If an analytic $f$ is injective on $\partial D$ then $f$ is injective on $D$

Question

Let $f(z)$ be analytic in a simply connected domain $D$ and on its boundary, the simple closed contour $C$. If $f(z)$ is injective on $C$, then $f(z)$ is injective on $D$.

If $z_0\in D$ is a point such that $f(z_0)\notin f(C)$ then we can use an argument principle to conclude $f(z_0) = f(z)$ then $z =z_0$ for $z\in D$. This is because by assumption, $f(C)$ is also a simple closed curve. But if there is a point $z\in C$ such that $f(z_0) =f(z)$ then the integrand in the argument principle is not well-defined so I can't use this method. I think the open mapping theorem and maximum modulus principle imply this cannot happen but I can't see the explicit contradiction. Please help.

Edit: There was a comment given by Aphelli which is incomplete but I wonder if it can be resolved: So the idea is using the argument principle with a slight modification on $C$. Since $f$ is holomorphic on $\overline{D}$, there is an open set $U\supset\overline{D}$ such that $f\in\mathcal{H}(U)$. Since $C$ is compact, we can find a simple close curve $C'\subset U$ enclosing $C$ as close as we want. Once we can show there is such $C'$ that $f$ is injective on $C'$ then we can apply the argument principle as before to prove the statement. The problem is the existence of such $C'$. Showing the existence of $C'$ such that $f'\neq 0$ on $C'$ is easy but this only ensures local injectivity on $C'$.

Answer 1

Here's a way to prove the remaining step using (a slightly stronger version of) the Jordan curve theorem, so it may not be what you are looking for.

By the Jordan curve theorem, $f(C)$ divides the plane into two connected regions; call them $U$ and $V$. By the open mapping theorem, $f(\mathrm{int}\,D)$ is open. We claim that $f(\mathrm{int}\,D)$ is contained completely in either $U$ or $V$. Proving this claim should solve your problem since it demonstrates that $f(\mathrm{int}\,D)$ and $f(C)$ are disjoint.

Suppose the claim is false. We invoke a slightly stronger version of the Jordan curve theorem, which states that $f(C)$ is the boundary of both $U$ and $V$. In particular, any ball centered at a point in $f(C)$ must intersect both $U$ and $V$. Hence, assuming the claim is false implies that $f(\mathrm{int}\,D)$ intersects both $U$ and $V$.

Consider the intersections $f(\mathrm{int}\,D) \cap U$ and $f(\mathrm{int}\,D) \cap V$, which we will view as subspaces of $U$ and $V$, respectively. Clearly, both these sets are relatively open as subspaces. Because $f(D)$ is closed (as it is compact), writing $$ f(\mathrm{int}\,D) \cap U = f(D) \cap U \quad \text{and} \quad f(\mathrm{int}\,D) \cap V = f(D) \cap V $$ shows that they are also relatively closed as subspaces. By connectedness, it follows that $f(\mathrm{int}\,D) \cap U = U$ and $f(\mathrm{int}\,D) \cap V = V$, and hence $f(D) = \mathbb{C}$. This is impossible because $f(D)$ is compact.

Edit: After looking into it, the version of the Jordan curve theorem I used above is not too much stronger than its usual statement. It follows quickly from the "Jordan arc theorem," see Exercise 63 here. But it seems like proving the Jordan arc theorem is not any more difficult than proving the Jordan curve theorem. In fact, the usual homological proof uses the Jordan arc theorem to prove the Jordan curve theorem (see, for instance, Proposition 2B.1 in Hatcher).

Answer 2

It turns out that the conclusion can follow rather elementarily from your observations, without needing analyticity outside $C$ or otherwise changing $C$ (the approach outlined in my comment – deleted as it was false [perhaps it could be fixed but I think the following is much better]).

It’s enough to show that there cannot exist distinct points $z \in D^{\circ}, u \in D$ such that $f(z)=f(u)$.

We already know that if $b \notin f(C)$, then $f^{-1}(b) \cap D$ has cardinality at most one.

Assume that such $z,u$ exist: then there are disjoint open subsets $U, V\subset D^{\circ}$ such that $u \in \overline{U}$ and $z \in V$. Then $f(V)$, $f(U)$ are open subsets of $\mathbb{C}$ with $f(z)=f(u) \in \overline{f(U)}$, $f(z) \in f(V)$. So $W=f(U) \cap f(V)$ is a non-empty open subset of $\mathbb{C}$ such that for all $b \in W$, $f^{-1}(b)$ contains at least two points. So $W \subset f(C)$. But that’s impossible because $f(C)$ is a closed simple curve (eg, removing any two points makes $f(C)$ disconnected; but this property is always false for a connected set with nonempty interior).

Answer 3

Let's put an easy proof that requires only continuity (and injectivity) of $f$ on the boundary, but uses a more advanced result, namely Caratheodory theorem, which shows that given any Jordan curve $J$ with inner domain $D$, any Riemann map from the unit disc to $D$ (analytic isomorphism which exists since $D$ is simply connected and not the plane) extends to a continuous injective map from the closed unit disc onto $\bar D$

Given this, one can take $f(J)=C$ and $D, \Omega$ the inner domains of $C,J$ and consider $\mathbb D \to D \to \Omega \to \mathbb D$ given by $g=\phi \circ f \circ \psi$ where $\phi, \psi$ are appropriate Riemann maps that by Caratheodory theorem extend continuously and injectively to the boundary.

Then $g$ is a continuous self-map from the closed unit disc to itself, analytic on the unit disc and sending the unit circle to itself injectively. By usual arguments, $g$ must be a finite Blaschke product ($|g(z)| \to 1, |z| \to 1$ means finitely many zeroes, take them out with $B$ finite Blaschke product and use maximum modulus of $g/B, B/g$ to show it's constant) and injectivity of $g$ on the circle means $B$ is degree $1$ hence disc automorphism, so by composing back to get $f$ we are done!