# Conformal map of the unit disk onto itself which is not 1 to 1

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 $g=\psi\circ\pi\circ\phi$.

• I think adding an image would help people seeing how it works. – Daniel Fischer Aug 23 '15 at 21:01
• @DanielFischer You're definitely right about that. But I have to draw it somehow... – David C. Ullrich Aug 23 '15 at 21:03
• Yep. Rather you than me ;) – Daniel Fischer Aug 23 '15 at 21:05
• @DanielFischer LLOL. (Literally laughing out loud...) – David C. Ullrich Aug 23 '15 at 21:05

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.

• Seems to me there are such maps, haven't been able to come up with a good explanation without drawing pictures. Regarding what you wrote, I don't get it. Yes, $g$ is injective in some disk $D(0,r)$. So the restriction of $g$ to that disk has an inverse $h_0$. But the domain of $h_0$ is not $D(0,r)$,, so I don't follow the part about continuing $h_0$ along any curve. (Hmm. If $\gamma$ is a curve starting at $0$ then maybe we can continue $h_0$ along the curve $g\circ \gamma$. But curves of that form do not give "any" curve...) ??? – David C. Ullrich Aug 23 '15 at 19:46
• Ok, we want to talk about Riemann surfaces lying over the disk great. Why does no branch points imply it splits into disjoint sheets? That's exactly what seems somewhat not so to me. The point is that the multivalued $g^{-1}$ need not be a covering map! In the example I have in mind there are points of the disk lying below just one point of the surface, but which are limits of point of the disk lying below two points of the surface... – David C. Ullrich Aug 23 '15 at 19:56
• Hmm, your comment seems to have disappeared. Presumably you deleted it. Possibly if you think about why it needed to be deleted that will lead you to the example I have in mind... – David C. Ullrich Aug 23 '15 at 19:59
• Consider the answer I just posted - explains what I have in mind at least somewhat. – David C. Ullrich Aug 23 '15 at 20:41
• Indeed, it need not be the case that $g^{-1}$ can be analytically continued along each curve, the continuation "$g^{-1}\circ \gamma$" may reach the boundary before $\gamma$ ends. – Daniel Fischer Aug 23 '15 at 20:58