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If you have a holomorphic function $f$ from the unit disc $D$ to a unit square $U$, such that:

1) $f$ has a continuous extension to the boundary of $D$,

2) $f$ restricted to the boundary of $D$ is a bijection between boundary of $D$ and boundary of $U$.

Then WTS $f$ is a biholomorphism between $D$ and $U$.

This was a question my professor mentioned as an example of something that might be on the final tomorrow. My hypothesis may be not quite correct, specifically I'm not sure if we are assuming 2 or if we can conclude it. I think we have to assume it?

Anyway, so I want to do this problem by applying the argument principle to the function $g(z) = f(z) - w$ for any $w \in U$. Showing that $\int_C \frac{g'(z)}{g(z)}dz=2\pi i$ will imply that $g(z)$ has exactly one zero (since $f(z)$ is holomorphic on $D$ so is $g(z)$, so it has no poles and we conclude 1 zero from the argument principle) so therefore $f$ is a bijection and we are done.

So I'm not quite sure how to handle this integral. Using contour integration, we get $$\int_C \frac{f'(z)}{f(z)-w} dz = \int_0^{2\pi} \frac{f'(e^{i\theta})}{f(e^{i\theta})-w}i\cdot e^{i\theta}d\theta.$$ And from here I want to use a logarithm, but I am getting hung up on branch cuts and where to go from here.

Could anyone walk me through how to complete this integral?

Thank you!

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What does WTS mean? – Mariano Suárez-Alvarez Dec 13 '11 at 2:21
It means Want to Show – Rakalakalili Dec 13 '11 at 2:36
Were you in such an enormous hurry that you could not write that out? :P – Mariano Suárez-Alvarez Dec 13 '11 at 2:44
I was under the assumption that it was extremely standard notation. It has been in every course I have taken, my apologies. – Rakalakalili Dec 13 '11 at 4:01
Closely related :… – Kalim Dec 13 '11 at 14:49

Let me try again :

First note that the result is true if $f$ is holomorphic in an open set containing the closure of the unit disk $\overline{D}$. Indeed, in this case, for each $w$ in the square $U$, we can use the argument principle to count the number of preimages of $w$ by evaluating the integral. But the integral is just the winding number of the image curve $f(\partial D)$ around $w$, which is $1$.

However, in our case, we can't use that argument directly because $f$ is a priori only continuous on the boundary. We can proceed as follows :

Let $\phi : U \rightarrow D$ be a conformal mapping of the square onto the unit disk. By Carathéodory's theorem on the boundary behaviour of conformal mappings, $\phi$ extends to a homeomorphism of $\overline{U}$ onto $\overline{D}$ (because the boundary of the square is a Jordan curve). Consider $$h:=\phi \circ f.$$ Then $h : D \rightarrow D$ is holomorphic, continuous on $\overline{D}$ and maps the unit circle onto itself bijectively. By Schwarz's reflection principle, $h$ extends to be holomorphic on a neighbourhood of $\overline{D}$. By the remark at the beginning of the proof, $h$ is conformal. Since $f=\phi^{-1} \circ h$ and $\phi$ is conformal, so is $f$.

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I don't think you can just "use the argument principle" to show that the number of preimages is continuous on $\overline{U}$. First of all, even in $U$ it would be the number counted by multiplicity that is constant. More importantly, continuity at points on the boundary is not so clear. For example, if $U$ is the upper half plane and $f(z) = -z$, points of $U$ have no preimages in $\overline{U}$ but points of the boundary have one. – Robert Israel Dec 13 '11 at 1:33
By $n(w)$, I meant the number of preimages of $w$ counting multiplicites. It wasn't clear, so I edited accordingly. As for countinuity on the boundary, you're right... $n$ is continuous in $U$ but not necessarily in $\overline{U}$... I have to think how to fix this. Thank you for the comment. – Kalim Dec 13 '11 at 2:10
Is attempting to compute that integral explicitly a method that would work? – Rakalakalili Dec 13 '11 at 2:37
How do you know that $h = \phi \circ f$ maps the unit circle onto iself bijectively without already knowing that $f$ is bijective? – Sam Dec 15 '11 at 14:10
we know that $f$ maps the unit circle onto the boundary of the unit square bijectively, it's hypothesis 2... – Kalim Dec 15 '11 at 16:03

Let $\gamma_r(t) = re^{it}$ for $t\in [0,2\pi]$. Then given any $w \in Int(U)$, the preimage $f^{-1}(w)$ is discrete in $\overline D$ (because no point on the boundary of $D$ can be an accumulation point of $f^{-1}(w)$ by assumption on $f$), hence finite.

So let $f^{-1}(w) = \{z_1, \ldots, z_n\}$ (where we allow $z_i = z_j$ according to multiplicity).

By choosing $0<r<1$ big enough -- to be concrete let's choose $r = \frac12(1+\max(|z_1|, \ldots, |z_n|)$ -- the path $\gamma_r$ will go around each one of the $z_i$ and $$H: [r,1]\times [0,2\pi]\to U, \quad H(s,t) = f(\gamma_s(t))$$ will be a homotopy between $f\circ\gamma_r$ and $f\circ\gamma_1$ in $U\setminus \{w\}$.

Therefore the winding numbers of $f(\gamma_r(t))$ and $f(\gamma_1(t))$ around $w$ must be equal, i.e. $$Ind(f\circ \gamma_r; w) = Ind(f\circ \gamma_1; w)$$ But $f\circ \gamma_1$ is a path which goes around $\partial U$ exactly once, because $f$ is bijective onto $\partial U$ when restricted to $\partial D$. So it follows by the argument principle that

$$n = Ind(f\circ \gamma_r;w) = Ind(f\circ \gamma_1;w) = \pm 1$$

Since the left hand side is $\ge 0$, the right hand side must be $=1$ and therefore we have $n = 1$.

Proving that $f$ is bijective on all of $D$.

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+1, nice answer! – Kalim Dec 15 '11 at 19:34

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