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I am looking at problem $13.7$ in Apostol's Mathematical Analysis.

Let $D=\{|z|<1\}$ and let $f=u+iv$ be such that

$(\rm i)$ $u,v\in{\mathcal C}^1(D)$

$(\rm ii)$ $f$ is continuous on $\overline D$

$(\rm iii)$ $f(z)=z$ on $\partial D$.

$(\rm iv)$ $J_f(z)>0$ if $z\in D$.

Show that

$(1)$ For every open $\Omega\subseteq D$, $f(\Omega)$ is open in $f(D)$.

$(2)$ $f(D)$ is an open ball of radius $1$.

$(3)$ For each $w\in f(D)$, $D\cap f^{-1}(w)$ is finite.

I have only proven $(1)$:

PROOF1 Since $u,v$ are of class $\mathcal C^1$ over $D$, so is $f$, and since the Jacobian doesn't vanish, the inverse function theorem says that for each $z\in D$ there exists an open nbhd $N_z$ such that $f\mid_{N_z}$ is a diffeomorphism of class $C_1$, thus $f\mid_{N_z}$ is a homeomorphism, so it is an open map. Let $\Omega\subseteq D$ be open, and choose $z\in\Omega$. Then there exists $N_z$ as above. Let $\widetilde N_z=N_z\cap \Omega$. This is open and lies inside $N_z$ so its image under $f\mid_{N_z}$ is open by the above. Then $f(\Omega)=f(\Omega)\cap f(D)=\bigcup_{z\in\Omega}f\left(N_z\right)\cap f(D)$ is the intersection of the union of open sets with $f(D)$, so it is open in $f(D)$. $\blacktriangle$.

PROOF2 Suppose that $f(D)\not\subseteq D$. Then certainly $f(\bar D)\not\subseteq \bar D$. Thus, there exists $q\in\bar D$ such that $|f(q)|>1$. But $|f|$ is continuous on the compact $\bar D$; thus it attains its maximum on some $p\in\bar D$. But we cannot have $p\in\partial D$ since we would have $|f(p)|=1$, so it must be the case $p\in D$. Let $K=\overline B$ where $B=B(0,r)\; ;\; r=|f(p)|$. Then $f(\bar D)\subseteq \bar B$, and $f(p)\in\partial B$. Let $N_p$ be an open nbhd of $p$ in $D$. Then we see $f(N_p)$ is not open in $f(D)$, contrary to what was proven in $(1)$. It follows that $f(D)\subseteq D$.

PROOF3 Suppose for the sake of a contradiction that there exists $w\in f(D)=D$such that $D \cap f^{-1}(w)$ is infinite. Since this is a subset of the compact $\bar D$, it must have an accumulation point in $\alpha \in \bar D$. Suppose first that $\alpha\in D$. Then for each $\epsilon >0$ the ball $B(\alpha;\epsilon)$ contains infinitely many points of $\{z\in D:f(z)=w\}$, in contradiction with the fact $f$ is a local diffeomorphism in $\alpha$. Suppose $\alpha\in \partial D$. Then $f(\alpha)=\alpha$, and by continuity, $f(\alpha)=w$, which is absurd since $D\cap\partial D=\varnothing$. $\blacktriangle$.

Chapter $13$ is "Implicit functions and extremum problems". I am not sure what theorems to use to prove $(2)$, and I haven't thought about $(3)$, so I will worry about $(3)$ later. I want to prove $(2)$ now. Any hints?

ADD If $A$ is any set, $\partial A$ denotes its boundary and $\overline A$ its closure.

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2. Since $f$ is a homeomorphism, and $D$ is simply connected, $f(D)$ is also simply connected. Now apply the boundary condition. – Gyu Eun Lee May 29 '13 at 2:59
@proximal Hmmm, simple connectedness has not been mentioned so far in the book. – Pedro Tamaroff May 29 '13 at 3:01
3. The idea here is that a differentiable function can be partially reconstructed from its image. If (3) were false, then "too many" points would map into $w$ for $f$ to be nontrivial. If you know about analytic continuation, or the identity theorem for power series, then that's the idea at play. – Gyu Eun Lee May 29 '13 at 3:04
@TedShifrin Whoops, you're right. Sometimes I forget that $u$ and $v$ are functions of $2$ variables. – Gyu Eun Lee May 29 '13 at 3:19
Winding numbers and degree theory are the appropriate generalizations of the intermediate value theorem to higher dimensions. Why not use compactness and also part (1)? What you know about $\overline{f(D)}$? – Ted Shifrin May 29 '13 at 3:28
up vote 7 down vote accepted

Here's an approach to prove $f(D) = D$. First show that $f(D) \subset D$ as follows: The fact that $f(D)$ is open means that $|f|$ cannot attain a maximum on $D$ and must therefore attain a maximum on $\partial D$; use condition (iii) to conclude that $|f(z)|< 1$ for $z\in D$. Then show that $D\subset f(D)$, using the fact that $f(\overline{D})$ is a compact set that is at the same time equal to $f(D)\cup \partial D$. (Therefore, $\partial f(D)\subset \partial D$, meaning that $f(D)$ must contain every point of $D$.)

