I am currently working on the following problem and have run into a bit of trouble:

Consider an entire function $f$ s.t. $\overline{B_1(0)}\subset f(\mathbb{C}).$ Show that V, a component of $f^{-1}(B_1(0)),$ is simply connected.

I figure the best way to achieve this result was by showing that any two curves with the same endpoints are homotopic. So I considered two parametrized closed curves $\gamma(t)$ and $\kappa(t)$ where $t\in[0,1]$ s.t. $$\gamma(0)=a=\kappa(0) \ \textrm{ and } \ \gamma(1)=b=\kappa(1).$$ Now $f(\gamma(t))$ and $f(\kappa(t))$ are parameterizations of closed curves joining the points $f(a)$ and $f(b).$ Now we can deform these to one another via the family of functions $$d_s(t)=(1-s)f(\gamma(t))+sf(\kappa(t)), \textrm{ where} \ s\in[0,1].$$

The problem is from here I have no where to go, because I am not guaranteed anything about $f^{-1}(d_s(t))$ since I only have that $f$ is entire.

In an effort to fix this I thought I could apply the inverse function theorem to get some local inverses, but I still don't believe I have enough information to obtain the result I want.

After this didn't pan out, I thought maybe I could show that the integral over any closed curve in $V$ with respect to $f(z)$ was 0, but I haven't had any luck with that either.

I know that there are several equivalencies to the statement ''V is simply connected,'' but I don't see how the others would work out in this case. I also saw on one website a comment that said this follows from the Maximum Modulus Principle, but I don't see how.

Any helpful hints are appreciated. Thanks

Note: I was also wondering if a component needs to be connected, because if not I don't see how the case of $f$ being constant would work.

Progress Update: So I may have gotten a tad closer to the solution. I know that $f'$ can only have countably many isolated singularities (otherwise it would be 0 implying $f\equiv0$). So choose $z_0\in V\setminus Z[f']$ and use the inverse function theorem to guarantee that $f$ is invertible in a small neighborhood, and so this neighborhood would be simply connected. Since we have this for every point except those in $Z[f']$ I feel like it would give us the result.


This follows from the maximum principle. First, note that $|f|=1$ on $\partial V$ because if $|f(\zeta)|<1$ for some $\zeta\in\partial V$, then a neighborhood of $\zeta$ would also belong to $V$, by the maximal property of a connected component. This contradicts $\zeta$ being a boundary point.

If $\mathbb{C}\setminus V$ had a bounded connected component, we would have $|f|\equiv 1$ there, violating the maximum principle. Thus, $\mathbb{C}\setminus V$ does not have a bounded connected component, which implies $V$ is simply-connected. A reference for the latter implication is Function Theory of One Complex Variable by Greene and Krantz (here's a relevant excerpt).

  • $\begingroup$ So I'm still not quite getting it I am afraid. If you wouldn't a couple follow up questions I'd be most appreciative. 1. A component of the pre-image of the ball is just a subset of the set which maps to the ball, correct? (We never explicit defined component.) If so wouldn't $|f|\leq 1$ on $\partial V$? 2. I am not familiar with the result $\mathbb{C}\setminus V$ not having a bounded connected component implying V is simply connected. Would you mind explaining more or referring me to a proof somewhere? I apologize for being so needy, I am just trying to understand the material well. $\endgroup$ – Scott Jul 20 '15 at 17:54
  • $\begingroup$ A component means a connected component here. I expanded the answer. $\endgroup$ – user147263 Jul 20 '15 at 18:12
  • $\begingroup$ I appreciate the references and the extra detail. Thanks. $\endgroup$ – Scott Jul 20 '15 at 18:20

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