphi :\mathbb R^{N}\times \mathbb R^M\to \mathbb R^N\times \mathbb R^M$ a bijective function defined as $$\varphi(x,y)=(x,F(x,y)),$$

where $F:\mathbb R^N\times \mathbb R^M\to \mathbb R^M$. Denote $\psi=\varphi^{-1}$.

Why is there a function $G:\mathbb R^N\times \mathbb R^M\to \mathbb R^M$ s.t. $$\psi(u,v)=(u,G(u,v)) \ \ ?$$

I denote $\varphi=(\varphi_1,\dots,\varphi_N,\dots,\varphi_{N+M})$. Since $\varphi_i(x,y)=x_i$ for all $i=1,\dots,N$ it's clear that $\psi_i(u,v)=u_i$ (since it's the identity), but if $i>N$ we indeed have that $\varphi_i(x,y)=F_i(x,y)$, but why will we have a function $G$ s.t. $\psi_i(u,v)=G_i(u,v)$? Is $G_i$, the inverse function of $F_i$? If yes, I really don't see why.

Answer 1

First off, constructing $G$ is simple, if we have $\psi$: just take $G(u,v) = \pi_2 \circ \psi(u,v)$, where $\pi_2$ is the projection to the second set of coordinates (in your case, the last $M$).

To answer your second question $G_i$ is not the inverse of $F_i$: we have that $F_i, G_i : \mathbb{R}^N \times \mathbb{R}^M \to \mathbb{R}$, so the domain of $G_i$ is not equal to the range of $F_i$.

Answer 2

If $\psi = (\psi_1,\dots,\psi_N,\psi_{N+1},\dots,\psi_{N+M})$, then each $\psi_i$ is a function of the coordinates $(u,v)$ where $u \in \mathbb{R}^N$ and $v \in \mathbb{R}^M$. Simply denote the function $\psi_{N+i} : \mathbb{R}^N \times \mathbb{R}^M \to \mathbb{R}$ by $G_i$. Then, we can define a function $G : \mathbb{R}^N \times \mathbb{R}^M \to \mathbb{R}^M$ by $G(u,v) = (G_1(u,v),\dots,G_M(u,v))$.

You have already observed that if $\varphi_i = x_i$ for $i = 1,\dots,N$, then $\psi_i = u_i$ for $i = 1,\dots,N$. So, the inverse function $\psi$ is of the form $\psi(u,v) = (u_1,\dots,u_n,G_1(u,v),\dots,G_M(u,v)) = (u,G(u,v))$.

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Basically, each of the coordinate functions of $\psi$ are some arbitrary functions of the coordinates $(u,v) \in \mathbb{R}^N \times \mathbb{R}^M$. However, we can explicitly write down what the first $N$ coordinates functions are because we know exactly what the first $N$ coordinate functions of $\varphi$ are. The remaining $M$ coordinate functions are arbitrary, so call them $G_i$, and club them together as the coordinate functions of a map $G: \mathbb{R}^N \times \mathbb{R}^M \to \mathbb{R}^M$.

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If the above is clear, then it should answer your second question as well. $G_i$ is not the inverse function of $F_i$ because $F_i$ and $G_i$ are both maps from $\mathbb{R}^N \times \mathbb{R}^M$ to $\mathbb{R}$, so the compositions $F_i \circ G_i$ and $G_i \circ F_i$ are both not compatible, and so cannot even be defined.

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Additionally, one small typo in your question details. Instead of

but if $i>N$ we indeed have that $\varphi_i(x,y)=F_i(x,y)$. . .

it should be either

if $i > N$ we indeed have that $\varphi_i(x,y) = F_{i-N}(x,y)$. . .

or

if $j = i + N$ for $i > N$ we indeed have that $\varphi_j(x,y) = F_i(x,y)$. . .

This is just some minor bookkeeping regarding the indices of the coordinate functions of $\varphi$ and $F$.

Answer 3

Just see what happen with $\varphi:\mathbb R^2\longrightarrow \mathbb R^2$

$$\varphi(x,y)=(\varphi_1(x,y),\varphi_2(x,y))=(x,\varphi_2(x,y)).$$ Set $$\psi(u,v)=(\psi_1(u,v),\psi_2(u,v))$$ that exist. You know that $$(u,v)=\varphi(\psi(u,v))=(\psi_1(u,v),K(u,v))$$ where $K(u,v)=F(\psi_1(u,v),\psi_2(u,v))$. In particular, $\psi_1(u,v)=u$ and thus, if you set $G(u,v)=\psi_2(u,v) you get $$\psi(u,v)=(u,G(u,v))$$ as expected.