Here is a little more detail about the proof of $D\subset f(D)$, as requested in the comments. Note that $f(\overline{D}) = f(D)\cup \partial D$ is compact, hence closed, so it contains $\overline{f(D)} = f(D)\cup\partial f(D)$. As $f(D)$ is open, it is disjoint from its boundary, so $f(D)\cup \partial f(D)\subset f(D)\cup \partial D$ implies $\partial f(D)\subset \partial D$. Then $f(D)\cap D = \overline{f(D)}\cap D$, meaning that $f(D)\cap D$ is both closed and open relative to $D$; since $D$ is connected and $f(D)\cap D$ is nonempty, it must be the case that $f(D)\cap D = D$, or equivalently that $D\subset f(D)$.

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I'm not sure the reasoning in the last sentence in the parenthesis is rigorous, though it seems intuitively correct. Could you provide more detail? – Potato May 29 '13 at 5:00
@Potato Sure: Note that $D\setminus \overline{f(D)} = D\setminus (f(D)\cup \partial D) = D\setminus f(D)$ is both relatively open and closed in $D$; since $D$ is connected and $f(D)$ intersects $D$ nontrivially, $D\setminus f(D)$ must be empty. – Nick Strehlke May 29 '13 at 5:21
@NickStrehlke By continuity I am getting $f(\overline D)\subseteq \overline{f(D)}$ so $f(D)\cup \partial D \subseteq f(D)\cup \partial f(D)$. This gives me $\partial D\subseteq \partial f(D)$. Is that what you intended in your last line? I see the reverse inclusion. – Pedro Tamaroff May 29 '13 at 15:41
@PeterTamaroff Actually, what I meant can be rephrased as follows: $\overline{f(D)} \subset f(\overline{D})$ because $f(\overline{D})$ is a compact (hence closed) set containing $f(D)$ and $\overline{f(D)}$ is by definition the smallest closed set containing $f(D)$. It follows that $\partial f(D) \subset \partial D$. – Nick Strehlke May 29 '13 at 18:00
@NickStrehlke Oh, I see. All in all, $f(\bar D)=\overline{f(D)}$, so $\partial D=\partial f(D)$. – Pedro Tamaroff May 29 '13 at 18:07

Part 2

We know that $f(\bar D)$ is compact subset of the closed unit ball, so it is closed, and its complement in the closed unit ball is open. This follows because $f$ is an open mapping, so it obeys a maximum principle: its maximum modulus on a compact region cannot be taken in the interior. So we must have $|f|<1$ for all $z\in D$.

We can write $f(\bar D)^c$ in $\bar D$ as $\bar D\cap U$ for some open set $U$ in $\mathbb R^2$. Then $f(\bar D)^c\cap D= U\cap D$ is an open set of $D$ in the subspace topology. So the set of points in $D$ omitted by $f(D)$ in is open in the subspace topology on $D$, as the boundary points of $\bar D$ are all mapped to the boundary of $D$.

But this set is also closed. We know $f(D)$ is open by part 1, and its complement in the plane is closed. So $f(D)^c\cap D$ is a closed set of $D$ in the subspace topology.

We see the omitted set must be empty by connectedness, as it is both open and closed.

Part 3

If the inverse image of some point $w$ were infinite, the set of inverse images would have a limit point in the closed unit ball by compactness. If $p$ is the limit point, then by continuity $f(p)=w$. If $p$ is in the open unit ball, then no neighborhood of the limit point is mapped diffeomorphically onto an open set, contradicting the third hypothesis and the inverse function theorem.

If the limit point is on the boundary, we get a contradiction, as $w$ must lie in the open unit ball and $f(p)$ on the boundary.

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"We know that $f(\bar D)$ is compact subset of the closed unit ball," Why? – Pedro Tamaroff May 29 '13 at 19:44
@PeterTamaroff We know $f$ is a continuous map, and that continuous maps sends compact sets to compact sets. So $f(\bar D)$ is compact. It is a subset of the closed unit ball because the boundary is mapped to $|z|=1$ and $D$ is mapped to the open unit ball (by a maximum principle argument similar to the other answer, since $f$ is an open map). – Potato May 29 '13 at 19:50
@PeterTamaroff Sorry, I wrote it up a little too quickly. It's there now. – Potato May 29 '13 at 19:53
What is the proof of "Since $f(D)$ is open $|f|$ cannot attain it's maximum on $D$"? (this claim is made by Nick) Also, you say "$D$ is mapped to the open unit ball": that is what we want to prove. – Pedro Tamaroff May 29 '13 at 20:00
@PeterTamaroff Sorry, I didn't get pinged for that comment and just saw it. Since $f(z)$ is open, around any point in the image we can find a small ball in the plane. In this ball you can find a point with strictly greater modulus. (I can provide a reference if needed -- this is a standard proof in complex variables.) – Potato May 30 '13 at 21:09

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