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I've looked at a couple of answers to a similar question; I get how to prove bijectivity if we know that the coefficients are real and $ad-bc > 0$. That's actually fairly simple.

What I don't get -- and what I suspect I'm overthinking -- is how to prove the converse, i.e. --

that a fractional linear transformation $$\frac{az + b}{cz + d}$$ has real coefficients up to multiplication by a constant, with $ad - bc > 0$ on top and bottom if it is bijective UHP $\rightarrow$ UHP.

At this point I've thrown just about everything at the wall and am spinning my wheels. I did prove that $ad - bc\neq 0$ (and thus one can multiply top and bottom by constants whose product is $\overline{ad-bc}$, but I lost my notes on how I did it.

As for the second part:

  • First I tried assuming injectivity, surjectivity, and the existence of an inverse and tried to derive a contradiction, but that led me mostly to (apparently) useless results. For example, if $d\neq 0$ and one takes the limit of the functionas $z\rightarrow 0$ then we have that $b / d$ has to be in the closure of the UHP, i.e., $Im(b\overline{d})$ has to has to have strictly negative imaginary part. Or, e.g., in order for the function to be a function we have to have $-d/c$ not in the upper half-plane. Not terribly helpful.
  • I tried making "real up to multiplication by a constant" mathematically precise, but that just sent me into a bunch of logical circles.
  • Finally, I tried using the fact that the function has to map the real line onto the real line. If we multiply the denominator by its conjugate then we get that a polynomial in $|z|$, $z$, and $\overline{z}$ has to be real for all real values of $z$. That, too, doesn't do much much good.

Edit Since I know more advanced complex analysis, I reasoned that the desired function would map the boundary, $\mathbb{R}$, to itself; i.e., it would map three distinct real points $a$, $b$, and $c$ to $0$, $1$ and $\infty$ respectively. So in that way we obtain a FLT with real coefficients, although I used the result that $ad - bc\neq 0$ in the proof.

However, I found this problem in Ch 1 of Greene & Krantz's book and so I suspect that there is a much, MUCH simpler answer that I just overlooked. Can anyone help?

matty_k_walrus
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    without loss of generality, you can restrict to $ad-bc= 1$ – reuns Sep 22 '16 at 23:01
  • @user1952009 I might need a more specific hint. I get why this is true but don't see the connection to bijectivity without using more advanced properties of FLTs. – matty_k_walrus Sep 26 '16 at 21:09
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    I wrote a proof for the unit disk yesterday, so adapt it, or more direct, pick a Möbius biholomorphism $\phi : H \to D$, and use that for every Möbius biholomorphism $\gamma : H \to H$, $\ \ \phi \circ \gamma \circ \phi^{-1}$ is a Möbius biholomorphism $D \to D$ – reuns Sep 26 '16 at 21:34
  • And $ab-cd \ne 0$ is obvious, since otherwise it is not invertible (the inverse is $\frac{Az+B}{Cz+D}$ where $\begin{pmatrix} A&B \C& D \end{pmatrix} = \begin{pmatrix} a&b \c& d \end{pmatrix}^{-1} =\frac{1}{ad-bc}\begin{pmatrix} d&-b \-c& a\end{pmatrix}$ ) – reuns Sep 26 '16 at 21:42

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