Is it necessarily the case that $B(0,1)$ gets mapped inside $\delta B(0,1)$?

Question

I'm currently trying to solve this problem:

Suppose that f is a holomorphic function defined on an open set U of the complex plane containing $\bar{B} (0, 1)$. Let $S^1 = \{z ∈ C : |z| = 1\} = ∂B(0, 1)$. Show that if $f(S^1)$ is an ellipse and f restricted on $S^1$ is injective, then f is injective on $\bar{B} (0, 1)$. [Hint: Apply the argument principle to $f(z) − w_0$ for $w_0 \in C$. ]

My current proof fully relies on the assumption that $B(0,1)$ gets mapped inside $\delta B(0,1)$ but I cannot figure out how to prove this is true (or even whether it is true at all). My thinking is that if $B(0,1)$ is mapped outside $f(S^1)$ then either injectivity, holomorphism, or something else will break.

Here's my attempt below: I'll mark whenever I use the assumption that $B(0,1)$ gets mapped inside $\delta B(0,1)$ with a $^\dagger$

$\forall z_0 \in B(0,1), \ g_{z_0}(z)=f(z)-f(z_0)$ is holomorphic in U and thus has no poles in $\bar{B} (0, 1)$

We would like to prove that there are no zeros of $g_{z_0}$ on $\delta B(0,1)$ in order to apply the argument principle: Suppose for contradiction $\exists z_1 \in S^1 : g_{z_0}(z_1) = 0$. $z_1$ is unique on $S^1$ as f is injective on $S^1$. This means the image ellipse passes through the origin. Note: $B(0,1)$ is open so by the Open Mapping Theorem, $f(B(0,1))$ is open. Therefore $\exists r>0:B(0,r)\in f(B(0,1))$. However, it is impossible for $B(0,r)$ to be in the ellipse as the ellipse passes through the origin. (Contradicting that $B(0,1)$ gets mapped inside $\delta B(0,1)$$^\dagger$). Thus our assumption that $g_{z_0}(z_1) = 0$ was false.

Now we can apply the argument principle to $g_{z_0}$. Let $N_{z_0}$ be the number of zeros of $g_{z_0}$ in B(0,1). $N_{z_0}$ = the winding number of $f(S^1)$ around the origin = 1 (as f is injective on $S^1$ and $0=g_{z_0}(z_0)$ lies inside $f(S^1)$$^\dagger$).

Thus $f(z)-f(z_0) = g_{z_0}(z)= 0$ iff $z=z_0, \forall z_0\in B(0,1)$, proving f is injective on $B(0,1)$.

Thanks :)

Answer 1

This overall problem has been asked about here, albeit in a more general context.

However, the particular step where you're stuck is proving the following:

If $z_0 \in B(0, 1)$ and $z_1 \in S^1$, then $f(z_0) \neq f(z_1)$.

So let's clarify that.

---

First, let's establish some notation.

• $E$ denotes the ellipse that is $f(S^1)$.
• $\text{Int}(E)$ denotes the interior of the ellipse $E$.
• $\text{Ext}(E)$ denotes the exterior of the ellipse $E$.

Thus $\text{Int}(E)$ and $\text{Ext}(E)$ are open subsets of $\mathbb C$, and $\mathbb C = \text{Int}(E) \cup E \cup \text{Ext}(E)$.

Now, let's prove the following:

If $z \in B(0, 1)$, then $f(z) \notin \text{Ext}(E)$.

Proof: Pick any $w \in \text{Ext}(E)$, and let $N_w$ be the number of points $z \in B(0, 1)$ such that $f(z) = w$. Our objective is to show that $N_w = 0$.

By the argument principle, $N_w$ is equal to the winding number of the curve $\gamma: [0, 1] \to \mathbb C$ defined by $\gamma(t) = f(e^{2\pi i t})$ around the point $w$.

Notice that $\text{Image}(\gamma)$, the image of the curve $\gamma$, is precisely $f(S^1)$, which is the ellipse $E$. The point $w$ lies inside $\text{Ext}(E)$, which is the unbounded connected component of $\mathbb C \setminus \text{Image}(\gamma)$. So the winding number of $\gamma$ around $w$ must be zero, as required.

(The other answer I linked makes use of the Jordan curve theorem to describe the connected components of $\mathbb C \setminus f(S^1)$. You don't need to use the Jordan curve theorem, since you're explicitly assuming that $f(S^1)$ is an ellipse.)

As you observed, we can strengthen our result:

If $z \in B(0, 1)$, then $f(z) \in \text{Int}(E)$.

Indeed, we just showed that $f(B(0, 1)) \subset \text{Int}(E) \cup E$. But the open mapping theorem tells us that $f(B(0, 1))$ is an open set. So it must actually be the case that $f(B(0, 1)) \subset \text{Int}(E)$.