It seems that i've forgotten my highschool math. Think the case that I have a bijective function, so the inverse is a function and i'm trying to find this inverse function as an Expression.
The problem is, this function has two parameters. i.e: $z = f(x,y)$, $ f:\mathbb{R} \times \mathbb{R} \to \mathbb{R}$.
How can i have inverse of this function which will give me
$$x = g(z)$$
$$y = h(z)$$
Thanks
There exist bijective functions $f$, i.e., functions having an inverse, from two-dimensional domains to one-dimensional domains, e.g. bijective maps $f:\ [0,1]^2\to[0,1]$, but they are not easy to construct and certainly cannot be given as simple "expressions", nor can their inverses.
A "typical" bijective map is between domains of the same dimension. In dimension $1$ an example is the map $$\sin:\quad\bigl]-{\pi\over2},{\pi\over2}\bigr[\ \to\ ]{-1},1[\ ,\qquad t\mapsto\sin t\ .$$ In any dimension $n\geq 1$ one may think of regular linear maps. Such a map is defined by an $(n\times n)$-matrix $A=[a_{ik}]$ of determinant $\ne0$ and appears as $$(x_1,\ldots, x_n)\mapsto (y_1,\ldots, y_n),\qquad y_i:=\sum_{k=1}^n a_{ik} x_k\quad(1\leq i\leq n)\ .$$ Its inverse has the same explicit form with the matrix $A^{-1}$ in place of $A$.
In dimension $2$ a vast set of examples is given by conformal mappings. Their description is greatly simplified by introducing the complex coordinate $z:=x+iy$ in place of the real coordinates $(x,y)$. An example: The map $$f:\quad z\mapsto w:=e^z:=\sum_{k=0}^\infty{1\over k!}\>z^k\ ,$$ restricted to the rectangle $R:=\bigl]-a,a[\ \times\ \bigl]-{\pi\over2},{\pi\over2}\bigr[$ in the $z$-plane maps this rectangle bijectively onto the annular sector $R':=\bigl\{w\>\bigm|\> e^{-a}< |w|< e^a, \ {\rm Re}(w)>0\bigr\}$. In this example the inverse function can be given by simple expressions in terms of elementary functions, namely $$f^{-1}(w)=\log|w| + i{\rm Arg}(w)\ ,$$ where ${\rm Arg}(w)$ denotes the polar angle of $w$.
In the general case the inverse of a bijective map $f:\ {\mathbb R}^n\to{\mathbb R}^n$ given in terms of "elementary expressions" cannot be expected to be expressible in a simple way.
You can consider the digit expansion for example but in $\Bbb R$, it's not trivial to formalise since $0.\overline{9}=1$. So I'll take integers but the idea remains the same.
$f:\begin{array}{l}\Bbb N \times \Bbb N \to \Bbb N\\(x,y)=\left(\sum\limits_{k=0}^{+\infty}x_k10^k,\sum\limits_{k=0}^{+\infty}y_k10^k\right)\mapsto \sum\limits_{k=0}^{+\infty}x_k10^{2k}+y_k10^{2k+1}\end{array}$
The idea is that you take all the digits and then put one from $x$, then one from $y$ (and you fill the blanks with $0$s) untill you have used all the digits of both numbers. So you easily get that
$f^{-1}:\begin{array}{l} \Bbb N \to \Bbb N \times \Bbb N\\x=\sum\limits_{k=0}^{+\infty}x_k10^k\mapsto \left(\sum\limits_{k=0}^{+\infty}x_{2k}10^k,\sum\limits_{k=0}^{+\infty}x_{2k+1k}10^k\right)\end{array}$
Regarding your comment:
You have $f_1:S \times T \to S$ and $f_2:S \times T \to A$ and you want $f_3:S \to A$ so that $f_2 = f_3 \circ f_1$ or $f_4:T \times S$ so that $f_2=(s,t)\mapsto f_4(t,f_1(s,t))$.
A sufficient condition for $f_3$ to exist would be $f_1$ injective. If you think of your machine as a graph, that would mean you never have two arrows pointing to the same state. This condition is not necessary because if $f_2$ is a constant, we can take $f_3=f_2$ even if $f_1$ isn't injective.
An necessary and sufficient condition for $f_3$ to exist is $\forall (x,y)\in (S \times T)^2,f_1(x)=f_1(y)\implies f_2(x)=f_2(y)$. In the previous sufficient but not necessary condition we used $\forall (x,y)\in (S \times T)^2,f_1(x)=f_1(y)\implies x = y \implies f_2(x)=f_2(y)$ which implies our necessary and sufficient (that is, equivalent) condition by transitivity of $\implies$.
The existance if basically the same thing except that we replace injectivity on $S \times T$, by injectivity with respect to the first variable on $S$ for all $t\in T$ which is a strictly weaker proposition.
