Is composition of analytic function with invertible function analytic?

Question

The Cayley transform $\psi(z)=i\frac{1-z}{1+z}$ maps the unit disk $\mathbb{D}=\{z\in \mathbb{C}\colon |z|<1\}$ conformally onto the upper half plane $\mathbb{U}=\{z\in \mathbb{C}\colon \textrm{Im}(z)>0\}$. The map $\psi$ is invertible. On the other hand if we consider a function $f\colon \mathbb{U}\to \mathbb{C}$ which is analytic. My question is: Will $f\circ \psi$ be an analytic function?

Answer 1

Edit: in an edit, OP substantially changed the question so my answer below no longer answers the question. To answer the new question: $\psi$ is conformal, so in particular it is holomorphic and analytic. Hence the composition $f\circ \psi$ is a composition of two analytic functions, and therefore again analytic. This however has everything to do with $\psi$ being analytic, and nothing to do with it being invertible.

Original answer:

On $\mathbb{C}$ the map $z\mapsto \bar{z}$ is invertible. If $f\colon\Omega\to\mathbb{C}$ is analytic on $\Omega$ it is holomorphic, but the composition with $z\mapsto \bar{z}$ will not be holomorphic, and hence not analytic.

Answer 2

Take the following invertible function $f_w:\Bbb R\to\Bbb R$ with $w\in\Bbb Q\!\setminus\!\{0\}$:

$$f_w(x) = \begin{cases} x, & x\in\Bbb R\setminus\Bbb Q \\ x+w, & x\in\Bbb Q \end{cases}$$ then $f_1$ is invertible with $f_1^{-1} = f_{-1}$, and $f_1$ is not analytic. Take $g(x)=x$ as analytic function, then neither $f_1\circ g$ nor $g\circ f_1$ are analytic.

This works actually for any such $w$.