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Let $f$ be a bijective elementary function, elementary invertible or not.
Let $h$ be a bijective nonelementary function, elementary invertible or not.
Which of the compositions $h(f(x))$ and $f(h(x))$ can be elementary, and which cannot be elementary?

The elementary functions are defined in differential algebra. That are the functions $X\in\mathbb{C}\to Y\in\mathbb{C}$ that are composed of $\exp$, $\ln$ and/or unary or multiary univalued algebraic functions.

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I answer myself.

The elementary functions are closed regarding composition. That means, the composition of two elementary functions is elementary again.

Let $^{-1}$ denote the compositional inverse.

Let $g$ be an elementary function.

Keep in mind that $f$ and $g$ are elementary.

We formulate the problems as $h(f(x))=g(x)$ and $f(h(x))=g(x)$.

1.)

$$h(f(x))=g(x)$$

Because $h$ is invertible:

$$f(x)=h^{-1}(g(x))$$

If $h^{-1}$ is nonelementary, $h^{-1}(g(x))$ is elementary, and the equation is valid.

If $h^{-1}$ is elementary, $h^{-1}(g(x))$ is elementary, and the equation is valid.

A left inverse of $h$ is sufficient.

2.)

$$f(h(x))=g(x)$$

Because $f$ is invertible:

$$h(x)=f^{-1}(g(x))$$

If $f^{-1}$ is nonelementary, $f^{-1}(g(x))$ is nonelementary, and the equation is valid.

If $f^{-1}$ is elementary, $f^{-1}(g(x))$ is nonelementary only if $g$ is nonelementary.

A left inverse of $f$ is sufficient.
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Functions with a left or right inverse are treated in MathStackexchange: Are elementary compositions of nonelementary functions also nonelementary?

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