# The conjecture about the existence of closed-form inverses of functions

Is my proof draft below already a proof? How can the proof be completed?

Definition:
A unary complex function is a function from a subset of $$\mathbb{C}$$ into $$\mathbb{C}$$.
A binary complex function is a function from a subset of $$\mathbb{C}^2$$ into $$\mathbb{C}$$.
A finitary complex function is a function from a subset of $$\mathbb{C}^n$$ into $$\mathbb{C}$$ wherein $$n\in\mathbb{N}_{\ge 1}$$.

Conjecture:
Let $$k,\kappa\in\mathbb{N}_{\ge 1}$$,
$$F_1,...,F_k$$ and $$\Phi_1,...,\Phi_\kappa$$ unary transcendental or finitary algebraic complex functions,
$$F$$ the set of all functions that are generated by applying finite numbers of $$F_1,...,F_k$$,
$$\Phi$$ the set of all functions that are generated by applying finite numbers of $$\Phi_1,...,\Phi_\kappa$$,
and let
$$n\in\{1,2\}$$,
$$A$$ a unary or binary algebraic complex function,
$$f_1,f_2$$ unary transcendental complex functions
so that
$$f\colon z\mapsto A(f_1(z),...,f_n(z))$$ is a bijective complex function.
Let $$f^{-1}$$ be the inverse of $$f$$.
If $$f\in F$$ and $$f^{-1}\in \Phi$$, there exist $$m$$ functions $$f_1,...,f_m\in F$$ so that $$f=f_m\circ\ ...\circ\ f_1$$.

Proof draft:
Let $$^{-1}$$ denote the respective inverse.
If $$n=1$$, the assumption of the conjecture is obviously.
Assume $$n=2$$.
a) Assume $$f_1,f_2$$ are algebraically dependent.
Because $$f_1,f_2$$ are algebraically dependent, there is an algebraic function $$A_1$$ so that $$f_2(z)=A_1(f_1(z))$$ for all $$z\in \text{dom}(f)$$. Therefore there is an algebraic function $$A_2$$ so that $$f(z)=A_2(f_1(z))$$ for all $$z\in \text{dom}(f)$$, and the assumption of the conjecture follows.
b) Assume $$f_1,f_2$$ are algebraically independent.
Let's draw how the functions are applied for generating $$f$$:

$$f\colon\ \ \ z\ \ {^{\nearrow\ f_1(z)}_{\searrow\ f_2(z)}}^{\searrow}_{\nearrow}\ \ A(f_1(z),f_2(z))$$

Because $$f$$ is bijective, $$A$$ is bijective, and therefore $$A^{-1}$$ exists. Because $$A$$ is binary, $$A^{-1}$$ is a function into $$\mathbb{C}^2$$.
Let's draw how the inverse of $$f$$ is generated through that:

$$f^{-1}\colon\ \ \ A(f_1(z),f_2(z))\ \ {^{\nearrow\ f_1(z)}_{\searrow\ f_2(z)}}^{\searrow}_{\nearrow}\ \ z$$

We see from the figures: If $$f$$ is represented as above, $$A^{-1}$$, a function into $$\mathbb{C}^2$$, is needed for representing $$f^{-1}$$.
But $$\Phi$$ doesn't contain a function into $$\mathbb{C}^2$$. Therefore $$n\neq 2$$.
Because $$A$$ is unary or binary according to the preconditions, $$A$$ is unary, and the assumption of the conjecture follows.
That proofs the conjecture.
q.e.d.

Clearly, the conjecture and its proof can easily be generalized to

• finitary $$A$$,
• functions contained in $$F$$ with representations containing finitary $$A$$,
• $$f_1,f_2$$ finitary complex functions,
• arbitrary fields $$\mathbb{K}$$ instead of the field $$\mathbb{C}$$,
• arbitrary sets $$S$$ with arbitrary finitary operations into $$S$$ instead of the field $$\mathbb{C}$$.