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Say, by the Riemann Mapping Theorem, there exists a biholomorphic, conformal mapping from the upper half plane to the (open) unit disk (since the UHP is simply connected and is not the entire complex plane.)

This mapping, call it $f(z)$, is intuitively an onto-mapping, since we can shrink the points in the UHP by suitable scaling to fit into the unit disk in the $w$-plane.

But why would such a mapping necessarily be one-to-one?

The UHP is so much bigger than the unit disk that...shouldn't we run into the situation were $z_1 \ne z_2$, but $f(z_1) = f(z_2)$?

Thanks,

Edit: I'm wondering whether the condition $f'$ not zero (non-vanishing derivative) makes the mapping one-to-one, but I am doubtful; I think the non-vanishing derivative only makes the mapping locally invertible, by the inverse function theorem. But I guess we can use this argument for all of the UHP, and conclude the mapping from the UHP to the unit disk is invertible, one-to-one, and onto. Am I close?

Michael Hardy
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User001
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    Consider $f(x)=e^x$. It is locally conformal, yet globally, $f(x)=f(x+2\pi i)$, so it is not injective (and $f'(x)$ is never $0$). – robjohn Jul 01 '15 at 22:01
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    QUOTE This mapping, call it $f(z)$, is intuitively an onto-mapping, since we can shrink the points in the UHP by suitable scaling to fit into the unit disk in the $w$-plane. END QUOTE ${}\qquad{}$ I don't see how this makes sense. If you mean it's onto the disk, then showing that the image of the function is entirely within the disk doesn't do that. And how could a rescaling in the $z$-plane rather than the $w$-plane make it fit into the disk? ${}\qquad{}$ – Michael Hardy Jul 01 '15 at 22:15
  • Thanks for the example to consider, @RobJohn. – User001 Jul 01 '15 at 22:28
  • Good points, @Michael Hardy. hmmm... – User001 Jul 01 '15 at 22:29
  • Hi @robjohn, then what good is the assumption of f(z) having non-vanishing derivative? Is it a necessary but insufficient condition for injectivity of f? It guarantees the injectivity of f, locally, I think (since f is never constant, locally.) – User001 Jul 01 '15 at 22:55
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    @LebronJames: The injectivity is an added constraint. It is not given by the non-vanishing derivative. The non-vanishing derivative ensures that angles are preserved, thus giving the conformality (local preservation of shape). – robjohn Jul 01 '15 at 23:01
  • Thanks @robjohn. Can I ask one more question? As I mentioned below to Andreas Blass's answer, I am able to show that z-i/z+i is one-to-one, by a simple contradiction proof. Also, with the standard symmetry arguments, I know that the UHP maps to the inside of the unit disk in the w-plane, while the LHP maps to the outside. But even though I am able to see the injectivity of f, algebraically, on paper, I still find it hard to believe, like many results in complex analysis. Do you have a suggestion on how I can think of why the mapping is one-to-one, besides an algebraic proof? Thanks, – User001 Jul 01 '15 at 23:12
  • ... is it the scaling by the factor 1/z+i ? – User001 Jul 01 '15 at 23:15
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    @LebronJames: the only way to show injectivity of $f$ is to assume $f(z)=f(w)$ and show that $z=w$ (i.e. algebraically). Differentiability is a local property, whereas injectivity is a global property. This is why a non-vanishing derivative does not imply injectivity. – robjohn Jul 02 '15 at 00:29
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    @LebronJames: Any conformal map is locally a translation composed with a scaling and a rotation. This is because locally, a conformal map is given by $$f(z)=\color{#00A000}{f(w)+}\color{#C00000}{f'(w)}(z\color{#00A000}{-w})$$ where the green parts contribute to the translation and the red parts to the rotation and scaling. $|f'(w)|$ is the scaling and $\arg(f'(w))$ is the rotation. – robjohn Jul 02 '15 at 00:29
  • Ok, got it. Thanks so much for the help, @robjohn. Have a great night. – User001 Jul 02 '15 at 03:02

2 Answers2

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Isn't bijectivity part of the definition of "biholomorphic"? In the particular situation you described, mapping the upper half-plane to the unit disk, instead of invoking the Riemann mapping theorem, you might as well look at a specific such mapping, say $$ z\mapsto\frac{z-i}{z+i},$$ check by calculation that it's one-to-one, and draw some pictures to see how this "compression" of the half-plane into the disk works.

