# If the critical values of a complex polynomial lie in the unit disc then the preimage of the unit disc is connected.

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?

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 $\Omega$.

• I don’t think you can always assume that the deck transformation group is normal. For example, take $\frac{1}{4} (z^3 - 3z)$. This has critical values $\pm \frac{1}{2}$ lying in the unit disk, but its deck transformation group is trivial, so you cannot permute the fibers using a deck transformations. Commented Jul 29 at 22:32

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 $\overline{\mathbb C}\setminus p^{-1}(D)$.

• 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. Commented Oct 17, 2013 at 6:00
• @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 $p'$. Commented Oct 17, 2013 at 12:12

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.

• 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 $p_2$? Commented Sep 11, 2013 at 0:02
• 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. Commented Sep 14, 2013 at 9:36
• @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. Commented Sep 14, 2013 at 16:08
• @Pete: small modification: you need a biholomorphism between $C$ and $I$. A differeomorphism is not enough. Commented Sep 14, 2013 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... Commented Sep 19, 2013 at 3:06

It’s enough to show that the Riemann surface of the inverse polynomial defined on the unit disk is connected, since the image of this surface under the continuous map $$(z,P^{-1}(z)) \rightarrow P^{-1}(z)$$ will then be connected, and the image is exactly $$P^{-1}(D)$$.

Now draw branch cuts from each of the critical points to the boundary of the unit disk, making sure none cross, and continue to extend them to infinity.

Notice that at infinity $$P$$ is biholomorphically equivalent to $$z^d$$, so that any inverse at infinity is biholomorphically equivalent to $$z^{\frac{1}{d}}$$. Because of this, when we loop around infinity (thereby crossing every branch cut), any inverse must pick up a change in argument of $$\frac{2\pi }{d}$$ after every cut has been crossed in this fashion.

Let the sheets of the surface we are considering be each of the $$d$$ preimages of the Riemann sphere minus the cuts extended to infinity by each of the definable branches of the inverse (we can define such branches because the Sphere minus these cuts is simply connected).

So we can index the sheets of the inverse function according to this, starting with sheet 1, and then indexing the second sheet when we reach it upon making a loop around infinity, and so on until we reach the final sheet, it follows that each sheet can be accessed from any other sheet by crossing a sequence of cuts. So the Riemann surface of the inverse polynomial on the whole Riemann sphere is connected.

Now the exact same logic applies if we restrict our focus to the unit disk minus these cuts. We have a simply connected domain, so we have $$d$$ definable branches for the inverse, and the branching behaviour between sheets is exactly the same as for the case of the whole sphere, so our Riemann surface remains connected.

Thus the preimage of the disk must also be connected.