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For a real analytic function $f:(a,b)\to\Bbb R$ we have that if $f'(x)\neq 0$ for all $x$ in the interval $(a,b)$, then $f$ is injective. My question is whether similar holds if $f$ is a holomorphic function $$f:G\to\Bbb C$$ where $G\subseteq\Bbb C$ is a simply connected, open area and $f'(z)\neq 0$.

A simple counter example is $f:z\mapsto z^2$ if $G$ is simply connected such that $-1,1\in G$ but $0\notin G$. Then $f$ is not injective even though $f'(z)\neq 0$ everywhere. So the idea is to fix the preconditions somehow. A second take:

Question: Let $f:G\to S$ be holomorphic with $f'(z)\neq0$ and where $G,S\subseteq\Bbb C$ are open and simply connected areas with $f(G) = S$. Does this imply that $f$ is injective?

It seems to fix the case $f:z\mapsto z^2$ because if $-1,1\in G$, there is some path $\gamma$ that connects $-1$ and $1$. But then $f(\gamma)$ runs around $0$, and because $S=f(G)$ is simply connected, there must be $0\in S$. This means $0\in G$ and thus $f'(z)\neq0$ does not hold and the conjectured theorem does not apply.

emacs drives me nuts
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  • $e^z$ (on $G=\mathbb C$) is a counterexample. – David Jaramillo Jun 09 '22 at 16:20
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    @David Jaramillo: I am not sure because $\exp(\Bbb C)=\Bbb C!\setminus!{0}$ which means the image is not simply connected, as opposed to be required by the theorem. – emacs drives me nuts Jun 09 '22 at 16:29
  • You have a local inverse (at any point of the image). Now analytically continue. – Ted Shifrin Jun 09 '22 at 16:45
  • Oh yes you are completly right (I missed that you wanted the image to be simply connected as well). Your result is true. By riemmann's mapping theorem you can reduce your problem to a map form the upper half plane to itself. The only holomorphic maps there are dilations and translations, both of which are invertible. Which proves the result. – David Jaramillo Jun 09 '22 at 16:54
  • @David Jaramillo: As it appears that doesn't even require that $f'(z)\neq0$? – emacs drives me nuts Jun 09 '22 at 17:08
  • @David the result is not true and there are conformal surjective but non-injective maps between simply connected domains; see https://math.stackexchange.com/questions/185647/conformal-map-from-punctured-disc-to-disc (Step 2 in the answer) – Conrad Jun 09 '22 at 17:10
  • @Conrad: Isn't that question all about a punctured disc? At the end of the aswer they state that $z\mapsto z^2$ is oncformal $\Bbb D\to\Bbb D$, which is only true if the domain is the punctured disc. So it's a typo? (conformal means the derivative does not vanish). My confusion only getr greater... – emacs drives me nuts Jun 09 '22 at 17:26
  • @Conrad I don't think the examples you are using are actual counterexamples. Unless I am missing something obvious. The step two you are quoting doesn't actually construct the function they are claiming is easy to construct. And if you see at the answer they actually end up giving they do something completely different (which is basically the same idea they are using in your second reference). If you think your argument is true it might be worth it to write it in some detail because I and the op (I believe) are not convinced. – David Jaramillo Jun 11 '22 at 17:05
  • This was asked and answered (in the negative) several times in the past. There are even examples where the domain and the range are the entire complex plane. – Moishe Kohan Jun 11 '22 at 18:12

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While there is a negative answer to the OP question in Conformal map from punctured disc to disc, there seems to be some confusion about it so I will expand my comment above a bit. First it is very easy to show that there is $f: D^{*} \to D$ conformal and surjective if and only if there is $g:D\to D$ conformal, surjective and non injective as the latter implies that there is $g(z_1)=g(z_2)$ so $g$ restricted to $\mathbb D-z_1$ is still surjective, while if there is $f$ conformal, surjective and with an isolated singularity at $0$ say, then $\exp \circ f$ maps the left half plane $\Re w <0$ conformally, surjectively and non-injectively to the unit disc and hence pulling that back to the unit disc by a Mobius transform we get our $g$

Now the answer linked above sketches how to construct such conformal, surjective but non-injective map between simply connected domains, but a very easy construction is in https://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=complex;action=display;num=1234362106 the third post for example; later posts in the same forum give more examples and references to literature where such examples were given in print

So the answer to the OP question is negative but non-trivially so. Note that there are tons of surjective holomorphic maps from the unit disc to itself (eg Blaschke products, singular inner functions, etc), so without conformality the result is negative trivially

Conrad
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  • In that ocf answer.3 they use an annulus $A_r = { z : 1/r < |z| < r }$ as domain, no? And you very long sentence will take a time to settle... – emacs drives me nuts Jun 09 '22 at 17:59
  • no, they use its image as domain which is an ellipse, so simply connected; from that ellipse they take some polynomial $F$ and show that the image of the ellipse under $F$ is also simply connected by using the original annulus and function and some algebraic properties ; again the result you ask for is not true, but counterexamples are not trivial, so it may take a while to understand them – Conrad Jun 09 '22 at 19:44
  • I don't think that example actually works. The function $F:\phi(A_r)\to \phi(A_r^3)$ has a vanishing derivative derivative at $\pm 1$ (which are points inside $\phi(A_r)$ for any $r$). In their problem they want a function from a punctured disk to the ball so they simply remove this points and that's how they ensure $F'\neq 0$.. This is basically the example equivalent to the z^2 example the op quoted. Am I missing something? – David Jaramillo Jun 11 '22 at 16:52
  • Well those points are also image of other points so they can be removed - read the example carefully – Conrad Jun 11 '22 at 18:17
  • It is like we are talking about completely different things. I agree the function $F:\phi(A_r)-{\pm 1}\to \phi(A_{r^3})$ is a surjective function with not vanishing derivative anywhere. $\phi(A_r)-{\pm 1}$ is not simply connected. Your idea was to say that $F:\phi(A_r)\to \phi(A_{r^3})$ was a countrexample. it is not because $F'(\pm)=0$. This is the same as with the other example. $z\mapsto z^2$ composed with a möbious translation that takes $0\to 1/2$. This gives a function from the punctured disk to the disk. however if you extended it to include $1/2$, $F'(1/2)=0$. – David Jaramillo Jun 11 '22 at 18:58
  • @David while it is correct that $\phi(A_r) - \pm 1$ is not simple connected, it is topologically a disc minus two points, so its universal holomorphic cover is the unit disc hence there is a surjective conformal $G$ from the unit disc onto it, hence $F\circ G$ is the required conformal surjective but not injective map between two simply connected domains – Conrad Jun 12 '22 at 02:49
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    @David So any conformal surjective map from a bounded but not simply connected domain to a disc gives rise to a counterxample by composing it with the surjective conformal map from the unit disc to that domain that is given by the Uniformization theorem – Conrad Jun 12 '22 at 02:57
  • @Conrad I don't see how the uniformization theorem gives you the map you want. Anyway, Someone else put a link to a post where they answer the question. Will delete my answer, thanks for the comment. – David Jaramillo Jun 12 '22 at 13:42
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    @David the uniformization theorem gives conformal surjective maps from the unit disc to any bounded domain in the plane (highly non injective if the target domain is non simply connected) so any conformal surjection from such to the unit disc (eg from the unit disc minus two points) gives the required map by composition; not sure what is that hard to understand and see that the example linked above works – Conrad Jun 12 '22 at 14:32