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Let $p(z)$ be a complex polynomial such that $p'(z) = 0 \implies p(z) \in D,$ where $D$ is the open unit disc: $D= \{z\in \mathbb{C}\:\: |\:\:|z| < 1 \}.$ I want to prove that $p^{-1}(D)$ is connected.

This is a problem from a past qualifying exam that I was having a it of trouble with. I think I figured it out as I was typing this, but I'm not too sure.

Attempt:

First we note that if $p$ is constant, then $p^{-1}(D)$ is either all of $\mathbb{C}$ or is the empty set both of which are connected. So we assume that $p$ is a polynomial of degree greater than zero.

Let $A = \{z_1,...,z_{n-1}\}$ be the critical values of $p$.

Note that $D-A$ is connected.

By the inverse function theorem, each point in $D-A$ is contained in a neighborhood over which $p^{-1}$ is analytic and in particular continuous. Thus, (using the pasting lemma and the fact that the image of a connected set under a continuous map is connected) $p^{-1}(D-A)$ is connected.

Is this on the right track?

Bohring
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3 Answers3

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Here is a direct topological proof. A degree $n$ polynomial $p:\mathbb{C}\to\mathbb{C}$ is a degree $n$ covering branched over its critical points, so deleting the critical point set $C$ and its preimage gives an $n$-fold covering of $\mathbb{C} - C$. The deck group of this covering is a finite group with $n$ elements, generated by loops around each critical point. (It is a subgroup of the fundamental group of $\mathbb{C} - C$, which is free on $|C|$ generators.)

Suppose that all critical points lie in the unit disk $\Omega$. Let $z$ be a non-critical point of the unit disk and $\widetilde{z}$ be any lift of $z$.

Claim: any other lift $\widetilde{z}'$ can be connected to $\widetilde{z}$ by a path in $p^{-1}(\Omega)$.

Proof: Let $\gamma$ be an element of the deck group such that $\widetilde{z}' = \gamma\widetilde{z}$. Now $\gamma$ can be represented by a closed curve containing $z$. Since by assumption all critical points are contained in the unit disk, $\gamma$ is homotopic to a closed curve which is contained in the unit disk. This closed curve lifts to a curve connecting $\widetilde{z}$ and $\widetilde{z}'$.

Now if $z_1$ and $z_2$ are non-critical in the unit disk and have lifts $\widetilde{z}_1$, $\widetilde{z}_2$, connect $z_1$ to $z_2$ by a path $c$ in the disk, lift the path from $\widetilde{z}_1$ to some $\alpha\widetilde{z}_2$ (where $\alpha$ is in the deck group), concatenate $c$ with a curve in the disk representing $\alpha$, and then lift the concatenation of $\alpha$ and $c$ to get a curve connecting $\widetilde{z}_1$ and $\widetilde{z}_2$ in $p^{-1}(\Omega)$.

The key idea is that to get to other preimages in the covering, you have to go around and around the critical points. If the critical points lie in $\Omega$, this going-around-and-around can also be done within $\Omega$, hence the preimage of $\Omega$ is connected. However, if any critical point lies outside of $\Omega$, you'll have to loop around that critical point to get to some of your preimages. To do that, you'll have to leave $\Omega$.

Neal
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Let $p_{1},p_{2}$ be two points such that $f(p_{1})=f(p_{2})=z\in D$. Now $f$ is a complex polynomial and $f-z$ has double roots in $\mathbb{C}$.

Now if $f^{-1}(D)$ is not connected, then we can always find such two points $p_{1},p_{2}$ with a curve $C$ connecting them. The corresponding map sends $f-z$ to $0$ then back to itself. So using mean value theorem we know there must be a point $q$ such that $f'(q)=0$, and $q\in C$ outside of $f^{-1}(D)$. In particular $f(q)\not \in D$. But this contradicts with our hypothesis. This finishes the proof.

Bombyx mori
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  • Thanks for the reply. I might need a few minutes to think it over but doesn't this imply that there is a critical point on every curve connecting $p_1$ to $p_2$? – Bohring Sep 11 '13 at 00:02
  • Actually, I don't think this works. There isn't an analog of the mean value theorem for complex valued functions that would allow you to conclude that such a point $q$ exists. – Bohring Sep 14 '13 at 09:36
  • @Pete: You do not need an "analog". Have a differeomorphism between C and I, then translate the standard mean value theorem to C. Note complex valued does not change anything. Then you are done. – Bombyx mori Sep 14 '13 at 16:08
  • @Pete: small modification: you need a biholomorphism between $C$ and $I$. A differeomorphism is not enough. – Bombyx mori Sep 14 '13 at 19:45
  • Let $h$ be the map from $C$ to $I$ and let $g = f-z$ we see that $g(z_1) = g(z_2) = 0$ and would like to use the mvt on $h \circ g$ (??) but $h\circ g$ isn't even defined since $g(C)\not\subset C$. I don't think I see what function you are applying the mvt to... – Bohring Sep 19 '13 at 03:06
  • @Pete: There is no need for $g(C)$ to be included in $C$. The classical mean value theorem also has no such requirement. The real subtlety is the complex derivative. I will leave you to finish the details. For the precise statement you need to prove, check:http://math.stackexchange.com/questions/208609/mean-value-theorem-for-complex-functions – Bombyx mori Sep 23 '13 at 06:05
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Being unable to follow the answer by user32240, I'll give another one. Let $U_1,\dots,U_k$ be the connected components of $p^{-1}(D)$. They are simply connected domains (by maximum principle) with smooth boundaries. Let $n_j$ be the number of zeros of $p$ within $U_j$, counting multiplicities. By the argument principle, the winding number of $p_{|\partial U_j}$ with respect to $0$ is $n_j$. The image of tangent vector to $\partial U_j$, namely $p'(z)\,dz$, has the same winding number, because it's just $p(z)$ rotated by $90$ degrees and rescaled. Since $dz$ winds around $0$ once as we travel along $\partial U_j$, it follows that $p'_{|\partial U_j} $ winds around zero $n_j-1$ times. By the argument principle, $p'$ has $n_j-1$ zeros in $U_j$, counting multiplicities.

Since there are no zeros of $p'$ outside of $\bigcup U_j$, it follows that $$n-1=\sum_{j=1}^k (n_j-1) = n - k$$ hence $k=1$.

Another approach would be to invert $p^{1/n}$ in $\overline{\mathbb C}\setminus D$, using the monodromy theorem. But I got stuck trying to show that the holomorphic map obtained in this way is a biholomorphism between $\overline{\mathbb C}\setminus D$ and $\overline{\mathbb C}\setminus p^{-1}(D)$.

user98130
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  • Oooh. I like this. Maybe another way to see that $p'$ has $n_j-1$ zeros in $U_j$ is to note that the convex hull of the roots of a polynomial contains the critical points and that this has to hold for $p$ restricted to each $U_j$ since the critical values are all in the unit disc. – Bohring Oct 17 '13 at 06:00
  • @Pete I don't think that works, because "$p$ restricted to each $U_j$" still has the same zeros as $p$, including some outside of $U_j$. Unless by "restricted" you mean defining a new polynomial, for which the zeros are all in $U_j$. But that polynomial would have a derivative different from $p'$. – user98130 Oct 17 '13 at 12:12