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The fragment below is from the book 'Theory of functions of a complex variable', by Carathéodory.

The situation is this. We have an analytic function $f(z)$ in some region $G_z\subset\mathbb{C}$. Then $G_w=f(G_z)$ is a region in the $w$-plane. We assume that $G_w^*$ is a simply-connected subregion of $G_w$ that does not contain points $w=f(z)$ such that $f'(z)=0$. We know, by the inverse function theorem, that $f$ has a local inverse around such points. But then Carathéodory claims (highlighted below) that we can start at any such point $w_0$ and continue analytically such a local inverse along any "polygonal train" -- (i.e any polygonal path) starting from $w_0=f(z_0)$ (as long as we stay inside $G_w^*$). He then deduces -- by the monodromy theorem -- that $f$ has a single-valued analytic inverse defined on the whole of $G_w^*$.

Carathédory's argument says this:

If $f$ is an analytic function defined on a region $\Omega$ such that $f'(z)\neq 0$ for every $z\in\Omega$, then $f$ has a single-valued analytic inverse defined in any simply connected subregion of $f(\Omega)$.

As far as I know, there are examples of functions defined in simply connected regions that cannot be analytically continued.

So my question is this:

-- Is the statement above true? How does Carathédory know that we can extend a local inverse analytically along every polygonal path that lies in $G_w^*$?

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Added In order to clarify the question, here is the final comment Carathéodory makes:

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uniquesolution
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  • what is your statement of the monodromy theorem ? – mercio Jan 19 '21 at 19:52
  • @mercio "Let $z_0$ be a point of a simply-connected region $G$ and let $f_0(z)$ be an analytic function defined in a neighborhood $D\subset G$ of $z_0$. If $f_0(z)$ can be continued analytically along any path emanating from $z_0$ and contained in $G$, then there exists an analytic function $g(z)$ defined on $G$ such that $g(z)=f_0(z)$ for all $z\in D$" – uniquesolution Jan 19 '21 at 22:59
  • If $U$ is closed, $f:U\to f(U)=V$ analytic (on an open containing $U$), $f'$ doesn't vanish on $U$, and $f(\partial U)\subset \partial V$, then $f^{-1}$ is analytic on any simply connected domain $\subset V$. This is because the maximum modulus gives that $f'$ is bounded below, when continuing $f^{-1}$ along a curve $\gamma(t),t\in [0,1]$ we can look at the least $T$ where $f^{-1}$ is not analytic at $\gamma(T)$, then $f^{-1}(\gamma(T))$ is in $U$ thus $f^{-1}$ is in fact analytic at $\gamma(T)$. – reuns Jan 21 '21 at 03:20
  • This is actually a well-know issue in algebraic topology: A surjective local homeomorphism (even between 2-dimensional manifolds) need not be a covering map. What one additionally needs for a covering map is the "path lifting property" or properness. – Moishe Kohan Jan 26 '21 at 02:28

1 Answers1

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This is not true without additional hypotesis. If it were true, it would imply Caratheodory's claim about the existence of a single valued inverse, which is false in general:

Let $f(z)=z^2(z-1)$ on $\mathbb{C}$. This map is surjective, and it has only two singular points: $0$ and $\frac23$. However, if we remove this values from the domain, we get a surjective map from $\mathbb{C}-\{0,\frac23\}$ to $\mathbb{C}$ which is non injective. Using the fact that $\mathbb{C}-\{0,\frac23\}$ has $\mathbb{D}$ as its universal cover, we get a map $\varphi:\mathbb{D}\twoheadrightarrow \mathbb{C}$ such that $\varphi'\neq 0$, $\varphi$ is surjective and not injective: thus $\varphi$ does not admit a single valued inverse on $G_w^*=Gw=\mathbb{C}$ as such a function should be entire and bounded and thus by Liouville's theorem, constant.

One can also construct $\varphi:\mathbb{C}\twoheadrightarrow \mathbb{C}$ such that $\varphi'\neq 0$, $\varphi$ is surjective and non linear (and thus non injective, as the only injective entire functions are linear): see this mathoverflow post for such an example (the error function is an explicit example).

$\varphi$ does not admit a single valued inverse on $G_w^*=G'w=\mathbb{C}$, as such a function should be entire and it should avoid more than one point (since for almost every $w, \varphi^{-1}(w)$ is infinite) and thus by Picard's little theorem it should be constant.

If, however, we let $f$ satisfy a few more assumptions, the result follows. For example, if $f:G_z\to G_w^*$ is proper, it is a well known result that $f$ is actually a covering map. As $G_w^*$ is simply connected, it's its own universal cover, and thus the holomorphic lifting lemma implies the existence of an inverse and thus in particular the possibility of analytic continuation along polygonal paths.

  • what simply connected region not containing a critical value $G_w^*$ do you pick for your counter-example claim ? – mercio Jan 19 '21 at 19:52
  • @mercio $\mathbb{C}$ itself. If one wishes to restrict oneself to a proper subregion, I think that the plane minus a ray should work –  Jan 19 '21 at 20:32
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    A surjective and non-injective function is not a counter-example. Take $f(z)=e^z$, for example. Carathéodory does not claim that this function has a global inverse. He claims that you can find a simply-connected domain in its range in which you can define a single valued inverse, which is of course true, as the principal branch of the Logarithm shows. The question is about the analytic continuation, not about the existence of a global inverse. – uniquesolution Jan 19 '21 at 22:43
  • @uniquesolution I edited the answer to explain why such a single valued inverse does not exists in either of the examples –  Jan 19 '21 at 23:46
  • Thank you for your comments. I have elaborated a bit above to explain why this does not answer the question. The function being proper or not is irrelevant, as we are dealing with possibly unbounded domains. As for the inverse being entire, this is also not in the context of the question. See Caratheodory's final comment, which I've added just now. – uniquesolution Jan 20 '21 at 06:51
  • @uniquesolution There is no such a single valued inverse on $G_w^*$ in general, as my answer explains (contradicting Caratheodory's claim, which is why I do not see how this does not answer the question). Since the existence of a continuation along polygonal paths would imply the existence of such an inverse through the monodromy theorem, that too is false in general. Such a function exists if we add the properness hypothesis (to which the boundedness of the domain is irrelevant), e.g. for complex polynomials. –  Jan 20 '21 at 07:06
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    @uniquesolution The fact that the function is entire is in the context of the question: if you take $G_w^*=\mathbb{C}$ you get a counterexample to Caratheodory's claim, since there $\varphi$ does not have a single valued inverse –  Jan 20 '21 at 07:24
  • @Ceffeine Please read carefully Carathéodory's text. The region $G_w^$ is a simply connected subregion of $G_w$ that does not contain any of the points... Therefore your example is not a counter-example. You cannot take $\mathbb{C}$ as $G_w^$. – uniquesolution Jan 20 '21 at 09:07
  • @uniquesolution Yes it is. There are no such points $z_v$, since $\varphi'$ is never zero, and thus there are no $w_v$ to avoid. –  Jan 20 '21 at 09:32
  • @Caffeine , Yes, you are right, and apparently Carathéodory was wrong. – uniquesolution Jan 20 '21 at 09:39