Inverse function of a composition with a multiary or multivalued function and dependent arguments?

Question

$A$, $F$, $f_1$, $f_2$, $G$: functions with complex arguments and complex components of their values

$A$: an algebraic function

$z$: complex variable

$F(z)=A(f_1(z),f_2(z))=A(G(z))$

$f_1$ and $f_2$ are algebraic independent

Of wich kind are the functions $A$ and $G$? Are they 2-ary functions, or 2-valued functions?

What are the domains and codomains of $A$ and $G$?

Are there $A$, $G$ or $F$ possible which have inverse functions?

Is there any $F$ that can be represented as a composition of only unary univalued functions?

Is there an inverse function of any $F$ that can be represented as a composition of only unary univalued functions?

I have no problem with this composition if the functions $f_1$ and $f_2$ have arguments which are different independent variables. But now both functions are in dependence because their arguments are the same variable $z$.

Let us take an example:

$f_1(z)=z$, $f_2(z)=e^z$

$F(z)=A(z,e^z)=G(z)$

$A(z_1,z_2)=z_1+z_2$

$F(z)=z+e^z$

The functions $f_1$ and $f_2$ are algebraically independent. Therefore $A(z,e^z)$ cannot be reduced to $A_1(z)$, where $A_1$ is another algebraic function.

Let us restrict $z$ and $F(z)$ to the reals. $F$ is bijective then. (The inverse of $F$ is $F^{-1}$ with $F^{-1}(z)=z-LambertW(e^z)$.)

But what are $A$ and $G$ and their inverses?

Answer 1

You had given $F(z)=A(f1(z),f2(z))=A(G(z))$.

$A$ is a function with $A\colon (z_1,z_2)\mapsto A(z_1,z_2)$. It is a tuple-argumented function.

$G$ is therefore a function with $G\colon z\mapsto G(z)=(f_1(z),f_2(z))$. It is a tuple-valued function.

The inverse relation of $A$ is $A^{-1}$ with $A^{-1}\colon A(z_1,z_2)\mapsto (z_1,z_2)$. It is a tuple-valued function.

The inverse relation of $G$ is $G^{-1}$ with $G^{-1}\colon (z_1,z_2)\mapsto z$. It is a tuple-argumented function.

The inverse relation $F^{-1}$ of $F$ is $F^{-1}=G^{-1}\circ A^{-1}$. $F$ is a function if we define $G^{-1}$ as a function. A function can be only univalued. If we define $A^{-1}$ and $G^{-1}$ as unary functions, $F^{-1}$ is a composition of only unary univalued functions for any $F$.