Andreas Blass
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  • Hi, @AndreasBlass, I was able to write a simple proof by contradiction to show that this mapping is indeed one-to-one. And yes, one-to-one and onto, together with non-vanishing derivative are necessary but not sufficient conditions to imply that the mapping is biholomorphic and conformal, I think. Hmm...even though I am now able to show the one-to-one condition, it seems hard to believe, conceptually. How can I best think of this "compression" action? – User001 Jul 01 '15 at 22:49
  • Actually, sorry, I think f(z) one-to-one and onto, with non-vanishing derivative, suitably restricted, is a conformal mapping. – User001 Jul 01 '15 at 22:52
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    Well, both the unit disk and the upper half-plane have the same cardinality, so there's no problem there. The upper half-plane has infinite area, while the unit disk has finite area; but biholomorphic functions don't have to preserve area, so that's not a problem, either. If you're familiar with an order-preserving bijection between the open interval $(0,1)$ and the entire real line, this biholomorphic function acts much the same way. – Sophie Swett Jul 02 '15 at 00:20
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While it was shown by explicit example that a particular conformal mapping from the upper half plane to the unit disk is one-to-one, the more general question of "why conformal mappings are one-to-one" was not addressed in detail. It was mentioned in a comment that conformal mappings are "locally linear", heuristically implying that they are locally one-to-one. While this may be good intuition to keep in mind, it is not precise. In the edit of the question, it is stated that "I think the non-vanishing derivative only makes the mapping locally invertible". Indeed: consider the function $z \mapsto z^2$ restricted to the annulus $A = \{ z : 1 < |z| < 2 \}$. On $A$, this mapping is conformal, but not one-to-one. Let us prove the following:

If $f$ is conformal in an open set $G$, then for each $a \in G$ there exists an $r > 0$ such that the restriction of $f$ to $D(a;r)$ is one-to-one. [$D(a;r)$ represents the open disk centered at $a$ of radius $r$.]

Take $a \in G$ arbitrarily and find $R > 0$ such that 1) $\bar{D}(a;R) \subset G$ and 2) $f(z) - f(a)$ is never zero in the set $\bar{D}(a;R) \backslash \{ a \}$. We must be allowed point 2) by the identity theorem: if we could not find any such $R$ we would be able to construct a sequence $\{ z_n \}$, converging to $a \in G$, on which $f-f(a)$ vanishes. This would imply $f=f(a)$ in a neighborhood of $a$, in turn implying $f'(a) = 0$, which is not allowed by the conformality of $f$. In the following we'll use the result (Rouché's theorem):

Let $f$ and $g$ be holomorphic inside and on a contour $\gamma$ and suppose that $|f(z)| > |g(z)|$ on $\gamma$. Then $f$ and $f+g$ have the same number of zeroes inside $\gamma$.

With this in mind, define $m = \inf \{ |f(z)-f(a)| : z \in C(a;R) \} > 0$, where $C(a;R)$ represents the circle of radius $R$ around $a$. Rouché's theorem now implies that for any $w \in D(f(a);m)$, the functions $f(z)-f(a)$ and $f(z)-w$ have the same amount of zeroes in the domain $D(a;R)$, that is, precisely one. Finally, by continuity of $f$ at $a$, we can always find an $r > 0$ with $r < R$ such that $f(D(a;r)) \subset D(f(a);m)$. Then it is not hard to see that $f$ restricted to $D(a;r)$ is one-to-one.

NOTE: this proof is an exercise, 16.5, in Priestley's book Introduction to Complex Analysis. All the elements that went into the above proof are heavily inspired by the preceding chapters of the book.

Latrace
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