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Apart from the well known biholomorphic maps from $D=\{|z|<1\}$ onto itself of the form $f(z)=e^{i \theta}\frac{z-a}{1-\overline{a}z}$ ($|a|<1$, $\theta$ real), are there any holomorphic maps from $D$ onto $D$ with nowhere vanishing derivative?
So, I am looking for a holomorphic $g: D\to D$ such that $g(D)=D$, and $g'(z)\not=0$ for all $z\in D$, and $g$ is not of the form as the $f$ above. If there are any, they cannot be one-to-one, this is all I know.
There do exist such maps. I've been putting off posting this answer because I haven't been able to figure out how to give the proof without relying on a picture. But since there's been an answer posted asserting the contrary it seems I should post this anyway.
If $M$ is a Riemann surface lying "over" the plane let $\pi:M\to\Bbb C$ be the "projection" "down" to the plane. I claim this:
There exists a simply connected Riemann surface $M$ lying over the plane, such that $\pi(M)=O$ is a bounded simply connected open set, but such that $\pi$ is not injective.
If nobody believes that I'll find a way to post a picture. If you think it's obviously not so I suspect that you're consciously or unconsciously assuming that $\pi$ must be a covering map. In the example I have in mind $\pi^{-1}(p)$ has one, two or three points for various $p\in O$.
If you believe that we're done:
Note that $M$ does not have the plane or the sphere for its universal cover. Let $\phi:\Bbb D\to M$ be a conformal equivalence. Let $\psi:O\to\Bbb D$ be a conformal equivalence. Let $g=\psi\circ\pi\circ\phi$.
The argument below is incorrect. Indeed, following the same logic one could "prove" that a surjective entire function $f:\mathbb{C}\to\mathbb{C}$ with nonvanishing derivative must be linear; a claim disproved by the example of $f(z) = \int_0^z \exp(-\zeta^2)\,d\zeta$. This example illustrates the problem with the argument: trying to analytically continue the inverse (e.g., along the positive real axis), one runs out of the domain (into infinity) when approaching $\sqrt{\pi}/2$.
There are no such maps, by the Monodromy theorem. Indeed, $g$ has an inverse $g^{-1}$ in a neighborhood of $0$, which can be analytically continued along any path in $D$ because $g$ has no critical values. And since $D$ is simply-connected, these extensions define a holomorphic map $h:D\to D$ such that $g\circ h$ is the identity map. If the range of $h$ is all of $D$, then $g$ is bijective. Otherwise, $D\cap \partial (h(D))$ is nonempty, and by continuity $|g|=1$ on this set, contradicting the maximum principle.
