3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.