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I say that a function $f : \mathbb{R} \to \mathbb{R}$ is algebraic if it is a solution of a polynomial equation, that is there exists a polynomial $F(x,y)$ such that $F(x,f(x)) =0$ for every $x$. I say that a function is transcendental if it is not algebraic. This should be consistent with the standard terminology.

Now, is it true that if I precompose a transcendental function with a non-constant algebraic function, the result is transcendental? If not true, is something similar true?

In symbols, given $g : \mathbb{R} \to \mathbb{R}$ transcendental, $f : \mathbb{R} \to \mathbb{R}$ algebraic and non-constant, is $g \circ f$ transcendental?

If needed/helpful, we can put some hypothesis on $g$; for instance, it can be taken to be $C^\infty$, even analytic everywhere except a finite set of points. Also, maybe it helps if the domains are not the whole $\mathbb{R}$: I would expect everything works in the same way if instead of $\mathbb{R}$ one takes some connected interval (or maybe some connected open set in $\mathbb{C}$?).

(I am not sure what kind of tags to put on this questions -- suggestions are welcome)

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    $\begingroup$ For $C^\infty$ the answer is false: take $f(x) = x^2$, then you can take $g$ algebraic on $[0, \infty)$ and whatever you want on $(-\infty, -1]$ (and something smooth to connect). For analytic $g$, everything can be extended to almost $\mathbb C$, and then, as $(g \circ f)\circ f^{-1} = g$ is transecdental, $g \circ f$ can't be analytic (not sure if we can hand-wave around multi-value functions here, but probably can). $\endgroup$
    – mihaild
    Jun 7, 2022 at 16:33
  • $\begingroup$ Of course! I am happy to consider a local inverse of $f$, which is indeed algebraic where it is defined. $\endgroup$
    – fulges
    Jun 7, 2022 at 20:12
  • $\begingroup$ Maybe you are asking whether "algebraic" can be a "local" property? $\endgroup$ Jun 10, 2022 at 23:09

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Is it true that the result is transcendental if I precompose a transcendental function with a non-constant algebraic function?

Yes, it is true.

Statement:
Let
$X,Y\subseteq\mathbb{R}$ (or $\mathbb{C}$),
$A\colon X\to Y$ a bijective algebraic function,
$T\colon Y\to\mathbb{C}$ a transcendental function.
Then the function $F=T\circ A$ is transcendental.

Proof:
A function in $\mathbb{R}$ is a function $D\subseteq\mathbb{R}\to\mathbb{R}$.
A function in $\mathbb{C}$ is a function $D\subseteq\mathbb{C}\to\mathbb{C}$.
Let $^{-1}$ denote the inverse function.
Assume $T\circ A$ is algebraic. Then there is an algebraic function $A_1$ of one variable so that $\forall x\in X\colon\ T(A(x))=A_1(x)$.
$$\text{for all}\ y\in Y:$$ $$T(A(A^{-1}(y)))=A_1(A^{-1}(y))$$ $$T(y)=A_1(A^{-1}(y))\tag{1}$$ Because the bijective algebraic functions in $\mathbb{R}$ (or $\mathbb{C}$) are closed under inversion and under composition, $A_1\circ A^{-1}$ is an algebraic function. But $T$ is a transcendental function. So (1) is a contradiction, and therefore the assumption is wrong. This proves the statement.
q.e.d.

If $A$ is not injective, you can decompose it into different pieces of injective functions (e.g. the inverses of the partial inverses of $A$) and apply the statement.

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